# 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 ${\displaystyle n.}$

Definitions.[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

${\displaystyle G=\bigsqcup _{i=1}^{n}g_{i}H.}$

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

${\displaystyle G=\bigsqcup _{i=1}^{n}Hd_{i}.}$

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] Let ${\displaystyle G}$ be a non-abelian group with subgroup ${\displaystyle H}$. Then the inverse elements ${\displaystyle (g_{1}^{-1},\ldots ,g_{n}^{-1})}$ of a left transversal ${\displaystyle (g_{1},\ldots ,g_{n})}$ of ${\displaystyle H}$ in ${\displaystyle G}$ form a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$. Moreover, if ${\displaystyle H}$ is a normal subgroup of ${\displaystyle G}$, then any left transversal is also a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$.

Proof. Since the mapping ${\displaystyle x\mapsto x^{-1}}$ is an involution of ${\displaystyle G}$ we see that:
${\displaystyle G=G^{-1}=\bigsqcup _{i=1}^{n}(g_{i}H)^{-1}=\bigsqcup _{i=1}^{n}H^{-1}g_{i}^{-1}=\bigsqcup _{i=1}^{n}Hg_{i}^{-1}.}$
For a normal subgroup ${\displaystyle H}$ we have ${\displaystyle xH=Hx}$ for each ${\displaystyle x\in G}$.

We must check when the image of a transversal under a homomorphism is also a transversal.

Proposition. 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 ${\displaystyle n.}$ The following two conditions are equivalent:

• ${\displaystyle (\phi (g_{1}),\ldots ,\phi (g_{n}))}$ is a left transversal of the subgroup ${\displaystyle \phi (H)}$ in the image ${\displaystyle \phi (G)}$ with finite index ${\displaystyle (\phi (G):\phi (H))=n.}$
• ${\displaystyle \ker(\phi )\leq H.}$
Proof. As a mapping of sets ${\displaystyle \phi }$ maps the union to another union:
${\displaystyle \phi (G)=\phi \left(\bigcup _{i=1}^{n}g_{i}H\right)=\bigcup _{i=1}^{n}\phi (g_{i}H)=\bigcup _{i=1}^{n}\phi (g_{i})\phi (H),}$
but weakens the equality for the intersection to a trivial inclusion:
${\displaystyle \emptyset =\phi (\emptyset )=\phi (g_{i}H\cap g_{j}H)\subseteq \phi (g_{i}H)\cap \phi (g_{j}H)=\phi (g_{i})\phi (H)\cap \phi (g_{j})\phi (H),\qquad i\neq j.}$
Suppose for some ${\displaystyle 1\leq i\leq j\leq n}$:
${\displaystyle \phi (g_{i})\phi (H)\cap \phi (g_{j})\phi (H)\neq \emptyset }$
then there exists elements ${\displaystyle h_{i},h_{j}\in H}$ such that
${\displaystyle \phi (g_{i})\phi (h_{i})=\phi (g_{j})\phi (h_{j})}$
Then we have:
{\displaystyle {\begin{aligned}\phi (g_{i})\phi (h_{i})=\phi (g_{j})\phi (h_{j})&\Longrightarrow \phi (g_{j})^{-1}\phi (g_{i})\phi (h_{i})\phi (h_{j})^{-1}=1\\&\Longrightarrow \phi \left(g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\right)=1\\&\Longrightarrow g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\in \ker(\phi )\\&\Longrightarrow g_{j}^{-1}g_{i}h_{i}h_{j}^{-1}\in H&&\ker(\phi )\leq H\\&\Longrightarrow g_{j}^{-1}g_{i}\in H&&h_{i}h_{j}^{-1}\in H\\&\Longrightarrow g_{i}H=g_{j}H\\&\Longrightarrow i=j\end{aligned}}}
Conversely if ${\displaystyle \ker(\phi )\nsubseteq H}$ then there exists ${\displaystyle x\in G\setminus H}$ such that ${\displaystyle \phi (x)=1.}$ But the homomorphism ${\displaystyle \phi }$ maps the disjoint cosets ${\displaystyle x\cdot H\cap 1\cdot H=\emptyset }$ to equal cosets:
${\displaystyle \phi (x)\phi (H)\cap \phi (1)\phi (H)=1\cdot \phi (H)\cap 1\cdot \phi (H)=\phi (H).}$

Remark. We emphasize the important equivalence of the proposition in a formula:

${\displaystyle (1)\quad \ker(\phi )\leq H\quad \Longleftrightarrow \quad {\begin{cases}\phi (G)=\bigsqcup _{i=1}^{n}\phi (g_{i})\phi (H)\\(\phi (G):\phi (H))=n\end{cases}}}$

## Permutation representation

Suppose ${\displaystyle (g_{1},\ldots ,g_{n})}$ is a left transversal of a subgroup ${\displaystyle H}$ of finite index ${\displaystyle n}$ 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:

${\displaystyle (2)\quad \forall i\in \{1,\ldots ,n\}:\qquad xg_{i}H=g_{\pi _{x}(i)}H\Longrightarrow xg_{i}\in g_{\pi _{x}(i)}H.}$

