# Artin transfer (group theory)

In the mathematical field of group theory, an Artin transfer is a certain homomorphism from an arbitrary finite or infinite group to the commutator quotient group of a subgroup of finite index. Originally, such mappings arose as group theoretic counterparts of class extension homomorphisms of abelian extensions of algebraic number fields by applying Artin's reciprocity maps to ideal class groups and analyzing the resulting homomorphisms between quotients of Galois groups. However, independently of number theoretic applications, a partial order on the kernels and targets of Artin transfers has recently turned out to be compatible with parent-descendant relations between finite p-groups (with a prime number p), which can be visualized in descendant trees. Therefore, Artin transfers provide a valuable tool for the classification of finite p-groups and for searching and identifying particular groups in descendant trees by looking for patterns defined by the kernels and targets of Artin transfers. These strategies of pattern recognition are useful in purely group theoretic context, as well as for applications in algebraic number theory concerning Galois groups of higher p-class fields and Hilbert p-class field towers.

## Transversals of a subgroup

Let ${\displaystyle G}$ be a group and ${\displaystyle H\leq G}$ be a subgroup of finite index ${\displaystyle n=(G:H)\geq 1}$.

Definitions. [1]

1. A left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ is an ordered system ${\displaystyle (g_{1},\ldots ,g_{n})}$ of representatives for the left cosets of ${\displaystyle H}$ in ${\displaystyle G}$ such that ${\displaystyle G={\dot {\bigcup }}_{i=1}^{n}\,g_{i}H}$ is a disjoint union.
2. Similarly, a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$ is an ordered system ${\displaystyle (d_{1},\ldots ,d_{n})}$ of representatives for the right cosets of ${\displaystyle H}$ in ${\displaystyle G}$ such that ${\displaystyle G={\dot {\bigcup }}_{i=1}^{n}\,Hd_{i}}$ is a disjoint union.

Remark. For any transversal of ${\displaystyle H}$ in ${\displaystyle G}$, there exists a unique subscript ${\displaystyle 1\leq i_{0}\leq n}$ such that ${\displaystyle g_{i_{0}}\in H}$, resp. ${\displaystyle d_{i_{0}}\in H}$. Of course, this element with subscript ${\displaystyle i_{0}}$ which represents the principal coset (i.e., the subgroup ${\displaystyle H}$ itself) may be, but need not be, replaced by the neutral element ${\displaystyle 1}$.

Lemma. [2]

1. If ${\displaystyle G}$ is non-abelian and ${\displaystyle H}$ is not a normal subgroup of ${\displaystyle G}$, then we can only say that the inverse elements ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ of a left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$ form a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$.
2. However, if ${\displaystyle H\triangleleft G}$ is a normal subgroup of ${\displaystyle G}$, then any left transversal is also a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$.

For the proof click show on the right hand side.

Proof
1. Since the mapping ${\displaystyle G\to G,\ x\mapsto x^{-1}}$ is an involution, that is a bijection which is its own inverse, we see that ${\displaystyle G={\dot {\bigcup }}_{i=1}^{n}\,g_{i}H}$ implies ${\displaystyle G=G^{-1}={\dot {\bigcup }}_{i=1}^{n}\,(g_{i}H)^{-1}={\dot {\bigcup }}_{i=1}^{n}\,H^{-1}g_{i}^{-1}={\dot {\bigcup }}_{i=1}^{n}\,Hg_{i}^{-1}}$.
2. For a normal subgroup ${\displaystyle H\triangleleft G}$, we have ${\displaystyle xH=Hx}$ for each ${\displaystyle x\in G}$.

Let ${\displaystyle \phi :\,G\to K}$ be a group homomorphism and ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of a subgroup ${\displaystyle H}$ in ${\displaystyle G}$ with finite index ${\displaystyle n=(G:H)\geq 1}$. We must check whether the image of this transversal under the homomorphism is again a transversal.

Proposition. The following two conditions are equivalent.

1. ${\displaystyle (\phi (g_{1}),\ldots ,\phi (g_{n}))}$ is a left transversal of the subgroup ${\displaystyle \phi (H)}$ in the image ${\displaystyle \phi (G)}$ with finite index ${\displaystyle (\phi (G):\phi (H))=n}$.
2. ${\displaystyle \ker(\phi )\leq H}$.

We emphasize this important equivalence in a formula:

${\displaystyle (1)\qquad \phi (G)={\dot {\bigcup }}_{i=1}^{n}\,\phi (g_{i})\phi (H)}$ and ${\displaystyle (\phi (G):\phi (H))=n\quad \Longleftrightarrow \quad \ker(\phi )\leq H}$.

For the proof click show on the right hand side.

Proof

By assumption, we have the disjoint left coset decomposition ${\displaystyle G={\dot {\bigcup }}_{i=1}^{n}\,g_{i}H}$ which comprises two statements simultaneously.

Firstly, the group ${\displaystyle G=\bigcup _{i=1}^{n}\,g_{i}H}$ is a union of cosets, and secondly, any two distinct cosets have an empty intersection ${\displaystyle g_{i}H\bigcap g_{j}H=\emptyset }$, for ${\displaystyle i\neq j}$.

Due to the properties of the set mapping associated with ${\displaystyle \phi }$, the homomorphism ${\displaystyle \phi }$ maps the union to another union

${\displaystyle \phi (G)=\phi (\bigcup _{i=1}^{n}\,g_{i}H)=\bigcup _{i=1}^{n}\,\phi (g_{i}H)=\bigcup _{i=1}^{n}\,\phi (g_{i})\phi (H)}$,

but weakens the equality for the intersection to a trivial inclusion

${\displaystyle \emptyset =\phi (\emptyset )=\phi (g_{i}H\bigcap g_{j}H)\subseteq \phi (g_{i}H)\bigcap \phi (g_{j}H)=\phi (g_{i})\phi (H)\bigcap \phi (g_{j})\phi (H)}$, for ${\displaystyle i\neq j}$.

To show that the images of the cosets remain disjoint we need the property ${\displaystyle \ker(\phi )\leq H}$ of the homomorphism ${\displaystyle \phi }$.

Suppose that ${\displaystyle \phi (g_{i})\phi (H)\bigcap \phi (g_{j})\phi (H)\neq \emptyset }$ for some ${\displaystyle 1\leq i\leq j\leq n}$, then we have ${\displaystyle \phi (g_{i})\phi (h_{i})=\phi (g_{j})\phi (h_{j})}$ for certain elements ${\displaystyle h_{i},h_{j}\in H}$.

Multiplying by ${\displaystyle \phi (g_{j})^{-1}}$ from the left and by ${\displaystyle \phi (h_{j})^{-1}}$ from the right, we obtain

${\displaystyle \phi (g_{j}^{-1}g_{i}h_{i}h_{j}^{-1})=\phi (g_{j})^{-1}\phi (g_{i})\phi (h_{i})\phi (h_{j})^{-1}=1}$, that is, ${\displaystyle g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\in \ker(\phi )\leq H}$.

Since ${\displaystyle h_{i}h_{j}^{-1}\in H}$, this implies ${\displaystyle g_{j}^{-1}g_{i}\in H}$, resp. ${\displaystyle g_{i}H=g_{j}H}$, and thus ${\displaystyle i=j}$.

Conversely, we use contraposition.

If the kernel ${\displaystyle \ker(\phi )}$ of ${\displaystyle \phi }$ is not contained in the subgroup ${\displaystyle H}$, then there exists an element ${\displaystyle x\in G\setminus H}$ such that ${\displaystyle \phi (x)=1}$.

But then the homomorphism ${\displaystyle \phi }$ maps the disjoint cosets ${\displaystyle xH\bigcap 1\cdot H=\emptyset }$ to equal cosets ${\displaystyle \phi (x)\phi (H)\bigcap \phi (1)\phi (H)=1\cdot \phi (H)\bigcap 1\cdot \phi (H)=\phi (H)}$.

## Permutation representation

Suppose ${\displaystyle (g_{1},\ldots ,g_{n})}$ is a left transversal of a subgroup ${\displaystyle H\leq G}$ of finite index ${\displaystyle n=(G:H)\geq 1}$ in a group ${\displaystyle G}$. A fixed element ${\displaystyle x\in G}$ gives rise to a unique permutation ${\displaystyle \pi _{x}\in S_{n}}$ of the left cosets of ${\displaystyle H}$ in ${\displaystyle G}$ by left multiplication such that

${\displaystyle (2)\qquad xg_{i}H=g_{\pi _{x}(i)}H,\qquad {\text{ resp. }}xg_{i}\in g_{\pi _{x}(i)}H,\qquad {\text{ resp. }}u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H}$, for each ${\displaystyle 1\leq i\leq n}$.

Similarly, if ${\displaystyle (d_{1},\ldots ,d_{n})}$ is a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$, then a fixed element ${\displaystyle x\in G}$ gives rise to a unique permutation ${\displaystyle \rho _{x}\in S_{n}}$ of the right cosets of ${\displaystyle H}$ in ${\displaystyle G}$ by right multiplication such that

${\displaystyle (3)\qquad Hd_{i}x=Hd_{\rho _{x}(i)},\qquad {\text{ resp. }}d_{i}x\in Hd_{\rho _{x}(i)},\qquad {\text{ resp. }}w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H}$, for each ${\displaystyle 1\leq i\leq n}$.

