Editing Index of a subgroup (section)

==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.
{{collapse top|Proof}}
If ''Hca'' ⊂ ''Hc'' ∀ ''c'' ∈ ''G'' and likewise ''Hcb'' ⊂ ''Hc'' ∀ ''c'' ∈ ''G'', then ''Hcab'' ⊂ ''Hc'' ∀ ''c'' ∈ ''G''. If ''h''<sub>1</sub>''ca'' = ''h''<sub>2</sub>''c'' for all ''c'' ∈ ''G'' (with ''h''<sub>1</sub>, ''h''<sub>2</sub> ∈ H) then ''h''<sub>2</sub>''ca''<sup>−1</sup> = ''h''<sub>1</sub>''c'', so ''Hca''<sup>−1</sup> ⊂ ''Hc''.
{{collapse bottom|Proof}}

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''.

{{collapse top|Proof}}
First let us show that if ''b''{{sub|1}}∈''B'', then any other element ''b''{{sub|2}} of ''B'' equals ''ab''{{sub|1}} for some ''a''∈''A''. Assume that multiplying the coset ''Hc'' on the right by elements of ''B'' gives elements of the coset ''Hd''. If ''cb''<sub>1</sub> = ''d'' and ''cb''<sub>2</sub> = ''hd'', then ''cb''<sub>2</sub>''b''<sub>1</sub><sup>−1</sup> = ''hc'' ∈ ''Hc'', or in other words ''b''{{sub|2}}=''ab''{{sub|1}} 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''.
{{collapse bottom|Proof}}

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{{sup|−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{{sub|''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 S{{sub|5}}.

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''{{sub|7}} of the non-abelian group of order 21 is normal (see [[List of small groups#List of small non-abelian groups|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 {{Harv|Lam|2004}}.

=== Examples ===
The group '''O''' of chiral [[octahedral symmetry]] has 24 elements. It has a [[dihedral symmetry|dihedral]] D<sub>4</sub> 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 D<sub>2</sub> 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]] S<sub>3</sub>. All elements from any particular coset of ''A'' perform the same permutation of the cosets of ''H''.

On the other hand, the group T<sub>h</sub> of [[pyritohedral symmetry]] also has 24 members and a subgroup of index 3 (this time it is a D<sub>2h</sub> [[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 S<sub>3</sub> symmetric group.