## 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. Anautomorphismof a group G is an isomorphism G → G. The set of all automorphisms of G forms a group Aut(G) under composition. Anactionof 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 aninner 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

centreof 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 beconjugatesif 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, *S*_{5} 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

*A*

_{5}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

*A*

_{5}is a factor of 60, so we’ve just proven the following.

Theorem. The only normal subgroups ofA_{5}are {e} and itself, i.e.A_{5}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 ofouter automorphismsof 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 . Thesemidirect productis 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*as well as subgroup

_{n}*H*= {

*e*, (1,2)}. Since , we see that the symmetric group is the semidirect product of the groups

*A*and

_{n}*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**/*p*^{2} 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).

- No because the group has a unique subgroup of order p.
- (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*^{-1}*hy*,*k*) in semidirect product for ψ. This gives an isomorphism between the two semidirect products.