The elements ${\displaystyle u_{x}(i)}$, resp. ${\displaystyle w_{x}(i)}$, ${\displaystyle 1\leq i\leq n}$, of the subgroup ${\displaystyle H}$ are called the monomials associated with ${\displaystyle x}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$, resp. ${\displaystyle (d_{1},\ldots ,d_{n})}$.

Definitions. [1]

The mapping ${\displaystyle G\to S_{n},\ x\mapsto \pi _{x}}$, resp. ${\displaystyle G\to S_{n},\ x\mapsto \rho _{x}}$, is called the permutation representation of ${\displaystyle G}$ in the symmetric group ${\displaystyle S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$, resp. ${\displaystyle (d_{1},\ldots ,d_{n})}$.

The mapping ${\displaystyle G\to H^{n}\times S_{n},\ x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})}$, resp. ${\displaystyle G\to H^{n}\times S_{n},\ x\mapsto (w_{x}(1),\ldots ,w_{x}(n);\rho _{x})}$, is called the monomial representation of ${\displaystyle G}$ in ${\displaystyle H^{n}\times S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$, resp. ${\displaystyle (d_{1},\ldots ,d_{n})}$.

Lemma. For the special right transversal ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ associated to the left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$, we have the following relations between the monomials and permutations corresponding to an element ${\displaystyle x\in G}$:

${\displaystyle (4)\qquad w_{x^{-1}}(i)=u_{x}(i)^{-1}}$ for ${\displaystyle 1\leq i\leq n\qquad {\text{ and }}\qquad \rho _{x^{-1}}=\pi _{x}}$.

For the proof click show on the right hand side.

Proof

For the right transversal ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$, we have ${\displaystyle w_{x}(i)=g_{i}^{-1}xg_{\rho _{x}(i)}}$, for each ${\displaystyle 1\leq i\leq n}$. On the other hand, for the left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$, we have ${\displaystyle u_{x}(i)^{-1}=(g_{\pi _{x}(i)}^{-1}xg_{i})^{-1}=g_{i}^{-1}x^{-1}g_{\pi _{x}(i)}=g_{i}^{-1}x^{-1}g_{\rho _{x^{-1}}(i)}=w_{x^{-1}}(i)}$, for each ${\displaystyle 1\leq i\leq n}$. This relation simultaneously shows that, for any ${\displaystyle x\in G}$, the permutation representations and the associated monomials are connected by

${\displaystyle \rho _{x^{-1}}=\pi _{x}}$ and ${\displaystyle w_{x^{-1}}(i)=u_{x}(i)^{-1}}$ for each ${\displaystyle 1\leq i\leq n}$.

## Artin transfer

Let ${\displaystyle G}$ be a group and ${\displaystyle H\leq G}$ be a subgroup of finite index ${\displaystyle n=(G:H)\geq 1}$. Assume that ${\displaystyle (g_{1},\ldots ,g_{n})}$, resp. ${\displaystyle (d_{1},\ldots ,d_{n})}$, is a left, resp. right, transversal of ${\displaystyle H}$ in ${\displaystyle G}$ with associated permutation representation ${\displaystyle G\to S_{n},\ x\mapsto \pi _{x}}$, resp. ${\displaystyle \rho _{x}}$, such that ${\displaystyle u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H}$, resp. ${\displaystyle w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H}$, for ${\displaystyle 1\leq i\leq n}$.

Definitions. [2] [3]

The Artin transfer ${\displaystyle T_{G,H}:\ G\to H/H^{\prime }}$ from ${\displaystyle G}$ to the abelianization ${\displaystyle H/H^{\prime }}$ of ${\displaystyle H}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$, resp. ${\displaystyle (d_{1},\ldots ,d_{n})}$, is defined by

${\displaystyle (5)\qquad T_{G,H}^{(g)}(x):=\prod _{i=1}^{n}\,g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H^{\prime }\qquad {\text{ briefly }}T_{G,H}(x)=\prod _{i=1}^{n}\,u_{x}(i)\cdot H^{\prime }}$,

resp.

${\displaystyle (6)\qquad T_{G,H}^{(d)}(x):=\prod _{i=1}^{n}\,d_{i}xd_{\rho _{x}(i)}^{-1}\cdot H^{\prime }\qquad {\text{ briefly }}T_{G,H}(x)=\prod _{i=1}^{n}\,w_{x}(i)\cdot H^{\prime }}$,

for ${\displaystyle x\in G}$.

Remarks. Isaacs [4] calls the mapping ${\displaystyle P:\,G\to H}$, ${\displaystyle x\mapsto \prod _{i=1}^{n}\,u_{x}(i)}$, resp. ${\displaystyle x\mapsto \prod _{i=1}^{n}\,w_{x}(i)}$, the pre-transfer from ${\displaystyle G}$ to ${\displaystyle H}$. The pre-transfer can be composed with a homomorphism ${\displaystyle \phi :\,H\to A}$ from ${\displaystyle H}$ into an abelian group ${\displaystyle A}$ to define a more general version of the transfer ${\displaystyle (\phi \circ P):\,G\to A}$, ${\displaystyle x\mapsto \prod _{i=1}^{n}\,\phi (u_{x}(i))}$, resp. ${\displaystyle x\mapsto \prod _{i=1}^{n}\,\phi (w_{x}(i))}$, from ${\displaystyle G}$ to ${\displaystyle A}$ via ${\displaystyle \phi }$, which occurs in the book by Gorenstein. [5] Taking the natural epimorphism ${\displaystyle \phi :\,H\to H/H^{\prime }}$, ${\displaystyle v\mapsto vH^{\prime }}$, yields the preceding Definition of the Artin transfer ${\displaystyle T_{G,H}}$ in its original form by Schur [2] and by Emil Artin, [3] which has also been dubbed Verlagerung by Hasse. [6] Note that, in general, the pre-transfer is neither independent of the transversal nor a group homomorphism.

### Independence of the transversal

Assume that ${\displaystyle (\ell _{1},\ldots ,\ell _{n})}$ is another left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ such that ${\displaystyle G={\dot {\cup }}_{i=1}^{n}\,\ell _{i}H}$.

Proposition. [1] [2] [4] [5] [7] [8] [9] The Artin transfers with respect to ${\displaystyle (\ell )}$ and ${\displaystyle (g)}$ coincide, that is, ${\displaystyle T_{G,H}^{(\ell )}=T_{G,H}^{(g)}}$.

For the proof click show on the right hand side.

Proof

There exists a unique permutation ${\displaystyle \sigma \in S_{n}}$ such that ${\displaystyle g_{i}H=\ell _{\sigma (i)}H}$, for all ${\displaystyle 1\leq i\leq n}$. Consequently, ${\displaystyle h_{i}:=g_{i}^{-1}\ell _{\sigma (i)}\in H}$, resp. ${\displaystyle \ell _{\sigma (i)}=g_{i}h_{i}}$ with ${\displaystyle h_{i}\in H}$, for all ${\displaystyle 1\leq i\leq n}$. For a fixed element ${\displaystyle x\in G}$, there exists a unique permutation ${\displaystyle \lambda _{x}\in S_{n}}$ such that we have

${\displaystyle \ell _{\lambda _{x}(\sigma (i))}H=x\ell _{\sigma (i)}H=xg_{i}h_{i}H=xg_{i}H=g_{\pi _{x}(i)}H=g_{\pi _{x}(i)}h_{\pi _{x}(i)}H=\ell _{\sigma (\pi _{x}(i))}H}$,

for all ${\displaystyle 1\leq i\leq n}$. Therefore, the permutation representation of ${\displaystyle G}$ with respect to ${\displaystyle (\ell _{1},\ldots ,\ell _{n})}$ is given by ${\displaystyle \lambda _{x}\circ \sigma =\sigma \circ \pi _{x}}$, resp. ${\displaystyle \lambda _{x}=\sigma \circ \pi _{x}\circ \sigma ^{-1}\in S_{n}}$, for ${\displaystyle x\in G}$. Furthermore, for the connection between the elements ${\displaystyle v_{x}(i):=\ell _{\lambda _{x}(i)}^{-1}x\ell _{i}\in H}$ and ${\displaystyle u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H}$, we obtain

${\displaystyle v_{x}(\sigma (i))=\ell _{\lambda _{x}(\sigma (i))}^{-1}x\ell _{\sigma (i)}=\ell _{\sigma (\pi _{x}(i))}^{-1}xg_{i}h_{i}}$

${\displaystyle =(g_{\pi _{x}(i)}h_{\pi _{x}(i)})^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}}$,

for all ${\displaystyle 1\leq i\leq n}$. Finally, due to the commutativity of the quotient group ${\displaystyle H/H^{\prime }}$ and the fact that ${\displaystyle \sigma }$ and ${\displaystyle \pi _{x}}$ are permutations, the Artin transfer turns out to be independent of the left transversal:

${\displaystyle T_{G,H}^{(\ell )}(x)=\prod _{i=1}^{n}\,v_{x}(\sigma (i))\cdot H^{\prime }=\prod _{i=1}^{n}\,h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}\cdot H^{\prime }}$

${\displaystyle =\prod _{i=1}^{n}\,u_{x}(i)\prod _{i=1}^{n}\,h_{\pi _{x}(i)}^{-1}\prod _{i=1}^{n}\,h_{i}\cdot H^{\prime }}$

${\displaystyle =\prod _{i=1}^{n}\,u_{x}(i)\cdot 1\cdot H^{\prime }=\prod _{i=1}^{n}\,u_{x}(i)\cdot H^{\prime }=T_{G,H}^{(g)}(x)}$,

as defined in formula (5).

