# Index of a subgroup

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In mathematics, specifically group theory, the index of a subgroup H in a group G is the number of left cosets of H in G, or equivalently, the number of right cosets of H in G. The index is denoted ${\displaystyle |G:H|}$ or ${\displaystyle [G:H]}$ or ${\displaystyle (G:H)}$. Because G is the disjoint union of the left cosets and because each left coset has the same size as H, the index is related to the orders of the two groups by the formula

${\displaystyle |G|=|G:H||H|}$

(interpret the quantities as cardinal numbers if some of them are infinite). Thus the index ${\displaystyle |G:H|}$ measures the "relative sizes" of G and H.

For example, let ${\displaystyle G=\mathbb {Z} }$ be the group of integers under addition, and let ${\displaystyle H=2\mathbb {Z} }$ be the subgroup consisting of the even integers. Then ${\displaystyle 2\mathbb {Z} }$ has two cosets in ${\displaystyle \mathbb {Z} }$, namely the set of even integers and the set of odd integers, so the index ${\displaystyle |\mathbb {Z} :2\mathbb {Z} |}$ is 2. More generally, ${\displaystyle |\mathbb {Z} :n\mathbb {Z} |=n}$ for any positive integer n.

When G is finite, the formula may be written as ${\displaystyle |G:H|=|G|/|H|}$, and it implies Lagrange's theorem that ${\displaystyle |H|}$ divides ${\displaystyle |G|}$.

When G is infinite, ${\displaystyle |G:H|}$ is a nonzero cardinal number that may be finite or infinite. For example, ${\displaystyle |\mathbb {Z} :2\mathbb {Z} |=2}$, but ${\displaystyle |\mathbb {R} :\mathbb {Z} |}$ is infinite.

If N is a normal subgroup of G, then ${\displaystyle |G:N|}$ is equal to the order of the quotient group ${\displaystyle G/N}$, since the underlying set of ${\displaystyle G/N}$ is the set of cosets of N in G.

## Properties

• If H is a subgroup of G and K is a subgroup of H, then
${\displaystyle |G:K|=|G:H|\,|H:K|.}$
• If H and K are subgroups of G, then
${\displaystyle |G:H\cap K|\leq |G:H|\,|G:K|,}$
with equality if ${\displaystyle HK=G}$. (If ${\displaystyle |G:H\cap K|}$ is finite, then equality holds if and only if ${\displaystyle HK=G}$.)
• Equivalently, if H and K are subgroups of G, then
${\displaystyle |H:H\cap K|\leq |G:K|,}$
with equality if ${\displaystyle HK=G}$. (If ${\displaystyle |H:H\cap K|}$ is finite, then equality holds if and only if ${\displaystyle HK=G}$.)
• If G and H are groups and ${\displaystyle \varphi \colon G\to H}$ is a homomorphism, then the index of the kernel of ${\displaystyle \varphi }$ in G is equal to the order of the image:
${\displaystyle |G:\operatorname {ker} \;\varphi |=|\operatorname {im} \;\varphi |.}$
${\displaystyle |Gx|=|G:G_{x}|.\!}$
This is known as the orbit-stabilizer theorem.
• As a special case of the orbit-stabilizer theorem, the number of conjugates ${\displaystyle gxg^{-1}}$ of an element ${\displaystyle x\in G}$ is equal to the index of the centralizer of x in G.
• Similarly, the number of conjugates ${\displaystyle gHg^{-1}}$ of a subgroup H in G is equal to the index of the normalizer of H in G.
• If H is a subgroup of G, the index of the normal core of H satisfies the following inequality:
${\displaystyle |G:\operatorname {Core} (H)|\leq |G:H|!}$
where ! denotes the factorial function; this is discussed further below.
• As a corollary, if the index of H in G is 2, or for a finite group the lowest prime p that divides the order of G, then H is normal, as the index of its core must also be p, and thus H equals its core, i.e., it is normal.
• Note that a subgroup of lowest prime index may not exist, such as in any simple group of non-prime order, or more generally any perfect group.

