Automorphisms and Conjugations of G

We’ve seen how groups can act on sets via bijections. If the underlying set were endowed with a group structure, we can restrict our attention to bijections which preserve the group operation.

Definition. An automorphism of a group G is an isomorphism G → G. The set of all automorphisms of G forms a group Aut(G) under composition. An action of a group G on H is a group homomorphism G → Aut(H).

One useful automorphism of a group is the conjugation map we saw in the previous post: given , the conjugation map . In the case of abelian groups, the conjugation map is trivial since .

Definition. The conjugation map is called an inner automorphism. Let the group of  be given by Inn(G).

Now let’s focus our attention on the conjugation action of G on itself.

Proposition. The homomorphism , corresponding to the conjugation action has kernel:

.

This is called the centre of G.

The proof is straightforward. It implies that the group of inner automorphisms is (canonically) isomorphic to G/Z(G).

Definition. Two elements of a group are said to be conjugates if they belong to the same orbit under the conjugancy map. Explicitly, this means for some element g of G.

In the case where , we’ve already seen that two permutations are conjugate iff they have the same cycle structure. For example, S₅ has 7 conjugancy classes, represented by e, (1, 2), (1, 2, 3), (1, 2, 3, 4), (1, 2, 3, 4, 5), (1, 2)(3, 4) and (1, 2)(3, 4, 5). Check that these are of sizes 1, 10, 20, 30, 24, 15 and 20 respectively.

If we consider the alternating group A_n, then this is no longer true. E.g. check that the group A₅ has 5 conjugancy classes, represented by e, (1, 2, 3), (1, 2)(3, 4), (1, 2, 3, 4, 5) and (1, 2, 3, 5, 4). These have sizes 1, 20, 15, 12 and 12 respectively. Since a normal subgroup is closed under conjugation, it is a union of conjugancy classes. However, no non-trivial union of conjugancy classes of A₅ is a factor of 60, so we’ve just proven the following.

Theorem. The only normal subgroups of A₅ are {e} and itself, i.e. A₅ is simple.

Now, we bring in results from the previous two posts regarding group actions on sets. First, we partition the set G into orbits under conjugancy to obtain:

since the centre Z(G) is precisely the set of fixed points under conjugation, and x runs over the conjugancy classes with >1 element. Now is the isotropy group of an element x.

If G is a p-group, then since , each summand must be divisible by p. In particular |Z(G)| is divisible by p, and so is non-trivial.

In summary: a p-group has non-trivial centre. 

Outer Automorphisms

First, we claim that the group of inner automorphisms (i.e. conjugate maps) forms a normal subgroup of Aut(G).

Proof. Take automorphisms and . Then:

.

Thus is also an inner automorphism, so the first assertion is done. ♦

Definition. The quotient group Out(G) := Aut(G)/Inn(G) is called the group of outer automorphisms of G.

[ Beware that an element of Out(G) is not an automorphism, but an entire class of automorphisms. ]

The group Out(G) is critical for solving the extension problem for groups. Recall that this is the problem where you’re given groups N and H and must classify all G containing N for which . We’ll illustrate one example by answering an earlier question regarding products of subgroups.

To summarise: let H, K ≤ G be subgroups such that H is normal in G. Also, H ∩ K = {e}. Then HK is a subgroup and the product map H × K → HK which takes (h, k) to hk is bijective but not a homomorphism as elements of H do not commute with those of K in general. However, we’ll use the following expression:

Note: upon swapping k and h’, we have replaced h’ by its conjugate with k, while all other terms remain unchanged.

The “conjugation-by-k” map thus gives rise to an automorphism . Note that since k doesn’t belong to H in general, the map is not an inner automorphism. The mapping thus gives an action of K on H.

Now we reverse this construction.

Definition. Suppose H and K are abstract groups, and K acts on H via . The semidirect product is the set , together with the product:

If the action is assumed to be known, we just write .

Having motivated the definition of the semidirect product, it’s quite trivial to verify that the above construction indeed gives a group. Note that the identity of is just (e, e) and the inverse of (h, k) is given by . Also,

• the subset of comprising of elements (e, k) is a subgroup;
• the subset comprising of (h, e) is a normal subgroup since the second component of the resulting product of is the identity;
• these two subgroups intersect trivially.

Example

The symmetric group S_n has normal subgroup A_n as well as subgroup H = {e, (1,2)}. Since , we see that the symmetric group is the semidirect product of the groups A_n and H.

Exercise 1 : consider the cyclic group Z/pq for primes p and q. If p≠q, we already know that this group is isomorphic to Z/p × Z/q and is hence a non-trivial direct product. In particular it is a semidirect product (the direct product is a special case where the action is trivial). Now suppose p=q. Is Z/p² a non-trivial semidirect product of two subgroups?

Exercise 2 : let K act on H in two ways via . This gives us semidirect products and . Prove that if projecting to Out(H) gives identical maps , then the two semidirect products are isomorphic.

Answers (highlight to read).

1. No because the group has a unique subgroup of order p.
2. (Sketch) Suppose ψ(x) = y·φ(x)·y^-1 for some y in H. The element (h, k) in semidirect product for φ then corresponds to the element (y^-1hy, k) in semidirect product for ψ. This gives an isomorphism between the two semidirect products.