It is clear that a similar proof shows that the Artin transfer is independent of the choice between two different right transversals. It remains to show that the Artin transfer with respect to a right transversal coincides with the Artin transfer with respect to a left transversal.

For this purpose, we select the special right transversal ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ associated to the left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$.

Proposition. The Artin transfers with respect to ${\displaystyle (g^{-1})}$ and ${\displaystyle (g)}$ coincide, that is, ${\displaystyle T_{G,H}^{(g^{-1})}=T_{G,H}^{(g)}}$.

For the proof click show on the right hand side.

Proof

Using the commutativity of ${\displaystyle H/H^{\prime }}$ and formula (4), we consider the expression

${\displaystyle T_{G,H}^{(g^{-1})}(x)=\prod _{i=1}^{n}\,g_{i}^{-1}xg_{\rho _{x}(i)}\cdot H^{\prime }=\prod _{i=1}^{n}\,w_{x}(i)\cdot H^{\prime }}$

${\displaystyle =\prod _{i=1}^{n}\,u_{x^{-1}}(i)^{-1}\cdot H^{\prime }=(\prod _{i=1}^{n}\,u_{x^{-1}}(i)\cdot H^{\prime })^{-1}}$

${\displaystyle =(T_{G,H}^{(g)}(x^{-1}))^{-1}=T_{G,H}^{(g)}(x)}$.

The last step is justified by the fact that the Artin transfer is a homomorphism. This will be shown in the following section.

### Artin transfers as homomorphisms

Let${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$.

Theorem. [1] [2] [4] [5] [7] [8] [9] The Artin transfer ${\displaystyle T_{G,H}:\,G\to H/H^{\prime },\ x\mapsto \prod _{i=1}^{n}\,g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H^{\prime }}$ and the permutation representation ${\displaystyle G\to S_{n},\ x\mapsto \pi _{x}}$ are group homomorphisms:

${\displaystyle (7)\qquad T_{G,H}(xy)=T_{G,H}(x)\cdot T_{G,H}(y){\text{ and }}\pi _{xy}=\pi _{x}\circ \pi _{y}{\text{ for }}x,y\in G}$.

For the proof click show on the right hand side.

Proof

Let ${\displaystyle x,y\in G}$ be two elements with transfer images

${\displaystyle T_{G,H}(x)=\prod _{i=1}^{n}\,g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H^{\prime }}$ and

${\displaystyle T_{G,H}(y)=\prod _{j=1}^{n}\,g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H^{\prime }}$.

Since ${\displaystyle H/H^{\prime }}$ is abelian and ${\displaystyle \pi _{y}}$ is a permutation, we can change the order of the factors in the following product:

${\displaystyle T_{G,H}(x)\cdot T_{G,H}(y)=\prod _{i=1}^{n}\,g_{\pi _{x}(i)}^{-1}xg_{i}H^{\prime }\cdot \prod _{j=1}^{n}\,g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H^{\prime }}$

${\displaystyle =\prod _{j=1}^{n}\,g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}H^{\prime }\cdot \prod _{j=1}^{n}\,g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H^{\prime }}$

${\displaystyle =\prod _{j=1}^{n}\,g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H^{\prime }}$

${\displaystyle =\prod _{j=1}^{n}\,g_{(\pi _{x}\circ \pi _{y})(j))}^{-1}xyg_{j}\cdot H^{\prime }=T_{G,H}(xy)}$.

This relation simultaneously shows that the Artin transfer ${\displaystyle T_{G,H}}$ and the permutation representation ${\displaystyle G\to S_{n},\ x\mapsto \pi _{x}}$ are homomorphisms, since ${\displaystyle T_{G,H}(xy)=T_{G,H}(x)\cdot T_{G,H}(y)}$ and ${\displaystyle \pi _{xy}=\pi _{x}\circ \pi _{y}}$.

It is illuminating to restate the homomorphism property of the Artin transfer in terms of the monomial representation. The images of the factors ${\displaystyle x,y}$ are given by ${\displaystyle T_{G,H}(x)=\prod _{i=1}^{n}\,u_{x}(i)\cdot H^{\prime }}$ and ${\displaystyle T_{G,H}(y)=\prod _{j=1}^{n}\,u_{y}(j)\cdot H^{\prime }}$. In the last proof, the image of the product ${\displaystyle xy}$ turned out to be

${\displaystyle T_{G,H}(xy)=\prod _{j=1}^{n}\,g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H^{\prime }=\prod _{j=1}^{n}\,u_{x}(\pi _{y}(j))\cdot u_{y}(j)\cdot H^{\prime }}$,

which is a very peculiar law of composition discussed in more detail in the following section.

The law reminds of the crossed homomorphisms ${\displaystyle x\mapsto u_{x}}$ in the first cohomology group ${\displaystyle \mathrm {H} ^{1}(G,M)}$ of a ${\displaystyle G}$-module ${\displaystyle M}$, which have the property ${\displaystyle u_{xy}=u_{x}^{y}\cdot u_{y}}$ for ${\displaystyle x,y\in G}$.

### Wreath product of H and S(n)

The peculiar structures which arose in the previous section can also be interpreted by endowing the cartesian product ${\displaystyle H^{n}\times S_{n}}$ with a special law of composition known as the wreath product ${\displaystyle H\wr S_{n}}$ of the groups ${\displaystyle H}$ and ${\displaystyle S_{n}}$ with respect to the set ${\displaystyle \lbrace 1,\ldots ,n\rbrace }$.

Definition. For ${\displaystyle x,y\in G}$, the wreath product of the associated monomials and permutations is given by

${\displaystyle (8)\qquad (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\cdot (u_{y}(1),\ldots ,u_{y}(n);\pi _{y})}$

${\displaystyle :=(u_{x}(\pi _{y}(1))\cdot u_{y}(1),\ldots ,u_{x}(\pi _{y}(n))\cdot u_{y}(n);\pi _{x}\circ \pi _{y})}$

${\displaystyle =(u_{xy}(1),\ldots ,u_{xy}(n);\pi _{xy})}$.

Theorem. [1] [7] This law of composition on ${\displaystyle H^{n}\times S_{n}}$ causes the monomial representation ${\displaystyle G\to H\wr S_{n},\ x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})}$ also to be a homomorphism. In fact, it is an injective homomorphism, also called a monomorphism or embedding, in contrast to the permutation representation, which cannot be injective if ${\displaystyle G}$ is infinite or at least of an order bigger than ${\displaystyle n!}$, the factorial.

For the proof click show on the right hand side.

Proof

The homomorphism property has been shown above already. For a homomorphism to be injective, it suffices to show the triviality of its kernel. The neutral element of the group ${\displaystyle H^{n}\times S_{n}}$ endowed with the wreath product is given by ${\displaystyle (1,\ldots ,1;1)}$, where the last ${\displaystyle 1}$ means the identity permutation. If ${\displaystyle (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})=(1,\ldots ,1;1)}$, for some ${\displaystyle x\in G}$, then ${\displaystyle \pi _{x}=1}$ and consequently ${\displaystyle 1=u_{x}(i)=g_{\pi _{x}(i)}^{-1}xg_{i}=g_{i}^{-1}xg_{i}}$, for all ${\displaystyle 1\leq i\leq n}$. Finally, an application of the inverse inner automorphism with ${\displaystyle g_{i}}$ yields ${\displaystyle x=1}$, as required for injectivity.

Whereas Huppert[1] uses the monomial representation for defining the Artin transfer, we prefer to give the immediate definitions in formulas (5) and (6) and to merely illustrate the homomorphism property of the Artin transfer with the aid of the monomial representation.

### Composition of Artin transfers

Let ${\displaystyle G}$ be a group with nested subgroups ${\displaystyle K\leq H\leq G}$ such that the indices ${\displaystyle (G:H)=n}$, ${\displaystyle (H:K)=m}$ and ${\displaystyle (G:K)=(G:H)\cdot (H:K)=n\cdot m}$ are finite.

Theorem. [1] [7] Then the Artin transfer ${\displaystyle T_{G,K}}$ is the compositum of the induced transfer ${\displaystyle {\tilde {T}}_{H,K}:\ H/H^{\prime }\to K/K^{\prime }}$ and the Artin transfer ${\displaystyle T_{G,H}}$, that is,

${\displaystyle (9)\qquad T_{G,K}={\tilde {T}}_{H,K}\circ T_{G,H}}$.

For the proof click show on the right hand side.

Proof

This claim can be seen in the following manner.

