# Inner automorphisms (need clarification)

1. Apr 20, 2010

### bennyska

1. The problem statement, all variables and given/known data

Show that the inner automorphisms of a group G form a normal subgroup of the group of all automorphisms of G under function composition.

2. Relevant equations

in the previous problem, i showed that all automorphisms of a group G form a group under function composition

3. The attempt at a solution

so i need help understanding the question. in the previous problem, assuming i did it correctly, i let A be the set of all automorphisms and showed it was a group in G. in this problem, am i to let say I be the set of all inner automorphisms and show it is a normal subgroup of A?

here is my previous proof.

: Claim: All automorphisms of a group G form a group under function composition.
Proof: Let A be the set of automorphisms of a group G, and let μ(g) and σ(g) be in A, with g in G. Since μ(g) and σ(g) are automorphisms, it follows that the mapping of μ(g)◦σ(g) defined as μ(σ(g)) is closed under function composition.
Let λ(g) be in A. Then
(μ(g)◦σ(g))◦λ(g) = μ(σ(g))◦λ(g)
= μ(g)◦σ(g)◦λ(g)
= μ(g)◦(σ(g)◦λ(g))
so A is associative.
Consider μ:G→G such that μ(g) = g. Then μ is the identity.
Consider σ:G→G such that σ(g) = a. Now consider λ:G→G such that λ(a) = g. Then σ(λ(a)) = σ(g) = a, and λ is the inverse of σ.
Since A is the set of automorphisms of G, it follows that for any σ in A, then σ is homomorphic. Hence, the set of automorphisms of a group G forms a group under function composition.

2. Apr 20, 2010

### VeeEight

You are asked to show that the set of inner automorphisms is a normal subgroup of the group of automorphisms of G.
An inner automorphism is a specific type of automorphism: http://en.wikipedia.org/wiki/Inner_automorphism, so it is natural that it will form a subset. If you've already shown that the set of automorphisms on group G is a group, then you have already showed closure (under function composition).

You are correct in showing the identity element, but you haven't used the specific property of conjugation that is the defining property of your set of inner automorphisms. You must show invertibility to show it is a subgroup and then that every left coset is a right coset, to show it's normal.