Using this we define a set of elements called the monomials associated with ${\displaystyle x}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H.}$

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:

${\displaystyle (3)\quad \forall i\in \{1,\ldots ,n\}:\qquad Hd_{i}x=Hd_{\rho _{x}(i)}\Longrightarrow d_{i}x\in Hd_{\rho _{x}(i)}.}$

And we define the monomials associated with ${\displaystyle x}$ with respect to ${\displaystyle (d_{1},\ldots ,d_{n})}$:

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H.}$

Definition.[1] The mappings:

${\displaystyle {\begin{cases}G\to S_{n}\\x\mapsto \pi _{x}\end{cases}}\qquad {\begin{cases}G\to S_{n}\\x\mapsto \rho _{x}\end{cases}}}$

are called the permutation representation of ${\displaystyle G}$ in the symmetric group ${\displaystyle S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ and ${\displaystyle (d_{1},\ldots ,d_{n})}$ respectively.

Definition.[1] The mappings:

${\displaystyle {\begin{cases}G\to H^{n}\times S_{n}\\x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\end{cases}}\qquad {\begin{cases}G\to H^{n}\times S_{n}\\x\mapsto (w_{x}(1),\ldots ,w_{x}(n);\rho _{x})\end{cases}}}$

are called the monomial representation of ${\displaystyle G}$ in ${\displaystyle H^{n}\times S_{n}}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ and ${\displaystyle (d_{1},\ldots ,d_{n})}$ respectively.

Lemma. For the 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 ${\displaystyle x\in G}$:

${\displaystyle (4)\quad {\begin{cases}w_{x^{-1}}(i)=u_{x}(i)^{-1}&1\leq i\leq n\\\rho _{x^{-1}}=\pi _{x}\end{cases}}}$
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
${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i)^{-1}=\left(g_{\pi _{x}(i)}^{-1}xg_{i}\right)^{-1}=g_{i}^{-1}x^{-1}g_{\pi _{x}(i)}=g_{i}^{-1}x^{-1}g_{\rho _{x^{-1}}(i)}=w_{x^{-1}}(i).}$
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

Definitions.[2][3] Let ${\displaystyle G}$ be a group and ${\displaystyle H}$ a subgroup of finite index ${\displaystyle n.}$ Assume ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ is a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ with associated permutation representation ${\displaystyle \pi _{x}:G\to S_{n},}$ such that

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad u_{x}(i):=g_{\pi _{x}(i)}^{-1}xg_{i}\in H.}$

Similarly let ${\displaystyle (d)=(d_{1},\ldots ,d_{n})}$ be a right transversal of ${\displaystyle H}$ in ${\displaystyle G}$ with associated permutation representation ${\displaystyle \rho _{x}:G\to S_{n}}$ such that

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad w_{x}(i):=d_{i}xd_{\rho _{x}(i)}^{-1}\in H.}$

The Artin transfer ${\displaystyle T_{G,H}^{(g)}:G\to H/H'}$ with respect to ${\displaystyle (g_{1},\ldots ,g_{n})}$ is defined as:

${\displaystyle (5)\quad \forall x\in G:\qquad T_{G,H}^{(g)}(x):=\prod _{i=1}^{n}g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot H'.}$

Similarly we define:

${\displaystyle (6)\quad \forall x\in G:\qquad T_{G,H}^{(d)}(x):=\prod _{i=1}^{n}d_{i}xd_{\rho _{x}(i)}^{-1}\cdot H'=\prod _{i=1}^{n}w_{x}(i)\cdot H'.}$

Remarks. Isaacs[4] calls the mappings

${\displaystyle {\begin{cases}P:G\to H\\x\mapsto \prod _{i=1}^{n}u_{x}(i)\end{cases}}\qquad {\begin{cases}P:G\to H\\x\mapsto \prod _{i=1}^{n}w_{x}(i)\end{cases}}}$

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 from ${\displaystyle G}$ to ${\displaystyle A}$ via ${\displaystyle \phi }$, which occurs in the book by Gorenstein.[5]

${\displaystyle {\begin{cases}(\phi \circ P):G\to A\\x\mapsto \prod _{i=1}^{n}\phi (u_{x}(i))\end{cases}}\qquad {\begin{cases}(\phi \circ P):G\to A\\x\mapsto \prod _{i=1}^{n}\phi (w_{x}(i))\end{cases}}}$

Taking the natural epimorphism

${\displaystyle {\begin{cases}\phi :H\to H/H'\\v\mapsto vH'\end{cases}}}$

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

Proposition.[1][2][4][5][7][8][9] The Artin transfers with respect to any two left transversals of ${\displaystyle H}$ in ${\displaystyle G}$ coincide.