If ${\displaystyle (g_{1},\ldots ,g_{n})}$ is a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ and ${\displaystyle (h_{1},\ldots ,h_{m})}$ is a left transversal of ${\displaystyle K}$ in ${\displaystyle H}$,

that is ${\displaystyle G={\dot {\cup }}_{i=1}^{n}\,g_{i}H}$ and ${\displaystyle H={\dot {\cup }}_{j=1}^{m}\,h_{j}K}$, then

${\displaystyle G={\dot {\cup }}_{i=1}^{n}\,{\dot {\cup }}_{j=1}^{m}\,g_{i}h_{j}K}$

is a disjoint left coset decomposition of ${\displaystyle G}$ with respect to ${\displaystyle K}$.

Given two elements ${\displaystyle x\in G}$ and ${\displaystyle y\in H}$, there exist unique permutations ${\displaystyle \pi _{x}\in S_{n}}$, and ${\displaystyle \sigma _{y}\in S_{m}}$, such that

${\displaystyle u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H}$, for each ${\displaystyle 1\leq i\leq n}$, and

${\displaystyle v_{y}(j):=h_{\sigma _{y}(j)}^{-1}yh_{j}\in K}$, for each ${\displaystyle 1\leq j\leq m}$.

Then, anticipating the definition of the induced transfer, we have

${\displaystyle T_{G,H}(x)=\prod _{i=1}^{n}\,u_{x}(i)\cdot H^{\prime }}$, and ${\displaystyle {\tilde {T}}_{H,K}(y\cdot H^{\prime })=T_{H,K}(y)=\prod _{j=1}^{m}\,v_{y}(j)\cdot K^{\prime }}$.

For each pair of subscripts ${\displaystyle 1\leq i\leq n}$ and ${\displaystyle 1\leq j\leq m}$, we put ${\displaystyle y_{i}:=u_{x}(i)}$, and we obtain

${\displaystyle xg_{i}h_{j}=g_{\pi _{x}(i)}g_{\pi _{x}(i)}^{-1}xg_{i}h_{j}=g_{\pi _{x}(i)}u_{x}(i)h_{j}=g_{\pi _{x}(i)}y_{i}h_{j}}$

${\displaystyle =g_{\pi _{x}(i)}h_{\sigma _{y_{i}}(j)}h_{\sigma _{y_{i}}(j)}^{-1}y_{i}h_{j}=g_{\pi _{x}(i)}h_{\sigma _{y_{i}}(j)}v_{y_{i}}(j)}$,

resp. ${\displaystyle h_{\sigma _{y_{i}}(j)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{j}=v_{y_{i}}(j)}$.

Therefore, the image of ${\displaystyle x}$ under the Artin transfer ${\displaystyle T_{G,K}}$ is given by

${\displaystyle T_{G,K}(x)=\prod _{i=1}^{n}\,\prod _{j=1}^{m}\,v_{y_{i}}(j)\cdot K^{\prime }=\prod _{i=1}^{n}\,\prod _{j=1}^{m}\,h_{\sigma _{y_{i}}(j)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{j}\cdot K^{\prime }}$

${\displaystyle =\prod _{i=1}^{n}\,\prod _{j=1}^{m}\,h_{\sigma _{y_{i}}(j)}^{-1}u_{x}(i)h_{j}\cdot K^{\prime }=\prod _{i=1}^{n}\,\prod _{j=1}^{m}\,h_{\sigma _{y_{i}}(j)}^{-1}y_{i}h_{j}\cdot K^{\prime }}$

${\displaystyle =\prod _{i=1}^{n}\,{\tilde {T}}_{H,K}(y_{i}\cdot H^{\prime })={\tilde {T}}_{H,K}(\prod _{i=1}^{n}\,y_{i}\cdot H^{\prime })={\tilde {T}}_{H,K}(\prod _{i=1}^{n}\,u_{x}(i)\cdot H^{\prime })}$

${\displaystyle ={\tilde {T}}_{H,K}(T_{G,H}(x))}$.

Finally, we want to emphasize the structural peculiarity of the monomial representation

${\displaystyle G\to K^{n\cdot m}\times S_{n\cdot m}}$, ${\displaystyle x\mapsto (k_{x}(1,1),\ldots ,k_{x}(n,m);\gamma _{x})}$,

which corresponds to the composite of Artin transfers, defining ${\displaystyle k_{x}(i,j):=((gh)_{\gamma _{x}(i,j)})^{-1}x(gh)_{(i,j)}\in K}$ for a permutation ${\displaystyle \gamma _{x}\in S_{n\cdot m}}$, and using the symbolic notation ${\displaystyle (gh)_{(i,j)}:=g_{i}h_{j}}$ for all pairs of subscripts ${\displaystyle 1\leq i\leq n}$, ${\displaystyle 1\leq j\leq m}$.

The preceding proof has shown that ${\displaystyle k_{x}(i,j)=h_{\sigma _{y_{i}}(j)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{j}}$. Therefore, the action of the permutation ${\displaystyle \gamma _{x}}$ on the set ${\displaystyle \lbrack 1,n\rbrack \times \lbrack 1,m\rbrack }$ is given by ${\displaystyle \gamma _{x}(i,j)=(\pi _{x}(i),\sigma _{u_{x}(i)}(j))}$. The action on the second component ${\displaystyle j}$ depends on the first component ${\displaystyle i}$ (via the permutation ${\displaystyle \sigma _{u_{x}(i)}\in S_{m}}$), whereas the action on the first component ${\displaystyle i}$ is independent of the second component ${\displaystyle j}$. Therefore, the permutation ${\displaystyle \gamma _{x}\in S_{n\cdot m}}$ can be identified with the multiplet

${\displaystyle (\pi _{x};\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)})\in S_{n}\times S_{m}^{n},}$

which will be written in twisted form in the next section.

### Wreath product of S(m) and S(n)

The permutations ${\displaystyle \gamma _{x}}$, which arose as second components of the monomial representation

${\displaystyle G\to K\wr S_{n\cdot m}}$, ${\displaystyle x\mapsto (k_{x}(1,1),\ldots ,k_{x}(n,m);\gamma _{x})}$,

in the previous section, are of a very special kind. They belong to the stabilizer of the natural equipartition of the set ${\displaystyle \lbrack 1,n\rbrack \times \lbrack 1,m\rbrack }$ into the ${\displaystyle n}$ rows of the corresponding matrix (rectangular array). Using the peculiarities of the composition of Artin transfers in the previous section, we show that this stabilizer is isomorphic to the wreath product ${\displaystyle S_{m}\wr S_{n}}$ of the symmetric groups ${\displaystyle S_{m}}$ and ${\displaystyle S_{n}}$ with respect to the set ${\displaystyle \lbrace 1,\ldots ,n\rbrace }$, whose underlying set ${\displaystyle S_{m}^{n}\times S_{n}}$ is endowed with the following law of composition

${\displaystyle (10)\qquad \gamma _{x}\cdot \gamma _{z}=(\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})\cdot (\sigma _{u_{z}(1)},\ldots ,\sigma _{u_{z}(n)};\pi _{z})}$

${\displaystyle =(\sigma _{u_{x}(\pi _{z}(1))}\circ \sigma _{u_{z}(1)},\ldots ,\sigma _{u_{x}(\pi _{z}(n))}\circ \sigma _{u_{z}(n)};\pi _{x}\circ \pi _{z})}$

${\displaystyle =(\sigma _{u_{xz}(1)},\ldots ,\sigma _{u_{xz}(n)};\pi _{xz})}$

${\displaystyle =\gamma _{xz}}$ for all ${\displaystyle x,z\in G}$.

This law reminds of the chain rule ${\displaystyle D(g\circ f)(x)=D(g)(f(x))\circ D(f)(x)}$ for the Fréchet derivative in ${\displaystyle x\in E}$ of the compositum of differentiable functions ${\displaystyle f:\,E\to F}$ and ${\displaystyle g:\,F\to G}$ between complete normed spaces.

The above considerations establish a third representation, the stabilizer representation,

${\displaystyle G\to S_{m}\wr S_{n},\ x\mapsto (\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})}$

of the group ${\displaystyle G}$ in the wreath product ${\displaystyle S_{m}\wr S_{n}}$, similar to the permutation representation and the monomial representation. As opposed to the latter, the stabilizer representation cannot be injective, in general. For instance, certainly not, if ${\displaystyle G}$ is infinite. Formula (10) proves the following statement.

Theorem. The stabilizer representation ${\displaystyle G\to S_{m}\wr S_{n},\ x\mapsto \gamma _{x}=(\sigma _{u_{x}(1)},\ldots ,\sigma _{u_{x}(n)};\pi _{x})}$ of the group ${\displaystyle G}$ in the wreath product ${\displaystyle S_{m}\wr S_{n}}$ of symmetric groups is a group homomorphism.

### Cycle decomposition

Let ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of a subgroup ${\displaystyle H\leq G}$ of finite index ${\displaystyle n=(G:H)\geq 1}$ in a group ${\displaystyle G}$. Suppose the element ${\displaystyle x\in G}$ gives rise to the permutation ${\displaystyle \pi _{x}\in S_{n}}$ of the left cosets of ${\displaystyle H}$ in ${\displaystyle G}$ such that ${\displaystyle xg_{i}H=g_{\pi _{x}(i)}H}$, resp. ${\displaystyle g_{\pi _{x}(i)}^{-1}xg_{i}=:u_{x}(i)\in H}$, for each ${\displaystyle 1\leq i\leq n}$.