## Examples

• The alternating group ${\displaystyle A_{n}}$ has index 2 in the symmetric group ${\displaystyle S_{n},}$ and thus is normal.
• The special orthogonal group ${\displaystyle \operatorname {SO} (n)}$ has index 2 in the orthogonal group ${\displaystyle \operatorname {O} (n)}$, and thus is normal.
• The free abelian group ${\displaystyle \mathbb {Z} \oplus \mathbb {Z} }$ has three subgroups of index 2, namely
${\displaystyle \{(x,y)\mid x{\text{ is even}}\},\quad \{(x,y)\mid y{\text{ is even}}\},\quad {\text{and}}\quad \{(x,y)\mid x+y{\text{ is even}}\}}$.
• More generally, if p is prime then ${\displaystyle \mathbb {Z} ^{n}}$ has ${\displaystyle (p^{n}-1)/(p-1)}$ subgroups of index p, corresponding to the ${\displaystyle (p^{n}-1)}$ nontrivial homomorphisms ${\displaystyle \mathbb {Z} ^{n}\to \mathbb {Z} /p\mathbb {Z} }$.[citation needed]
• Similarly, the free group ${\displaystyle F_{n}}$ has ${\displaystyle (p^{n}-1)}$ subgroups of index p.
• The infinite dihedral group has a cyclic subgroup of index 2, which is necessarily normal.

## Infinite index

If H has an infinite number of cosets in G, then the index of H in G is said to be infinite. In this case, the index ${\displaystyle |G:H|}$ is actually a cardinal number. For example, the index of H in G may be countable or uncountable, depending on whether H has a countable number of cosets in G. Note that the index of H is at most the order of G, which is realized for the trivial subgroup, or in fact any subgroup H of infinite cardinality less than that of G.

## Finite index

An infinite group G may have subgroups H of finite index (for example, the even integers inside the group of integers). Such a subgroup always contains a normal subgroup N (of G), also of finite index. In fact, if H has index n, then the index of N can be taken as some factor of n!; indeed, N can be taken to be the kernel of the natural homomorphism from G to the permutation group of the left (or right) cosets of H.

A special case, n = 2, gives the general result that a subgroup of index 2 is a normal subgroup, because the normal subgroup (N above) must have index 2 and therefore be identical to the original subgroup. More generally, a subgroup of index p where p is the smallest prime factor of the order of G (if G is finite) is necessarily normal, as the index of N divides p! and thus must equal p, having no other prime factors.

An alternative proof of the result that subgroup of index lowest prime p is normal, and other properties of subgroups of prime index are given in (Lam 2004).

### Examples

The above considerations are true for finite groups as well. For instance, the group O of chiral octahedral symmetry has 24 elements. It has a dihedral D4 subgroup (in fact it has three such) of order 8, and thus of index 3 in O, which we shall call H. This dihedral group has a 4-member D2 subgroup, which we may call A. Multiplying on the right any element of a right coset of H by an element of A gives a member of the same coset of H (Hca = Hc). A is normal in O. There are six cosets of A, corresponding to the six elements of the symmetric group S3. All elements from any particular coset of A perform the same permutation of the cosets of H.

On the other hand, the group Th of pyritohedral symmetry also has 24 members and a subgroup of index 3 (this time it is a D2h prismatic symmetry group, see point groups in three dimensions), but in this case the whole subgroup is a normal subgroup. All members of a particular coset carry out the same permutation of these cosets, but in this case they represent only the 3-element alternating group in the 6-member S3 symmetric group.

## Normal subgroups of prime power index

Normal subgroups of prime power index are kernels of surjective maps to p-groups and have interesting structure, as described at Focal subgroup theorem: Subgroups and elaborated at focal subgroup theorem.

