Index of a subgroup

• 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:

${\displaystyle |G:\operatorname {ker} \;\varphi |=|\operatorname {im} \;\varphi |.}$ 

• Let G be a group acting on a set X, and let x ∈ X. Then the cardinality of the orbit of x under G is equal to the index of the stabilizer of x:

${\displaystyle |Gx|=|G:G_{x}|.\!}$ 

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.

• 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 ${\displaystyle (p^{n}-1)/(p-1)}$  subgroups of index p.
• The infinite dihedral group has a cyclic subgroup of index 2, which is necessarily normal.

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.

A subgroup H of finite index in a group G (finite or infinite) always contains a normal subgroup N (of G), also of finite index. In fact, if H has index n, then the index of N will be some divisor of n! and a multiple 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. Let us explain this in more detail, using right cosets:

The elements of G that leave all cosets the same form a group.

Let us call this group A. Let B be the set of elements of G which perform a given permutation on the cosets of H. Then B is a right coset of A.

What we have said so far applies whether the index of H is finite or infinte. Now assume that it is the finite number n. Since the number of possible permutations of cosets is finite, namely n!, then there can only be a finite number of sets like B. (If G is infinite, then all such sets are therefore infinite.) The set of these sets forms a group isomorphic to a subset of the group of permutations, so the number of these sets must divide n!. Furthermore, it must be a multiple of n because each coset of H contains the same number of cosets of A. Finally, if for some c ∈ G and a ∈ A we have ca = xc, then for any d ∈ G dca = dxc, but also dca = hdc for some h ∈ H (by the definition of A), so hd = dx. Since this is true for any d, x must be a member of A, so ca = xc implies that cac⁻¹ ∈ A and therefore A is a normal subgroup.

The index of the normal subgroup not only has to be a divisor of n!, but must satisfy other criteria as well. Since the normal subgroup is a subgroup of H, its index in G must be n times its index inside H. Its index in G must also correspond to a subgroup of the symmetric group S_n, the group of permutations of n objects. So for example if n is 5, the index cannot be 15 even though this divides 5!, because there is no subgroup of order 15 in S₅.

In the case of n = 2 this gives the rather obvious result that a subgroup H of index 2 is a normal subgroup, because the normal subgroup of H must have index 2 in G and therefore be identical to H. (We can arrive at this fact also by noting that all the elements of G that are not in H constitute the right coset of H and also the left coset, so the two are identical.) 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. For example, the subgroup Z₇ of the non-abelian group of order 21 is normal (see List of small non-abelian groups and Frobenius group#Examples).

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

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 T_h of pyritohedral symmetry also has 24 members and a subgroup of index 3 (this time it is a D_2h 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 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:

• E^p(G) is the intersection of all index p normal subgroups; G/E^p(G) is an elementary abelian group, and is the largest elementary abelian p-group onto which G surjects.
• A^p(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/A^p(G) is the largest abelian p-group (not necessarily elementary) onto which G surjects.
• O^p(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/O^p(G) is the largest p-group (not necessarily abelian) onto which G surjects. O^p(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.

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.

• Virtually
• Codimension

• Normality of subgroups of prime index at PlanetMath.
• "Subgroup of least prime index is normal" at Groupprops, The Group Properties Wiki