Theorem. [1] [3] [4] [5] [8] [9] If the permutation ${\displaystyle \pi _{x}}$ has the decomposition ${\displaystyle \pi _{x}=\prod _{j=1}^{t}\,\zeta _{j}}$ into pairwise disjoint (and thus commuting) cycles ${\displaystyle \zeta _{j}\in S_{n}}$ of lengths ${\displaystyle f_{j}\geq 1}$, which is unique up to the ordering of the cycles, more explicitly, if

${\displaystyle (11)\qquad (g_{j}H,g_{\zeta _{j}(j)}H,g_{\zeta _{j}^{2}(j)}H,\ldots ,g_{\zeta _{j}^{f_{j}-1}(j)}H)=(g_{j}H,xg_{j}H,x^{2}g_{j}H,\ldots ,x^{f_{j}-1}g_{j}H)}$,

for ${\displaystyle 1\leq j\leq t}$, and ${\displaystyle \sum _{j=1}^{t}\,f_{j}=n}$, then the image of ${\displaystyle x\in G}$ under the Artin transfer ${\displaystyle T_{G,H}}$ is given by

${\displaystyle (12)\qquad T_{G,H}(x)=\prod _{j=1}^{t}\,g_{j}^{-1}x^{f_{j}}g_{j}\cdot H^{\prime }}$.

For the proof click show on the right hand side.

Proof

The reason for this fact is that we obtain another left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ by putting ${\displaystyle \ell _{j,k}:=x^{k}g_{j}}$ for ${\displaystyle 0\leq k\leq f_{j}-1}$ and ${\displaystyle 1\leq j\leq t}$, since

${\displaystyle (13)\qquad G={\dot {\cup }}_{j=1}^{t}\,{\dot {\cup }}_{k=0}^{f_{j}-1}\,x^{k}g_{j}H}$

is a disjoint decomposition of ${\displaystyle G}$ into left cosets of ${\displaystyle H}$.

Let us fix a value of ${\displaystyle 1\leq j\leq t}$. For ${\displaystyle 0\leq k\leq f_{j}-2}$, we have

${\displaystyle x\ell _{j,k}=xx^{k}g_{j}=x^{k+1}g_{j}=\ell _{j,k+1}\in \ell _{j,k+1}H}$, resp. ${\displaystyle u_{x}(j,k):=\ell _{j,k+1}^{-1}x\ell _{j,k}=1\in H}$.

However, for ${\displaystyle k=f_{j}-1}$, we obtain

${\displaystyle x\ell _{j,f_{j}-1}=xx^{f_{j}-1}g_{j}=x^{f_{j}}g_{j}\in g_{j}H=\ell _{j,0}H}$, resp. ${\displaystyle u_{x}(j,f_{j}-1):=\ell _{j,0}^{-1}x\ell _{j,f_{j}-1}=g_{j}^{-1}x^{f_{j}}g_{j}\in H}$.

Consequently,

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}\,\prod _{k=0}^{f_{j}-1}\,u_{x}(j,k)\cdot H^{\prime }=\prod _{j=1}^{t}\,(\prod _{k=0}^{f_{j}-2}\,1)\cdot u_{x}(j,f_{j}-1)\cdot H^{\prime }=\prod _{j=1}^{t}\,g_{j}^{-1}x^{f_{j}}g_{j}\cdot H^{\prime }}$.

The cycle decomposition corresponds to a double coset decomposition ${\displaystyle G={\dot {\cup }}_{j=1}^{t}\,\langle x\rangle g_{j}H}$ of the group ${\displaystyle G}$ modulo the cyclic group ${\displaystyle \langle x\rangle }$ and modulo the subgroup ${\displaystyle H}$. It was actually this cycle decomposition form of the transfer homomorphism which was given by E. Artin in his original 1929 paper.[3]

### Transfer to a normal subgroup

Let ${\displaystyle H\triangleleft G}$ be a normal subgroup of finite index ${\displaystyle n=(G:H)\geq 1}$ in a group ${\displaystyle G}$. Then we have ${\displaystyle xH=Hx}$, for all ${\displaystyle x\in G}$, and there exists the quotient group ${\displaystyle G/H}$ of order ${\displaystyle n}$. For an element ${\displaystyle x\in G}$, we let ${\displaystyle f:=\mathrm {ord} (xH)}$ denote the order of the coset ${\displaystyle xH}$ in ${\displaystyle G/H}$, and we let ${\displaystyle (\ell _{1},\ldots ,\ell _{t})}$ be a left transversal of the subgroup ${\displaystyle \langle x,H\rangle }$ in ${\displaystyle G}$, where ${\displaystyle t=n/f}$.

Theorem. Then the image of ${\displaystyle x\in G}$ under the Artin transfer ${\displaystyle T_{G,H}}$ is given by

${\displaystyle (14)\qquad T_{G,H}(x)=\prod _{j=1}^{t}\,g_{j}^{-1}x^{f}g_{j}\cdot H^{\prime }}$.

For the proof click show on the right hand side.

Proof

${\displaystyle \langle xH\rangle }$ is a cyclic subgroup of order ${\displaystyle f}$ in ${\displaystyle G/H}$, and a left transversal ${\displaystyle (g_{1},\ldots ,g_{t})}$ of the subgroup ${\displaystyle \langle x,H\rangle }$ in ${\displaystyle G}$,

where ${\displaystyle t=n/f}$ and ${\displaystyle G={\dot {\cup }}_{j=1}^{t}\,g_{j}\langle x,H\rangle }$ is the corresponding disjoint left coset decomposition,

can be refined to a left transversal ${\displaystyle g_{j}x^{k}}$ ${\displaystyle (1\leq j\leq t,\ 0\leq k\leq f-1)}$ with disjoint left coset decomposition

${\displaystyle (15)\qquad G={\dot {\cup }}_{j=1}^{t}\,{\dot {\cup }}_{k=0}^{f-1}\,g_{j}x^{k}H}$

of ${\displaystyle H}$ in ${\displaystyle G}$. Hence, the formula for the image of ${\displaystyle x}$ under the Artin transfer ${\displaystyle T_{G,H}}$ in the previous section takes the particular shape

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}\,g_{j}^{-1}x^{f}g_{j}\cdot H^{\prime }}$

with exponent ${\displaystyle f}$ independent of ${\displaystyle j}$.

Corollary. In particular, the inner transfer of an element ${\displaystyle x\in H}$ is given as a symbolic power

${\displaystyle (16)\qquad T_{G,H}(x)=x^{\mathrm {Tr} _{G}(H)}\cdot H^{\prime }}$

with the trace element

${\displaystyle (17)\qquad \mathrm {Tr} _{G}(H)=\sum _{j=1}^{t}\,g_{j}\in \mathbb {Z} \lbrack G\rbrack }$

of ${\displaystyle H}$ in ${\displaystyle G}$ as symbolic exponent.

The other extreme is the outer transfer of an element ${\displaystyle x\in G\setminus H}$ which generates ${\displaystyle G}$ modulo ${\displaystyle H}$, that is ${\displaystyle G=\langle x,H\rangle }$.

It is simply an ${\displaystyle n}$th power

${\displaystyle (18)\qquad T_{G,H}(x)=x^{n}\cdot H^{\prime }}$.

For the proof click show on the right hand side.

Proof

The inner transfer of an element ${\displaystyle x\in H}$, whose coset ${\displaystyle xH=H}$ is the principal set in ${\displaystyle G/H}$ of order ${\displaystyle f=1}$, is given as the symbolic power

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}\,g_{j}^{-1}xg_{j}\cdot H^{\prime }=\prod _{j=1}^{t}\,x^{g_{j}}\cdot H^{\prime }=x^{\sum _{j=1}^{t}\,g_{j}}\cdot H^{\prime }}$

with the trace element

${\displaystyle \mathrm {Tr} _{G}(H)=\sum _{j=1}^{t}\,g_{j}\in \mathbb {Z} \lbrack G\rbrack }$

of ${\displaystyle H}$ in ${\displaystyle G}$ as symbolic exponent.

The outer transfer of an element ${\displaystyle x\in G\setminus H}$ which generates ${\displaystyle G}$ modulo ${\displaystyle H}$, that is ${\displaystyle G=\langle x,H\rangle }$,

whence the coset ${\displaystyle xH}$ is generator of ${\displaystyle G/H}$ with order${\displaystyle f=n}$, is given as the ${\displaystyle n}$th power

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{1}\,1^{-1}\cdot x^{n}\cdot 1\cdot H^{\prime }=x^{n}\cdot H^{\prime }}$.

Transfers to normal subgroups will be the most important cases in the sequel, since the central concept of this article, the Artin pattern, which endows descendant trees with additional structure, consists of targets and kernels of Artin transfers from a group ${\displaystyle G}$ to intermediate groups ${\displaystyle G^{\prime }\leq H\leq G}$ between ${\displaystyle G}$ and its commutator subgroup ${\displaystyle G^{\prime }}$. For these intermediate groups we have the following lemma.

Lemma. All subgroups ${\displaystyle H\leq G}$ of a group ${\displaystyle G}$ which contain the commutator subgroup ${\displaystyle G^{\prime }}$ are normal subgroups ${\displaystyle H\triangleleft G}$.

For the proof click show on the right hand side.