There are three important normal subgroups of prime power index, each being the smallest normal subgroup in a certain class:

• Ep(G) is the intersection of all index p normal subgroups; G/Ep(G) is an elementary abelian group, and is the largest elementary abelian p-group onto which G surjects.
• Ap(G) is the intersection of all normal subgroups K such that G/K is an abelian p-group (i.e., K is an index ${\displaystyle p^{k}}$ normal subgroup that contains the derived group ${\displaystyle [G,G]}$): G/Ap(G) is the largest abelian p-group (not necessarily elementary) onto which G surjects.
• Op(G) is the intersection of all normal subgroups K of G such that G/K is a (possibly non-abelian) p-group (i.e., K is an index ${\displaystyle p^{k}}$ normal subgroup): G/Op(G) is the largest p-group (not necessarily abelian) onto which G surjects. Op(G) is also known as the p-residual subgroup.

As these are weaker conditions on the groups K, one obtains the containments

${\displaystyle \mathbf {E} ^{p}(G)\supseteq \mathbf {A} ^{p}(G)\supseteq \mathbf {O} ^{p}(G).}$

These groups have important connections to the Sylow subgroups and the transfer homomorphism, as discussed there.

### Geometric structure

An elementary observation is that one cannot have exactly 2 subgroups of index 2, as the complement of their symmetric difference yields a third. This is a simple corollary of the above discussion (namely the projectivization of the vector space structure of the elementary abelian group

${\displaystyle G/\mathbf {E} ^{p}(G)\cong (\mathbf {Z} /p)^{k}}$,

and further, G does not act on this geometry, nor does it reflect any of the non-abelian structure (in both cases because the quotient is abelian).

However, it is an elementary result, which can be seen concretely as follows: the set of normal subgroups of a given index p form a projective space, namely the projective space

${\displaystyle \mathbf {P} (\operatorname {Hom} (G,\mathbf {Z} /p)).}$

In detail, the space of homomorphisms from G to the (cyclic) group of order p, ${\displaystyle \operatorname {Hom} (G,\mathbf {Z} /p),}$ is a vector space over the finite field ${\displaystyle \mathbf {F} _{p}=\mathbf {Z} /p.}$ A non-trivial such map has as kernel a normal subgroup of index p, and multiplying the map by an element of ${\displaystyle (\mathbf {Z} /p)^{\times }}$ (a non-zero number mod p) does not change the kernel; thus one obtains a map from

${\displaystyle \mathbf {P} (\operatorname {Hom} (G,\mathbf {Z} /p)):=(\operatorname {Hom} (G,\mathbf {Z} /p))\setminus \{0\})/(\mathbf {Z} /p)^{\times }}$

to normal index p subgroups. Conversely, a normal subgroup of index p determines a non-trivial map to ${\displaystyle \mathbf {Z} /p}$ up to a choice of "which coset maps to ${\displaystyle 1\in \mathbf {Z} /p,}$ which shows that this map is a bijection.

As a consequence, the number of normal subgroups of index p is

${\displaystyle (p^{k+1}-1)/(p-1)=1+p+\cdots +p^{k}}$

for some k; ${\displaystyle k=-1}$ corresponds to no normal subgroups of index p. Further, given two distinct normal subgroups of index p, one obtains a projective line consisting of ${\displaystyle p+1}$ such subgroups.

For ${\displaystyle p=2,}$ the symmetric difference of two distinct index 2 subgroups (which are necessarily normal) gives the third point on the projective line containing these subgroups, and a group must contain ${\displaystyle 0,1,3,7,15,\ldots }$ index 2 subgroups – it cannot contain exactly 2 or 4 index 2 subgroups, for instance.

## References

• Lam, T. Y. (March 2004), "On Subgroups of Prime Index", The American Mathematical Monthly, 111 (3): 256–258, JSTOR 4145135, alternative download {{citation}}: External link in |postscript= (help)CS1 maint: postscript (link)