Index of a subgroup

• 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| \le |G : H|\,|G : K|, :with equality if HK=G. (If |G:H\cap 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| \le |G:K|, :with equality if HK=G. (If |H:H\cap K| is finite, then equality holds if and only if HK=G.) • If G and H are groups and \varphi \colon G\to H is a homomorphism, then the index of the kernel of \varphi in G is equal to the order of the image: ::|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: ::|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 \in 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)| \le |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 A_n has index 2 in the symmetric group S_n, and thus is normal. • The special orthogonal group \operatorname{SO}(n) has index 2 in the orthogonal group \operatorname{O}(n), and thus is normal. • The free abelian group \Z\oplus \Z has three subgroups of index 2, namely ::\{(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 \Z^n has (p^n-1)/(p-1) subgroups of index p, corresponding to the (p^n-1) nontrivial homomorphisms \Z^n \to \Z/p\Z. • Similarly, the free group F_n has (p^n-1)/(p - 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 |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==

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. If Hca ⊂ Hc ∀ c ∈ G and likewise Hcb ⊂ Hc ∀ c ∈ G, then Hcab ⊂ Hc ∀ c ∈ G. If h1ca = h2c for all c ∈ G (with h1, h2 ∈ H) then h2ca−1 = h1c, so Hca−1 ⊂ Hc. Let us call this group A. Note that A is a subgroup of H, since Ha ⊂ H by the definition of 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. First let us show that if b∈B, then any other element b of B equals ab for some a∈A. Assume that multiplying the coset Hc on the right by elements of B gives elements of the coset Hd. If cb1 = d and cb2 = hd, then cb2b1−1 = hc ∈ Hc, or in other words b=ab for some a∈A, as desired. Now we show that for any b∈B and a∈A, ab will be an element of B. This is because the coset Hc is the same as Hca, so Hcb = Hcab. Since this is true for any c (that is, for any coset), it shows that multiplying on the right by ab makes the same permutation of cosets as multiplying by b, and therefore ab∈B. 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 ∈ 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, 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 S. 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 Z 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 . Examples 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 p^k normal subgroup that contains the derived group [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 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 :\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. ==See also==