Proof

Let ${\displaystyle G^{\prime }\leq H\leq G}$. If ${\displaystyle H}$ were not a normal subgroup of ${\displaystyle G}$, then we had ${\displaystyle x^{-1}Hx\not \subseteq H}$ for some element ${\displaystyle x\in G\setminus H}$. This would imply the existence of elements ${\displaystyle h\in H}$ and ${\displaystyle y\in G\setminus H}$ such that ${\displaystyle x^{-1}hx=y}$, and consequently the commutator ${\displaystyle \lbrack h,x\rbrack =h^{-1}x^{-1}hx=h^{-1}y}$ would be an element in ${\displaystyle G\setminus H}$ in contradiction to ${\displaystyle G^{\prime }\leq H}$.

Explicit implementations of Artin transfers in the simplest situations are presented in the following section.

## Computational implementation

### Abelianization of type (p,p)

Let ${\displaystyle G}$ be a p-group with abelianization ${\displaystyle G/G^{\prime }}$ of elementary abelian type ${\displaystyle (p,p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:H_{i})=p}$. In this particular case, the Frattini subgroup ${\displaystyle \Phi (G):=\bigcap _{i=1}^{p+1}\,H_{i}}$, which is defined as the intersection of all maximal subgroups, coincides with the commutator subgroup ${\displaystyle G^{\prime }=\lbrack G,G\rbrack }$, since the latter contains all pth powers ${\displaystyle G^{\prime }\geq G^{p}}$, and thus we have ${\displaystyle \Phi (G)=G^{p}\cdot G^{\prime }=G^{\prime }}$.

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{i}:\,G\to H_{i}/H_{i}^{\prime }}$ be the Artin transfer homomorphism from ${\displaystyle G}$ to the abelianization of ${\displaystyle H_{i}}$. According to Burnside's basis theorem, the group ${\displaystyle G}$ has generator rank ${\displaystyle d(G)=2}$ and can therefore be generated as ${\displaystyle G=\langle x,y\rangle }$ by two elements ${\displaystyle x,y}$ such that ${\displaystyle x^{p},y^{p}\in G^{\prime }}$. For each of the maximal subgroups ${\displaystyle H_{i}}$, which are normal subgroups ${\displaystyle H_{i}\triangleleft G}$ by the Lemma in the preceding section, we need a generator ${\displaystyle h_{i}}$ with respect to ${\displaystyle G^{\prime }}$, and a generator ${\displaystyle t_{i}}$ of a transversal ${\displaystyle (1,t_{i},t_{i}^{2},\ldots ,t_{i}^{p-1})}$ such that ${\displaystyle H_{i}=\langle h_{i},G^{\prime }\rangle }$ and ${\displaystyle G=\langle t_{i},H_{i}\rangle ={\dot {\bigcup }}_{j=0}^{p-1}\,t_{i}^{j}H_{i}}$.

A convenient selection is given by

${\displaystyle (19)\qquad h_{1}=y,\ t_{1}=x,{\text{ and }}h_{i}=xy^{i-2},\ t_{i}=y,{\text{ for all }}2\leq i\leq p+1}$.

Then, for each ${\displaystyle 1\leq i\leq p+1}$, it is possible to implement the inner transfer by

${\displaystyle (20)\qquad T_{i}(h_{i})=h_{i}^{\mathrm {Tr} _{G}(H_{i})}\cdot H_{i}^{\prime }=h_{i}^{1+t_{i}+t_{i}^{2}+\cdots +t_{i}^{p-1}}\cdot H_{i}^{\prime }}$,

according to equation (16) in the last Corollary, which can also be expressed by a product of two pth powers,

${\displaystyle (21)\qquad T_{i}(h_{i})=h_{i}\cdot t_{i}^{-1}h_{i}t_{i}\cdot t_{i}^{-2}h_{i}t_{i}^{2}\cdots t_{i}^{-p+1}h_{i}t_{i}^{p-1}\cdot H_{i}^{\prime }=(h_{i}t_{i}^{-1})^{p}t_{i}^{p}\cdot H_{i}^{\prime }}$,

and to implement the outer transfer as a complete pth power by

${\displaystyle (22)\qquad T_{i}(t_{i})=t_{i}^{p}\cdot H_{i}^{\prime }}$,

according to equation (18) in the preceding Corollary. The reason is that ${\displaystyle \mathrm {ord} (h_{i}H_{i})=1}$ and ${\displaystyle \mathrm {ord} (t_{i}H_{i})=p}$ in the quotient group ${\displaystyle G/H_{i}}$.

It should be pointed out that the complete specification of the Artin transfers ${\displaystyle T_{i}}$ also requires explicit knowledge of the derived subgroups ${\displaystyle H_{i}^{\prime }}$. Since ${\displaystyle G^{\prime }}$ is a normal subgroup of index ${\displaystyle p}$ in ${\displaystyle H_{i}}$, a certain general reduction is possible by ${\displaystyle H_{i}^{\prime }=\lbrack H_{i},H_{i}\rbrack =\lbrack G^{\prime },H_{i}\rbrack =(G^{\prime })^{h_{i}-1}}$, [10] but a presentation of ${\displaystyle G}$ must be known for determining generators of ${\displaystyle G^{\prime }=\langle s_{1},\ldots ,s_{n}\rangle }$, whence

${\displaystyle (23)\qquad H_{i}^{\prime }=(G^{\prime })^{h_{i}-1}=\langle \lbrack s_{1},h_{i}\rbrack ,\ldots ,\lbrack s_{n},h_{i}\rbrack \rangle }$.

### Abelianization of type (p2,p)

Let ${\displaystyle G}$ be a p-group with abelianization ${\displaystyle G/G^{\prime }}$ of non-elementary abelian type ${\displaystyle (p^{2},p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:H_{i})=p}$ and ${\displaystyle p+1}$ subgroups ${\displaystyle U_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:U_{i})=p^{2}}$.

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{1,i}:\,G\to H_{i}/H_{i}^{\prime }}$, resp. ${\displaystyle T_{2,i}:\,G\to U_{i}/U_{i}^{\prime }}$, be the Artin transfer homomorphism from ${\displaystyle G}$ to the abelianization of ${\displaystyle H_{i}}$, resp. ${\displaystyle U_{i}}$. Burnside's basis theorem asserts that the group ${\displaystyle G}$ has generator rank ${\displaystyle d(G)=2}$ and can therefore be generated as ${\displaystyle G=\langle x,y\rangle }$ by two elements ${\displaystyle x,y}$ such that ${\displaystyle x^{p^{2}},y^{p}\in G^{\prime }}$.

We begin by considering the first layer of subgroups. For each of the normal subgroups ${\displaystyle H_{i}\triangleleft G}$ ${\displaystyle (1\leq i\leq p)}$, we select a generator

${\displaystyle (24)\qquad h_{i}=xy^{i-1}}$ such that ${\displaystyle H_{i}=\langle h_{i},G^{\prime }\rangle }$.

These are the cases where the factor group ${\displaystyle H_{i}/G^{\prime }}$ is cyclic of order ${\displaystyle p^{2}}$. However, for the distinguished maximal subgroup ${\displaystyle H_{p+1}}$, for which the factor group ${\displaystyle H_{p+1}/G^{\prime }}$ is bicyclic of type ${\displaystyle (p,p)}$, we need two generators

${\displaystyle (25)\qquad h_{p+1}=y}$ and ${\displaystyle h_{0}=x^{p}}$ such that ${\displaystyle H_{p+1}=\langle h_{p+1},h_{0},G^{\prime }\rangle }$.

Further, a generator ${\displaystyle t_{i}}$ of a transversal must be given such that ${\displaystyle G=\langle t_{i},H_{i}\rangle }$, for each ${\displaystyle 1\leq i\leq p+1}$. It is convenient to define

${\displaystyle (26)\qquad t_{i}=y}$, for ${\displaystyle 1\leq i\leq p}$, and ${\displaystyle t_{p+1}=x}$.

Then, for each ${\displaystyle 1\leq i\leq p+1}$, we have the inner transfer

${\displaystyle (27)\qquad T_{1,i}(h_{i})=h_{i}^{\mathrm {Tr} _{G}(H_{i})}\cdot H_{i}^{\prime }=h_{i}^{1+t_{i}+t_{i}^{2}+\ldots +t_{i}^{p-1}}\cdot H_{i}^{\prime }}$,

which equals ${\displaystyle (h_{i}t_{i}^{-1})^{p}t_{i}^{p}\cdot H_{i}^{\prime }}$, and the outer transfer

${\displaystyle (28)\qquad T_{1,i}(t_{i})=t_{i}^{p}\cdot H_{i}^{\prime }}$,

since ${\displaystyle \mathrm {ord} (h_{i}H_{i})=1}$ and ${\displaystyle \mathrm {ord} (t_{i}H_{i})=p}$.