Proof. Let ${\displaystyle (\ell )=(\ell _{1},\ldots ,\ell _{n})}$ and ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ be two left transversals of ${\displaystyle H}$ in ${\displaystyle G}$. Then there exists a unique permutation ${\displaystyle \sigma \in S_{n}}$ such that:
${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad g_{i}H=\ell _{\sigma (i)}H.}$
Consequently:
${\displaystyle \forall i\in \{1,\ldots ,n\},\exists h_{i}\in H:\qquad g_{i}h_{i}=\ell _{\sigma (i)}.}$
For a fixed element ${\displaystyle x\in G}$, there exists a unique permutation ${\displaystyle \lambda _{x}\in S_{n}}$ such that:
${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad \ell _{\lambda _{x}(\sigma (i))}H=x\ell _{\sigma (i)}H=xg_{i}h_{i}H=xg_{i}H=g_{\pi _{x}(i)}H=g_{\pi _{x}(i)}h_{\pi _{x}(i)}H=\ell _{\sigma (\pi _{x}(i))}H.}$
Therefore, the permutation representation of ${\displaystyle G}$ with respect to ${\displaystyle (\ell _{1},\ldots ,\ell _{n})}$ is given by ${\displaystyle \lambda _{x}\circ \sigma =\sigma \circ \pi _{x}}$ which yields: ${\displaystyle \lambda _{x}=\sigma \circ \pi _{x}\circ \sigma ^{-1}\in S_{n}.}$ Furthermore, for the connection between the two elements:
{\displaystyle {\begin{aligned}v_{x}(i)&:=\ell _{\lambda _{x}(i)}^{-1}x\ell _{i}\in H\\u_{x}(i)&:=g_{\pi _{x}(i)}^{-1}xg_{i}\in H\end{aligned}}}
we have:
${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad v_{x}(\sigma (i))=\ell _{\lambda _{x}(\sigma (i))}^{-1}x\ell _{\sigma (i)}=\ell _{\sigma (\pi _{x}(i))}^{-1}xg_{i}h_{i}=\left(g_{\pi _{x}(i)}h_{\pi _{x}(i)}\right)^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}g_{\pi _{x}(i)}^{-1}xg_{i}h_{i}=h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}.}$
Finally since ${\displaystyle H/H'}$ is abelian and ${\displaystyle \sigma }$ and ${\displaystyle \pi _{x}}$ are permutations, the Artin transfer turns out to be independent of the left transversal:
${\displaystyle T_{G,H}^{(\ell )}(x)=\prod _{i=1}^{n}v_{x}(\sigma (i))\cdot H'=\prod _{i=1}^{n}h_{\pi _{x}(i)}^{-1}u_{x}(i)h_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\prod _{i=1}^{n}h_{\pi _{x}(i)}^{-1}\prod _{i=1}^{n}h_{i}\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot 1\cdot H'=\prod _{i=1}^{n}u_{x}(i)\cdot H'=T_{G,H}^{(g)}(x),}$
as defined in formula (5).

Proposition. The Artin transfers with respect to any two right transversals of ${\displaystyle H}$ in ${\displaystyle G}$ coincide.

Proof. Similar to the previous proposition.

Proposition. The Artin transfers with respect to ${\displaystyle (g^{-1})=(g_{1}^{-1},\ldots ,g_{n}^{-1})}$ and ${\displaystyle (g)=(g_{1},\ldots ,g_{n})}$ coincide.

Proof. Using formula (4) and ${\displaystyle H/H'}$ being abelian we have:
${\displaystyle T_{G,H}^{(g^{-1})}(x)=\prod _{i=1}^{n}g_{i}^{-1}xg_{\rho _{x}(i)}\cdot H'=\prod _{i=1}^{n}w_{x}(i)\cdot H'=\prod _{i=1}^{n}u_{x^{-1}}(i)^{-1}\cdot H'=\left(\prod _{i=1}^{n}u_{x^{-1}}(i)\cdot H'\right)^{-1}=\left(T_{G,H}^{(g)}\left(x^{-1}\right)\right)^{-1}=T_{G,H}^{(g)}(x).}$
The last step is justified by the fact that the Artin transfer is a homomorphism. This will be shown in the following section.

Corollary. The Artin transfer is independent of the choice of transversals and only depends on ${\displaystyle H}$ and ${\displaystyle G}$.

### Artin transfers as homomorphisms

Theorem.[1][2][4][5][7][8][9] Let ${\displaystyle (g_{1},\ldots ,g_{n})}$ be a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$. The Artin transfer

${\displaystyle {\begin{cases}T_{G,H}:G\to H/H'\\x\mapsto \prod _{i=1}^{n}g_{\pi _{x}(i)}^{-1}xg_{i}\cdot H'\end{cases}}}$

and the permutation representation:

${\displaystyle {\begin{cases}G\to S_{n}\\x\mapsto \pi _{x}\end{cases}}}$

are group homomorphisms:

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

Let ${\displaystyle x,y\in G}$:

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

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

{\displaystyle {\begin{aligned}\prod _{i=1}^{n}g_{\pi _{x}(i)}^{-1}xg_{i}H'\cdot \prod _{j=1}^{n}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H'&=\prod _{j=1}^{n}g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}H'\cdot \prod _{j=1}^{n}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H'\\&=\prod _{j=1}^{n}g_{\pi _{x}(\pi _{y}(j))}^{-1}xg_{\pi _{y}(j)}g_{\pi _{y}(j)}^{-1}yg_{j}\cdot H'\\&=\prod _{j=1}^{n}g_{(\pi _{x}\circ \pi _{y})(j))}^{-1}xyg_{j}\cdot H'\\&=T_{G,H}(xy)\end{aligned}}}

This relation simultaneously shows that the Artin transfer and the permutation representation are homomorphisms.

