# Group acts as automorphisms by conjugation

This article gives the statement, and possibly proof, of a basic fact in group theory.

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This article describes a pivotal group action: a group action on a set closely associated with the group. This action is important to understand and remember.

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## Contents

## Statement

Let be a group. For any , define the map:

given by:

(this is termed *conjugation* by or the *inner automorphism* induced by ).

Then, the following are true:

- For every , is an automorphism of (an automorphism arising this way is termed an inner automorphism).
- The map from to its automorphism group that sends an element to the automorphism , is a homomorphism of groups.

## Related facts

### Analogues in other structures

## Proof

### Proof that every conjugation is an automorphism

*Given*: A group , an element

*To prove*: The map is an automorphism of

*Proof*: is a map from to . We need to thus prove three things:

The key idea for this is the fact that . Formally:

The key idea for this is the fact that . Formally:

Here, we use the fact that the inverse of a product of elements is the product of their inverses, in reverse order (i.e. the inverse map is involutive). Thus, we have:

Note that for the proof to go through we need to use the fact that and are left and right inverses of each other.

NOTE: It actually suffices to prove only the first of these three things, because to test whether a map between groups is a homomorphism of groups, it suffices to check whether it sends products to products. However, when working in somewhat greater generality than groups, it becomes important to check the other conditions, and they're explained here for illustrative purposes.

Thus, every is a homomorphism. It remains to show that this homomorphism is injective and surjective. Injectivity is clear, because . For surjectivity, note that given any , setting:

yields . (This also becomes clearer from the fact that the map is *itself* a homomorphism of groups).

Thus, is an automorphism.

### Proof that the map is a homomorphism

*Given*: A group

*To prove*: The map , is a homomorphism from to

*Proof*: We'll check three things:

The proof again uses the reversal law for inverses (inverse map is involutive):

- is the identity map:

This follows from the first two facts, because which is the identity map, and the same holds for composition in the other order.