Now we continue by considering the second layer of subgroups. For each of the normal subgroups ${\displaystyle U_{i}\triangleleft G}$ ${\displaystyle (1\leq i\leq p+1)}$, we select a generator

${\displaystyle (29)\qquad u_{1}=y}$, ${\displaystyle u_{i}=x^{p}y^{i-1}}$ for ${\displaystyle 2\leq i\leq p}$, and ${\displaystyle u_{p+1}=x^{p}}$,

such that ${\displaystyle U_{i}=\langle u_{i},G^{\prime }\rangle }$. Among these subgroups, the Frattini subgroup ${\displaystyle U_{p+1}=\langle x^{p},G^{\prime }\rangle =G^{p}\cdot G^{\prime }}$ is particularly distinguished. A uniform way of defining generators ${\displaystyle t_{i},w_{i}}$ of a transversal such that ${\displaystyle G=\langle t_{i},w_{i},U_{i}\rangle }$, is to set

${\displaystyle (30)\qquad t_{i}=x,w_{i}=x^{p}}$, for ${\displaystyle 1\leq i\leq p}$, and ${\displaystyle t_{p+1}=x,w_{p+1}=y}$.

Since ${\displaystyle \mathrm {ord} (u_{i}U_{i})=1}$, but on the other hand ${\displaystyle \mathrm {ord} (t_{i}U_{i})=p^{2}}$ and ${\displaystyle \mathrm {ord} (w_{i}U_{i})=p}$, for ${\displaystyle 1\leq i\leq p+1}$, with the single exception that ${\displaystyle \mathrm {ord} (t_{p+1}U_{p+1})=p}$, we obtain the following expressions for the inner transfer

${\displaystyle (31)\qquad T_{2,i}(u_{i})=u_{i}^{\mathrm {Tr} _{G}(U_{i})}\cdot U_{i}^{\prime }=u_{i}^{\sum _{j=0}^{p-1}\,\sum _{k=0}^{p-1}\,w_{i}^{j}t_{i}^{k}}\cdot U_{i}^{\prime }=\prod _{j=0}^{p-1}\,\prod _{k=0}^{p-1}\,(w_{i}^{j}t_{i}^{k})^{-1}u_{i}w_{i}^{j}t_{i}^{k}\cdot U_{i}^{\prime }}$,

and for the outer transfer

${\displaystyle (32)\qquad T_{2,i}(t_{i})=t_{i}^{p^{2}}\cdot U_{i}^{\prime }}$,

exceptionally

${\displaystyle (33)\qquad T_{2,p+1}(t_{p+1})=(t_{p+1}^{p})^{1+w_{p+1}+w_{p+1}^{2}+\ldots +w_{p+1}^{p-1}}\cdot U_{p+1}^{\prime }}$,

and

${\displaystyle (34)\qquad T_{2,i}(w_{i})=(w_{i}^{p})^{1+t_{i}+t_{i}^{2}+\ldots +t_{i}^{p-1}}\cdot U_{i}^{\prime }}$,

for ${\displaystyle 1\leq i\leq p+1}$. Again, it should be emphasized that the structure of the derived subgroups ${\displaystyle H_{i}^{\prime }}$ and ${\displaystyle U_{i}^{\prime }}$ must be known to specify the action of the Artin transfers completely.

## Transfer kernels and targets

Let ${\displaystyle G}$ be a group with finite abelianization ${\displaystyle G/G^{\prime }}$. Suppose that ${\displaystyle (H_{i})_{i\in I}}$ denotes the family of all subgroups ${\displaystyle H_{i}\triangleleft G}$ which contain the commutator subgroup ${\displaystyle G^{\prime }}$ and are therefore necessarily normal, enumerated by means of the finite index set ${\displaystyle I}$. For each ${\displaystyle i\in I}$, let ${\displaystyle T_{i}:=T_{G,H_{i}}}$ be the Artin transfer from ${\displaystyle G}$ to the abelianization ${\displaystyle H_{i}/H_{i}^{\prime }}$.

Definition. [11]

The family of normal subgroups ${\displaystyle \varkappa _{H}(G)=(\ker(T_{i}))_{i\in I}}$ is called the transfer kernel type (TKT) of ${\displaystyle G}$ with respect to ${\displaystyle (H_{i})_{i\in I}}$, and the family of abelianizations (resp. their abelian type invariants) ${\displaystyle \tau _{H}(G)=(H_{i}/H_{i}^{\prime })_{i\in I}}$ is called the transfer target type (TTT) of ${\displaystyle G}$ with respect to ${\displaystyle (H_{i})_{i\in I}}$. Both families are also called multiplets whereas a single component will be referred to as a singulet.

Important examples for these concepts are provided in the following two sections.

## Abelianization of type (p,p)

Let ${\displaystyle G}$ be a p-group with abelianization ${\displaystyle G/G^{\prime }}$ of elementary abelian type ${\displaystyle (p,p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:H_{i})=p}$. For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{i}:\,G\to H_{i}/H_{i}^{\prime }}$ be the Artin transfer homomorphism from ${\displaystyle G}$ to the abelianization of ${\displaystyle H_{i}}$.

Definition.

The family of normal subgroups ${\displaystyle \varkappa _{H}(G)=(\ker(T_{i}))_{1\leq i\leq p+1}}$ is called the transfer kernel type (TKT) of ${\displaystyle G}$ with respect to ${\displaystyle H_{1},\ldots ,H_{p+1}}$.

Remarks.

• For brevity, the TKT is identified with the multiplet ${\displaystyle (\varkappa (i))_{1\leq i\leq p+1}}$, whose integer components are given by ${\displaystyle \varkappa (i)={\begin{cases}0&{\text{if }}\ker(T_{i})=G,\\j&{\text{if }}\ker(T_{i})=H_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$ Here, we take into consideration that each transfer kernel ${\displaystyle \ker(T_{i})}$ must contain the commutator subgroup ${\displaystyle G^{\prime }}$ of ${\displaystyle G}$, since the transfer target ${\displaystyle H_{i}/H_{i}^{\prime }}$ is abelian. However, the minimal case ${\displaystyle \ker(T_{i})=G^{\prime }}$ cannot occur.
• A renumeration of the maximal subgroups ${\displaystyle K_{i}=H_{\pi (i)}}$ and of the transfers ${\displaystyle V_{i}=T_{\pi (i)}}$ by means of a permutation ${\displaystyle \pi \in S_{p+1}}$ gives rise to a new TKT ${\displaystyle \lambda _{K}(G)=(\ker(V_{i}))_{1\leq i\leq p+1}}$ with respect to ${\displaystyle K_{1},\ldots ,K_{p+1}}$, identified with ${\displaystyle (\lambda (i))_{1\leq i\leq p+1}}$, where ${\displaystyle \lambda (i)={\begin{cases}0&{\text{ if }}\ker(V_{i})=G,\\j&{\text{ if }}\ker(V_{i})=K_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$ It is adequate to view the TKTs ${\displaystyle \lambda _{K}(G)\sim \varkappa _{H}(G)}$ as equivalent. Since we have ${\displaystyle K_{\lambda (i)}=\ker(V_{i})=\ker(T_{\pi (i)})=H_{\varkappa (\pi (i))}=K_{{\tilde {\pi }}^{-1}(\varkappa (\pi (i)))}}$, the relation between ${\displaystyle \lambda }$ and ${\displaystyle \varkappa }$ is given by ${\displaystyle \lambda ={\tilde {\pi }}^{-1}\circ \varkappa \circ \pi }$. Therefore, ${\displaystyle \lambda }$ is another representative of the orbit ${\displaystyle \varkappa ^{S_{p+1}}}$ of ${\displaystyle \varkappa }$ under the operation ${\displaystyle (\pi ,\mu )\mapsto {\tilde {\pi }}^{-1}\circ \mu \circ \pi }$ of the symmetric group ${\displaystyle S_{p+1}}$ on the set of all mappings from ${\displaystyle \lbrace 1,\ldots ,p+1\rbrace }$ to ${\displaystyle \lbrace 0,\ldots ,p+1\rbrace }$, where the extension ${\displaystyle {\tilde {\pi }}\in S_{p+2}}$ of the permutation ${\displaystyle \pi \in S_{p+1}}$ is defined by ${\displaystyle {\tilde {\pi }}(0)=0}$, and formally ${\displaystyle H_{0}=G}$, ${\displaystyle K_{0}=G}$.

Definition.

The orbit ${\displaystyle \varkappa (G)=\varkappa ^{S_{p+1}}}$ of any representative ${\displaystyle \varkappa }$ is an invariant of the p-group ${\displaystyle G}$ and is called its transfer kernel type, briefly TKT.

Remark.

Let ${\displaystyle \#{\mathcal {H}}_{0}(G):=\#\lbrace 1\leq i\leq p+1\mid \varkappa (i)=0\rbrace }$ denote the counter of total transfer kernels ${\displaystyle \ker(T_{i})=G}$, which is an invariant of the group ${\displaystyle G}$. In 1980, S. M. Chang and R. Foote [12] proved that, for any odd prime ${\displaystyle p}$ and for any integer ${\displaystyle 0\leq n\leq p+1}$, there exist metabelian p-groups ${\displaystyle G}$ having abelianization ${\displaystyle G/G^{\prime }}$ of type ${\displaystyle (p,p)}$ such that ${\displaystyle \#{\mathcal {H}}_{0}(G)=n}$. However, for ${\displaystyle p=2}$, there do not exist non-abelian ${\displaystyle 2}$-groups ${\displaystyle G}$ with ${\displaystyle G/G^{\prime }\simeq (2,2)}$, which must be metabelian of maximal class, such that ${\displaystyle \#{\mathcal {H}}_{0}(G)\geq 2}$. Only the elementary abelian ${\displaystyle 2}$-group ${\displaystyle G=C_{2}\times C_{2}}$ has ${\displaystyle \#{\mathcal {H}}_{0}(G)=3}$. See Figure 5.