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

${\displaystyle T_{G,H}(x)=\prod _{i=1}^{n}u_{x}(i)\cdot H'\quad {\text{and}}\quad T_{G,H}(y)=\prod _{j=1}^{n}u_{y}(j)\cdot H'.}$

In the last proof, the image of the product ${\displaystyle xy}$ turned out to be

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

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

The law is reminiscent of 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 ${\displaystyle \{1,\ldots ,n\}.}$

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

${\displaystyle (8)\quad (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\cdot (u_{y}(1),\ldots ,u_{y}(n);\pi _{y}):=(u_{x}(\pi _{y}(1))\cdot u_{y}(1),\ldots ,u_{x}(\pi _{y}(n))\cdot u_{y}(n);\pi _{x}\circ \pi _{y})=(u_{xy}(1),\ldots ,u_{xy}(n);\pi _{xy}).}$

Theorem.[1][7] With this law of composition on ${\displaystyle H^{n}\times S_{n}}$ the monomial representation

${\displaystyle {\begin{cases}G\to H\wr S_{n}\\x\mapsto (u_{x}(1),\ldots ,u_{x}(n);\pi _{x})\end{cases}}}$

is an injective homomorphism.

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

${\displaystyle \forall i\in \{1,\ldots ,n\}:\qquad 1=u_{x}(i)=g_{\pi _{x}(i)}^{-1}xg_{i}=g_{i}^{-1}xg_{i}.}$

Finally, an application of the inverse inner automorphism with ${\displaystyle g_{i}}$ yields ${\displaystyle x=1}$, as required for injectivity.

Remark. The monomial representation of the theorem stands in contrast to the permutation representation, which cannot be injective if ${\displaystyle |G|>n!.}$

Remark. 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

Theorem.[1][7] Let ${\displaystyle G}$ be a group with nested subgroups ${\displaystyle K\leq H\leq G}$ such that ${\displaystyle (G:H)=n,(H:K)=m}$ and ${\displaystyle (G:K)=(G:H)\cdot (H:K)=nm<\infty .}$ Then the Artin transfer ${\displaystyle T_{G,K}}$ is the compositum of the induced transfer ${\displaystyle {\tilde {T}}_{H,K}:H/H'\to K/K'}$ and the Artin transfer ${\displaystyle T_{G,H}}$, that is:

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

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

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

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 {\begin{aligned}u_{x}(i)&:=g_{\pi _{x}(i)}^{-1}xg_{i}\in H&&{\text{for all }}1\leq i\leq n\\v_{y}(j)&:=h_{\sigma _{y}(j)}^{-1}yh_{j}\in K&&{\text{for all }}1\leq j\leq m\end{aligned}}}

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

{\displaystyle {\begin{aligned}T_{G,H}(x)&=\prod _{i=1}^{n}u_{x}(i)\cdot H'\\{\tilde {T}}_{H,K}(y\cdot H')&=T_{H,K}(y)=\prod _{j=1}^{m}v_{y}(j)\cdot K'\end{aligned}}}

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

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

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 {\begin{aligned}T_{G,K}(x)&=\prod _{i=1}^{n}\prod _{j=1}^{m}v_{y_{i}}(j)\cdot K'\\&=\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'\\&=\prod _{i=1}^{n}\prod _{j=1}^{m}h_{\sigma _{y_{i}}(j)}^{-1}u_{x}(i)h_{j}\cdot K'\\&=\prod _{i=1}^{n}\prod _{j=1}^{m}h_{\sigma _{y_{i}}(j)}^{-1}y_{i}h_{j}\cdot K'\\&=\prod _{i=1}^{n}{\tilde {T}}_{H,K}\left(y_{i}\cdot H'\right)\\&={\tilde {T}}_{H,K}\left(\prod _{i=1}^{n}y_{i}\cdot H'\right)\\&={\tilde {T}}_{H,K}\left(\prod _{i=1}^{n}u_{x}(i)\cdot H'\right)\\&={\tilde {T}}_{H,K}(T_{G,H}(x))\end{aligned}}}

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

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

which corresponds to the composite of Artin transfers, defining

${\displaystyle k_{x}(i,j):=\left((gh)_{\gamma _{x}(i,j)}\right)^{-1}x(gh)_{(i,j)}\in K}$

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 ${\displaystyle [1,n]\times [1,m]}$ 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

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

in the previous section, are of a very special kind. They belong to the stabilizer of the natural equipartition of the set ${\displaystyle [1,n]\times [1,m]}$ 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 ${\displaystyle \{1,\ldots ,n\}}$, whose underlying set ${\displaystyle S_{m}^{n}\times S_{n}}$ is endowed with the following law of composition:

{\displaystyle {\begin{aligned}(10)\quad \forall x,z\in G:\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})\\&=(\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})\\&=(\sigma _{u_{xz}(1)},\ldots ,\sigma _{u_{xz}(n)};\pi _{xz})\\&=\gamma _{xz}\end{aligned}}}

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,

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

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

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

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 ${\displaystyle H}$ of finite index ${\displaystyle n}$ in a group ${\displaystyle G}$ and ${\displaystyle x\mapsto \pi _{x}}$ be its associated permutation representation.

