# Index of a subgroup

In mathematics, specifically group theory, the index of a subgroup H in a group G is the "relative size" of H in G: equivalently, the number of "copies" (cosets) of H that fill up G. For example, if H has index 2 in G, then intuitively "half" of the elements of G lie in H. The index of H in G is usually denoted |G : H| or [G : H] or (G:H).

Formally, the index of H in G is defined as the number of cosets of H in G. (The number of left cosets of H in G is always equal to the number of right cosets.) For example, let Z be the group of integers under addition, and let 2Z be the subgroup of Z consisting of the even integers. Then 2Z has two cosets in Z (namely the even integers and the odd integers), so the index of 2Z in Z is two. To generalize,

$|{\mathbf {Z} }:n{\mathbf {Z} }|=n$ for any positive integer n.

If N is a normal subgroup of G, then the index of N in G is also equal to the order of the quotient group G / N, since this is defined in terms of a group structure on the set of cosets of N in G.

If G is infinite, the index of a subgroup H will in general be a non-zero cardinal number. It may be finite - that is, a positive integer - as the example above shows.

If G and H are finite groups, then the index of H in G is equal to the quotient of the orders of the two groups:

$|G:H|={\frac {|G|}{|H|}}.$ This is Lagrange's theorem, and in this case the quotient is necessarily a positive integer.

## Properties

• If H is a subgroup of G and K is a subgroup of H, then
$|G:K|=|G:H|\,|H:K|.$ • If H and K are subgroups of G, then
$|G:H\cap K|\leq |G:H|\,|G:K|,$ with equality if HK = G. (If |G : H ∩ K| is finite, then equality holds if and only if HK = G.)
• Equivalently, if H and K are subgroups of G, then
$|H:H\cap K|\leq |G:K|,$ with equality if HK = G. (If |H : H ∩ K| is finite, then equality holds if and only if HK = G.)
• If G and H are groups and φG → H is a homomorphism, then the index of the kernel of φ in G is equal to the order of the image:
$|G:\operatorname {ker} \;\varphi |=|\operatorname {im} \;\varphi |.$ $|Gx|=|G:G_{x}|.\!$ This is known as the orbit-stabilizer theorem.
• As a special case of the orbit-stabilizer theorem, the number of conjugates gxg−1 of an element x ∈ G is equal to the index of the centralizer of x in G.
• Similarly, the number of conjugates 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:
$|G:\operatorname {Core} (H)|\leq |G:H|!$ 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., 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

$\{(x,y)\mid x{\text{ is even}}\},\quad \{(x,y)\mid y{\text{ is even}}\},\quad {\text{and}}\quad \{(x,y)\mid x+y{\text{ is even}}\}$ .
• More generally, if p is prime then Zn has (pn − 1) / (p − 1) subgroups of index p, corresponding to the pn − 1 nontrivial homomorphisms Zn → Z/pZ.{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B=

{{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}

## 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 |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 group (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 Template:Harv.

### 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:

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

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

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

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

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

${\mathbf {P} }(\operatorname {Hom} (G,{\mathbf {Z} }/p)):=(\operatorname {Hom} (G,{\mathbf {Z} }/p))\setminus \{0\})/({\mathbf {Z} }/p)^{\times }$ to normal index p subgroups. Conversely, a normal subgroup of index p determines a non-trivial map to ${\mathbf {Z} }/p$ up to a choice of "which coset maps to $1\in {\mathbf {Z} }/p,$ which shows that this map is a bijection.

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

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

For $p=2,$ 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 $0,1,3,7,15,\ldots$ index 2 subgroups – it cannot contain exactly 2 or 4 index 2 subgroups, for instance.