In the following concrete examples for the counters ${\displaystyle \#{\mathcal {H}}_{0}(G)}$, and also in the remainder of this article, we use identifiers of finite p-groups in the SmallGroups Library by H. U. Besche, B. Eick and E. A. O'Brien .[13] [14]

For ${\displaystyle p=3}$, we have

• ${\displaystyle \#{\mathcal {H}}_{0}(G)=0}$ for the extra special group ${\displaystyle G=\langle 27,4\rangle }$ of exponent ${\displaystyle 9}$ with TKT ${\displaystyle \varkappa =(1111)}$ (Figure 6),
• ${\displaystyle \#{\mathcal {H}}_{0}(G)=1}$ for the two groups ${\displaystyle G\in \lbrace \langle 243,6\rangle ,\langle 243,8\rangle \rbrace }$ with TKTs ${\displaystyle \varkappa \in \lbrace (0122),(2034)\rbrace }$ (Figures 8 and 9),
• ${\displaystyle \#{\mathcal {H}}_{0}(G)=2}$ for the group ${\displaystyle G=\langle 243,3\rangle }$ with TKT ${\displaystyle \varkappa =(0043)}$ (Figure 4 in the article on descendant trees),
• ${\displaystyle \#{\mathcal {H}}_{0}(G)=3}$ for the group ${\displaystyle G=\langle 81,7\rangle }$ with TKT ${\displaystyle \varkappa =(2000)}$ (Figure 6),
• ${\displaystyle \#{\mathcal {H}}_{0}(G)=4}$ for the extra special group ${\displaystyle G=\langle 27,3\rangle }$ of exponent ${\displaystyle 3}$ with TKT ${\displaystyle \varkappa =(0000)}$ (Figure 6).

## Abelianization of type (p2,p)

Let ${\displaystyle G}$ be a p-group with abelianization ${\displaystyle G/G^{\prime }}$ of non-elementary abelian type ${\displaystyle (p^{2},p)}$. Then ${\displaystyle G}$ possesses ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:H_{i})=p}$, and ${\displaystyle p+1}$ subgroups ${\displaystyle U_{i} ${\displaystyle (1\leq i\leq p+1)}$ of index ${\displaystyle (G:U_{i})=p^{2}}$.

Assumption.

Suppose that ${\displaystyle H_{p+1}=\prod _{j=1}^{p+1}\,U_{j}}$ is the distinguished maximal subgroup which is the product of all subgroups of index ${\displaystyle p^{2}}$, and ${\displaystyle U_{p+1}=\cap _{j=1}^{p+1}\,H_{j}}$ is the distinguished subgroup of index ${\displaystyle p^{2}}$ which is the intersection of all maximal subgroups, that is the Frattini subgroup ${\displaystyle \Phi (G)}$ of ${\displaystyle G}$.

### First layer

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{1,i}:\,G\to H_{i}/H_{i}^{\prime }}$ be the Artin transfer homomorphism from ${\displaystyle G}$ to the abelianization of ${\displaystyle H_{i}}$.

Definition.

The family ${\displaystyle \varkappa _{1,H,U}(G)=(\ker(T_{1,i}))_{1\leq i\leq p+1}}$ is called the first layer transfer kernel type of ${\displaystyle G}$ with respect to ${\displaystyle H_{1},\ldots ,H_{p+1}}$ and ${\displaystyle U_{1},\ldots ,U_{p+1}}$, and is identified with ${\displaystyle (\varkappa _{1}(i))_{1\leq i\leq p+1}}$, where ${\displaystyle \varkappa _{1}(i)={\begin{cases}0&{\text{ if }}\ker(T_{1,i})=H_{p+1},\\j&{\text{ if }}\ker(T_{1,i})=U_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$

Remark.

Here, we observe that each first layer transfer kernel is of exponent ${\displaystyle p}$ with respect to ${\displaystyle G^{\prime }}$ and consequently cannot coincide with ${\displaystyle H_{j}}$ for any ${\displaystyle 1\leq j\leq p}$, since ${\displaystyle H_{j}/G^{\prime }}$ is cyclic of order ${\displaystyle p^{2}}$, whereas ${\displaystyle H_{p+1}/G^{\prime }}$ is bicyclic of type ${\displaystyle (p,p)}$.

### Second layer

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{2,i}:\,G\to U_{i}/U_{i}^{\prime }}$ be the Artin transfer homomorphism from ${\displaystyle G}$ to the abelianization of ${\displaystyle U_{i}}$.

Definition.

The family ${\displaystyle \varkappa _{2,U,H}(G)=(\ker(T_{2,i}))_{1\leq i\leq p+1}}$ is called the second layer transfer kernel type of ${\displaystyle G}$ with respect to ${\displaystyle U_{1},\ldots ,U_{p+1}}$ and ${\displaystyle H_{1},\ldots ,H_{p+1}}$, and is identified with ${\displaystyle (\varkappa _{2}(i))_{1\leq i\leq p+1}}$, where ${\displaystyle \varkappa _{2}(i)={\begin{cases}0&{\text{ if }}\ker(T_{2,i})=G,\\j&{\text{ if }}\ker(T_{2,i})=H_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$

### Transfer kernel type

Combining the information on the two layers, we obtain the (complete) transfer kernel type ${\displaystyle \varkappa _{H,U}(G)=(\varkappa _{1,H,U}(G);\varkappa _{2,U,H}(G))}$ of the p-group ${\displaystyle G}$ with respect to ${\displaystyle H_{1},\ldots ,H_{p+1}}$ and ${\displaystyle U_{1},\ldots ,U_{p+1}}$.

Remark.

The distinguished subgroups ${\displaystyle H_{p+1}}$ and ${\displaystyle U_{p+1}=\Phi (G)}$ are unique invariants of ${\displaystyle G}$ and should not be renumerated. However, independent renumerations of the remaining maximal subgroups ${\displaystyle K_{i}=H_{\tau (i)}}$ ${\displaystyle (1\leq i\leq p)}$ and the transfers ${\displaystyle V_{1,i}=T_{1,\tau (i)}}$ by means of a permutation ${\displaystyle \tau \in S_{p}}$, and of the remaining subgroups ${\displaystyle W_{i}=U_{\sigma (i)}}$ ${\displaystyle (1\leq i\leq p)}$ of index ${\displaystyle p^{2}}$ and the transfers ${\displaystyle V_{2,i}=T_{2,\sigma (i)}}$ by means of a permutation ${\displaystyle \sigma \in S_{p}}$, give rise to new TKTs ${\displaystyle \lambda _{1,K,W}(G)=(\ker(V_{1,i}))_{1\leq i\leq p+1}}$ with respect to ${\displaystyle K_{1},\ldots ,K_{p+1}}$ and ${\displaystyle W_{1},\ldots ,W_{p+1}}$, identified with ${\displaystyle (\lambda _{1}(i))_{1\leq i\leq p+1}}$, where ${\displaystyle \lambda _{1}(i)={\begin{cases}0&{\text{ if }}\ker(V_{1,i})=K_{p+1},\\j&{\text{ if }}\ker(V_{1,i})=W_{j}{\text{ for some }}1\leq j\leq p+1,\end{cases}}}$ and ${\displaystyle \lambda _{2,W,K}(G)=(\ker(V_{2,i}))_{1\leq i\leq p+1}}$ with respect to ${\displaystyle W_{1},\ldots ,W_{p+1}}$ and ${\displaystyle K_{1},\ldots ,K_{p+1}}$, identified with ${\displaystyle (\lambda _{2}(i))_{1\leq i\leq p+1}}$, where ${\displaystyle \lambda _{2}(i)={\begin{cases}0&{\text{ if }}\ker(V_{2,i})=G,\\j&{\text{ if }}\ker(V_{2,i})=K_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$ It is adequate to view the TKTs ${\displaystyle \lambda _{1,K,W}(G)\sim \varkappa _{1,H,U}(G)}$ and ${\displaystyle \lambda _{2,W,K}(G)\sim \varkappa _{2,U,H}(G)}$ as equivalent. Since we have ${\displaystyle W_{\lambda _{1}(i)}=\ker(V_{1,i})=\ker(T_{1,{\hat {\tau }}(i)})=U_{\varkappa _{1}({\hat {\tau }}(i))}=W_{{\tilde {\sigma }}^{-1}(\varkappa _{1}({\hat {\tau }}(i)))}}$, resp. ${\displaystyle K_{\lambda _{2}(i)}=\ker(V_{2,i})=\ker(T_{2,{\hat {\sigma }}(i)})=H_{\varkappa _{2}({\hat {\sigma }}(i))}=K_{{\tilde {\tau }}^{-1}(\varkappa _{2}({\hat {\sigma }}(i)))}}$, the relations between ${\displaystyle \lambda _{1}}$ and ${\displaystyle \varkappa _{1}}$, resp. ${\displaystyle \lambda _{2}}$ and ${\displaystyle \varkappa _{2}}$, are given by ${\displaystyle \lambda _{1}={$