Theorem.[1][3][4][5][8][9] Suppose the permutation ${\displaystyle \pi _{x}}$ decomposes into pairwise disjoint (and thus commuting) cycles ${\displaystyle \zeta _{1},\ldots ,\zeta _{t}\in S_{n}}$ of lengths ${\displaystyle f_{1},\ldots f_{t},}$ which is unique up to the ordering of the cycles. More explicitly, suppose

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

for ${\displaystyle 1\leq j\leq t}$, and ${\displaystyle f_{1}+\cdots +f_{t}=n.}$ Then the image of ${\displaystyle x\in G}$ under the Artin transfer is given by

${\displaystyle (12)\quad T_{G,H}(x)=\prod _{j=1}^{t}g_{j}^{-1}x^{f_{j}}g_{j}\cdot H'.}$
Proof

Define ${\displaystyle \ell _{j,k}:=x^{k}g_{j}}$ for ${\displaystyle 0\leq k\leq f_{j}-1}$ and ${\displaystyle 1\leq j\leq t}$. This is a left transversal of ${\displaystyle H}$ in ${\displaystyle G}$ since

${\displaystyle (13)\quad G=\bigsqcup _{j=1}^{t}\bigsqcup _{k=0}^{f_{j}-1}x^{k}g_{j}H}$

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

Fix a value of ${\displaystyle 1\leq j\leq t}$. Then:

{\displaystyle {\begin{aligned}x\ell _{j,k}&=xx^{k}g_{j}=x^{k+1}g_{j}=\ell _{j,k+1}\in \ell _{j,k+1}H&&\forall k\in \{0,\ldots ,f_{j}-2\}\\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\end{aligned}}}

Define:

{\displaystyle {\begin{aligned}u_{x}(j,k)&:=\ell _{j,k+1}^{-1}x\ell _{j,k}=1\in H&&\forall k\in \{0,\ldots ,f_{j}-2\}\\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\end{aligned}}}

Consequently,

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}\prod _{k=0}^{f_{j}-1}u_{x}(j,k)\cdot H'=\prod _{j=1}^{t}\left(\prod _{k=0}^{f_{j}-2}1\right)\cdot u_{x}(j,f_{j}-1)\cdot H'=\prod _{j=1}^{t}g_{j}^{-1}x^{f_{j}}g_{j}\cdot H'.}$

The cycle decomposition corresponds to a ${\displaystyle (\langle x\rangle ,H)}$ double coset decomposition of ${\displaystyle G}$:

${\displaystyle G=\bigsqcup _{j=1}^{t}\langle x\rangle g_{j}H}$

It was 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 ${\displaystyle H}$ be a normal subgroup of finite index ${\displaystyle n}$ 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 ${\displaystyle (g_{1},\ldots ,g_{t})}$ 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:

${\displaystyle (14)\quad T_{G,H}(x)=\prod _{j=1}^{t}g_{j}^{-1}x^{f}g_{j}\cdot H'}$.
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 ${\displaystyle G=\sqcup _{j=1}^{t}g_{j}\langle x,H\rangle }$ is the corresponding disjoint left coset decomposition, can be refined to a left transversal ${\displaystyle g_{j}x^{k}(1\leq j\leq t,\ 0\leq k\leq f-1)}$ with disjoint left coset decomposition:

${\displaystyle (15)\quad G=\sqcup _{j=1}^{t}\sqcup _{k=0}^{f-1}g_{j}x^{k}H}$

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

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}g_{j}^{-1}x^{f}g_{j}\cdot H'}$

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:

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

with the trace element

${\displaystyle (17)\quad \mathrm {Tr} _{G}(H)=\sum _{j=1}^{t}g_{j}\in \mathbb {Z} [G]}$

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 ${\displaystyle G/H}$, that is ${\displaystyle G=\langle x,H\rangle }$.

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

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

${\displaystyle T_{G,H}(x)=\prod _{j=1}^{t}g_{j}^{-1}xg_{j}\cdot H'=\prod _{j=1}^{t}x^{g_{j}}\cdot H'=x^{\sum _{j=1}^{t}g_{j}}\cdot H'}$

with the trace element

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

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

The outer transfer of an element ${\displaystyle x\in G\setminus H}$ which generates ${\displaystyle G/H}$, 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

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

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 ${\displaystyle G'\leq H\leq G}$ between ${\displaystyle G}$ and ${\displaystyle G'}$. For these intermediate groups we have the following lemma.

Lemma. All subgroups containing the commutator subgroup are normal.

Proof

Let ${\displaystyle G'\leq H\leq G}$. 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 ${\displaystyle [h,x]=h^{-1}x^{-1}hx=h^{-1}y}$ would be an element in ${\displaystyle G\setminus H}$ in contradiction to ${\displaystyle G'\leq H}$.

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 ${\displaystyle G/G'}$ of elementary abelian type ${\displaystyle (p,p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{1},\ldots ,H_{p+1}}$ of index ${\displaystyle p.}$

Lemma. In this particular case, the Frattini subgroup, which is defined as the intersection of all maximal subgroups coincides with the commutator subgroup.

Proof. To see this note that due to the abelian type of ${\displaystyle G/G'}$ the commutator subgroup contains all p-th powers ${\displaystyle G'\supset G^{p},}$ and thus we have ${\displaystyle \Phi (G)=G^{p}\cdot G'=G'}$.

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{i}:G\to H_{i}/H_{i}'}$ be the Artin transfer homomorphism. According to Burnside's basis theorem the group ${\displaystyle G}$ can therefore be generated by two elements ${\displaystyle x,y}$ such that ${\displaystyle x^{p},y^{p}\in G'.}$ For each of the maximal subgroups ${\displaystyle H_{i}}$, which are also normal we need a generator ${\displaystyle h_{i}}$ with respect to ${\displaystyle G'}$, and a generator ${\displaystyle t_{i}}$ of a transversal ${\displaystyle (1,t_{i},t_{i}^{2},\ldots ,t_{i}^{p-1})}$ such that

{\displaystyle {\begin{aligned}H_{i}&=\langle h_{i},G'\rangle \\G&=\langle t_{i},H_{i}\rangle =\bigsqcup _{j=0}^{p-1}t_{i}^{j}H_{i}\end{aligned}}}

A convenient selection is given by

${\displaystyle (19)\quad {\begin{cases}h_{1}=y\\t_{1}=x\\h_{i}=xy^{i-2}&2\leq i\leq p+1\\t_{i}=y&2\leq i\leq p+1\end{cases}}}$

Then, for each ${\displaystyle 1\leq i\leq p+1}$ we use equations (16) and (18) to implement the inner and outer transfers:

{\displaystyle {\begin{aligned}(20)\quad T_{i}(h_{i})&=h_{i}^{\mathrm {Tr} _{G}(H_{i})}\cdot H_{i}'=h_{i}^{1+t_{i}+t_{i}^{2}+\cdots +t_{i}^{p-1}}\cdot H_{i}'=h_{i}\cdot \left(t_{i}^{-1}h_{i}t_{i}\right)\cdot \left(t_{i}^{-2}h_{i}t_{i}^{2}\right)\cdots \left(t_{i}^{-p+1}h_{i}t_{i}^{p-1}\right)\cdot H_{i}'=\left(h_{i}t_{i}^{-1}\right)^{p}t_{i}^{p}\cdot H_{i}'\\(21)\quad T_{i}(t_{i})&=t_{i}^{p}\cdot H_{i}'\end{aligned}}},

The reason is that in ${\displaystyle G/H_{i},}$ ${\displaystyle \mathrm {ord} (h_{i}H_{i})=1}$ and ${\displaystyle \mathrm {ord} (t_{i}H_{i})=p.}$

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

${\displaystyle (22)\quad H_{i}'=(G')^{h_{i}-1}=\langle [s_{1},h_{i}],\ldots ,[s_{n},h_{i}]\rangle .}$

### Abelianization of type (p2,p)

Let ${\displaystyle G}$ be a p-group with abelianization ${\displaystyle G/G'}$ of non-elementary abelian type ${\displaystyle (p^{2},p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{1},\ldots ,H_{p+1}}$ of index ${\displaystyle p}$ and ${\displaystyle p+1}$ subgroups ${\displaystyle U_{1},\ldots ,U_{p+1}}$ of index ${\displaystyle p^{2}.}$ For each ${\displaystyle i\in \{1,\ldots ,p+1\}}$ let

{\displaystyle {\begin{aligned}T_{1,i}:G&\to H_{i}/H_{i}'\\T_{2,i}:G&\to U_{i}/U_{i}'\end{aligned}}}

be the Artin transfer homomorphisms. Burnside's basis theorem asserts that the group ${\displaystyle G}$ can be generated by two elements ${\displaystyle x,y}$ such that ${\displaystyle x^{p^{2}},y^{p}\in G'.}$

We begin by considering the first layer of subgroups. For each of the normal subgroups ${\displaystyle H_{i}}$, we select a generator

${\displaystyle (23)\quad h_{i}=xy^{i-1}}$

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

${\displaystyle (24)\quad {\begin{cases}h_{p+1}=y\\h_{0}=x^{p}\end{cases}}}$

such that ${\displaystyle H_{p+1}=\langle h_{p+1},h_{0},G'\rangle }$. 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

${\displaystyle (25)\quad {\begin{cases}t_{i}=y&1\leq i\leq p\\t_{p+1}=x\end{cases}}}$

Then, for each ${\displaystyle 1\leq i\leq p+1}$, we have inner and outer transfers:

{\displaystyle {\begin{aligned}(26)\quad T_{1,i}(h_{i})&=h_{i}^{\mathrm {Tr} _{G}(H_{i})}\cdot H_{i}'=h_{i}^{1+t_{i}+t_{i}^{2}+\ldots +t_{i}^{p-1}}\cdot H_{i}'=\left(h_{i}t_{i}^{-1}\right)^{p}t_{i}^{p}\cdot H_{i}'\\(27)\quad T_{1,i}(t_{i})&=t_{i}^{p}\cdot H_{i}'\end{aligned}}}

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 ${\displaystyle U_{i}}$, we select a generator

${\displaystyle (28)\quad {\begin{cases}u_{1}=y\\u_{i}=x^{p}y^{i-1}&2\leq i\leq p\\u_{p+1}=x^{p}\end{cases}}}$

such that ${\displaystyle U_{i}=\langle u_{i},G'\rangle }$. Among these subgroups, the Frattini subgroup ${\displaystyle U_{p+1}=\langle x^{p},G'\rangle =G^{p}\cdot G'}$ 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

${\displaystyle (29)\quad {\begin{cases}t_{i}=x&1\leq i\leq p\\w_{i}=x^{p}&1\leq i\leq p\\t_{p+1}=x\\w_{p+1}=y\end{cases}}}$

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 and outer transfers

{\displaystyle {\begin{aligned}(30)\quad T_{2,i}(u_{i})&=u_{i}^{\mathrm {Tr} _{G}(U_{i})}\cdot U_{i}'=u_{i}^{\sum _{j=0}^{p-1}\sum _{k=0}^{p-1}w_{i}^{j}t_{i}^{k}}\cdot U_{i}'=\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}'\\(31)\quad T_{2,i}(t_{i})&=t_{i}^{p^{2}}\cdot U_{i}'\end{aligned}}}

exceptionally

{\displaystyle {\begin{aligned}&(32)\quad T_{2,p+1}\left(t_{p+1}\right)=\left(t_{p+1}^{p}\right)^{1+w_{p+1}+w_{p+1}^{2}+\ldots +w_{p+1}^{p-1}}\cdot U_{p+1}'\\&(33)\quad T_{2,i}(w_{i})=\left(w_{i}^{p}\right)^{1+t_{i}+t_{i}^{2}+\ldots +t_{i}^{p-1}}\cdot U_{i}'&&1\leq i\leq p+1\end{aligned}}}

The structure of the derived subgroups ${\displaystyle H_{i}'}$ and ${\displaystyle U_{i}'}$ 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 ${\displaystyle G/G'}$. Suppose that ${\displaystyle (H_{i})_{i\in I}}$ denotes the family of all subgroups which contain ${\displaystyle G'}$ and are therefore necessarily normal, enumerated by a 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 ${\displaystyle H_{i}/H_{i}'}$.

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) ${\displaystyle \tau _{H}(G)=(H_{i}/H_{i}')_{i\in I}}$ 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 ${\displaystyle G/G'}$ of elementary abelian type ${\displaystyle (p,p)}$. Then ${\displaystyle G}$ has ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{1},\ldots ,H_{p+1}}$ of index ${\displaystyle p}$. For ${\displaystyle i\in \{1,\ldots ,p+1\}}$ let ${\displaystyle T_{i}:G\to H_{i}/H_{i}'}$ denote the Artin transfer homomorphism.

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 ${\displaystyle H_{1},\ldots ,H_{p+1}}$.

Remark. For brevity, the TKT is identified with the multiplet ${\displaystyle (\varkappa (i))_{1\leq i\leq p+1}}$, whose integer components are given by

${\displaystyle \varkappa (i)={\begin{cases}0&\ker(T_{i})=G\\j&\ker(T_{i})=H_{j}{\text{ for some }}1\leq j\leq p+1\end{cases}}}$

Here, we take into consideration that each transfer kernel ${\displaystyle \ker(T_{i})}$ must contain the commutator subgroup ${\displaystyle G'}$ of ${\displaystyle G}$, since the transfer target ${\displaystyle H_{i}/H_{i}'}$ is abelian. However, the minimal case ${\displaystyle \ker(T_{i})=G'}$ cannot occur.

Remark. 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

${\displaystyle \lambda (i)={\begin{cases}0&\ker(V_{i})=G\\j&\ker(V_{i})=K_{j}{\text{ for some }}1\leq j\leq p+1\end{cases}}}$

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 action ${\displaystyle (\pi ,\mu )\mapsto {\tilde {\pi }}^{-1}\circ \mu \circ \pi }$ of the symmetric group ${\displaystyle S_{p+1}}$ on the set of all mappings from ${\displaystyle \{1,\ldots ,p+1\}\to \{0,1,\ldots ,p+1\},}$ 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 ${\displaystyle H_{0}=G,K_{0}=G.}$

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 ${\displaystyle \#{\mathcal {H}}_{0}(G):=\#\{1\leq i\leq p+1\mid \varkappa (i)=0\}}$ 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 ${\displaystyle G/G'}$ 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 ${\displaystyle G/G'\simeq (2,2)}$, 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 ${\displaystyle G\in \{\langle 243,6\rangle ,\langle 243,8\rangle \}}$ with TKTs ${\displaystyle \varkappa \in \{(0122),(2034)\}}$ (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 ${\displaystyle G/G'}$ of non-elementary abelian type ${\displaystyle (p^{2},p).}$ Then ${\displaystyle G}$ possesses ${\displaystyle p+1}$ maximal subgroups ${\displaystyle H_{1},\ldots ,H_{p+1}}$ of index ${\displaystyle p}$ and ${\displaystyle p+1}$ subgroups ${\displaystyle U_{1},\ldots ,U_{p+1}}$ of index ${\displaystyle p^{2}.}$

Assumption. Suppose

${\displaystyle H_{p+1}=\prod _{j=1}^{p+1}U_{j}}$

is the distinguished maximal subgroup and

${\displaystyle U_{p+1}=\bigcap _{j=1}^{p+1}H_{j}}$

is the distinguished subgroup of index ${\displaystyle p^{2}}$ which as the intersection of all maximal subgroups, is the Frattini subgroup ${\displaystyle \Phi (G)}$ of ${\displaystyle G}$.

### First layer

For each ${\displaystyle 1\leq i\leq p+1}$, let ${\displaystyle T_{1,i}:G\to H_{i}/H_{i}'}$ denote the Artin transfer homomorphism.

Definition. The family ${\displaystyle \varkappa _{1,H,U}(G)=(\ker(T_{1,i}))_{i=1}^{p+1}}$ 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 ${\displaystyle (\varkappa _{1}(i))_{i=1}^{p+1}}$, where

${\displaystyle \varkappa _{1}(i)={\begin{cases}0&\ker(T_{1,i})=H_{p+1},\\j&\ker(T_{1,i})=U_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$

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

### Second layer

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

Definition. The family ${\displaystyle \varkappa _{2,U,H}(G)=(\ker(T_{2,i}))_{i=1}^{p+1}}$ 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 ${\displaystyle (\varkappa _{2}(i))_{i=1}^{p+1},}$ where

${\displaystyle \varkappa _{2}(i)={\begin{cases}0&\ker(T_{2,i})=G,\\j&\ker(T_{2,i})=H_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$

### 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 ${\displaystyle K_{i}=H_{\tau (i)}(1\leq i\leq p)}$ 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 ${\displaystyle W_{i}=U_{\sigma (i)}(1\leq i\leq p)}$ 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 ${\displaystyle \lambda _{1,K,W}(G)=(\ker(V_{1,i}))_{i=1}^{p+1}}$ with respect to ${\displaystyle K_{1},\ldots ,K_{p+1}}$ and ${\displaystyle W_{1},\ldots ,W_{p+1}}$, identified with ${\displaystyle (\lambda _{1}(i))_{i=1}^{p+1}}$, where

${\displaystyle \lambda _{1}(i)={\begin{cases}0&\ker(V_{1,i})=K_{p+1},\\j&\ker(V_{1,i})=W_{j}{\text{ for some }}1\leq j\leq p+1,\end{cases}}}$

and ${\displaystyle \lambda _{2,W,K}(G)=(\ker(V_{2,i}))_{i=1}^{p+1}}$ with respect to ${\displaystyle W_{1},\ldots ,W_{p+1}}$ and ${\displaystyle K_{1},\ldots ,K_{p+1}}$, identified with ${\displaystyle (\lambda _{2}(i))_{i=1}^{p+1},}$ where

${\displaystyle \lambda _{2}(i)={\begin{cases}0&\ker(V_{2,i})=G,\\j&\ker(V_{2,i})=K_{j}{\text{ for some }}1\leq j\leq p+1.\end{cases}}}$

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 {\begin{aligned}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)))}\\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)))}\end{aligned}}}

the relations between ${\displaystyle \lambda _{1}}$ and ${\displaystyle \varkappa _{1}}$, and ${\displaystyle \lambda _{2}}$ and ${\displaystyle \varkappa _{2}}$, are given by

${\displaystyle \lambda _{1}={\tilde {\sigma }}^{-1}\circ \varkappa _{1}\circ {\hat {\tau }}}$
${\displaystyle \lambda _{2}={\tilde {\tau }}^{-1}\circ \varkappa _{2}\circ {\hat {\sigma }}}$

Therefore, ${\displaystyle \lambda =(\lambda _{1},\lambda _{2})}$ is another representative of the orbit ${\displaystyle \varkappa ^{S_{p}\times S_{p}}}$ of ${\displaystyle \varkappa =(\varkappa _{1},\varkappa _{2})}$ under the action:

${\displaystyle ((\sigma ,\tau ),(\mu _{1},\mu _{2}))\mapsto \left({\tilde {\sigma }}^{-1}\circ \mu _{1}\circ {\hat {\tau }},{\tilde {\tau }}^{-1}\circ \mu _{2}\circ {\hat {\sigma }}\right)}$

of the product of two symmetric groups ${\displaystyle S_{p}\times S_{p}}$ on the set of all pairs of mappings ${\displaystyle \{1,\ldots ,p+1\}\to \{0,1,\ldots ,p+1\}}$, where the extensions ${\displaystyle {\hat {\pi }}\in S_{p+1}}$ and ${\displaystyle {\tilde {\pi }}\in S_{p+2}}$ of a permutation ${\displaystyle \pi \in S_{p}}$ are defined by