Clean and clever proofs to show the inner automorphism on G is a normal subgroup of the group of automorphisms of G

Question

I am continuing to practice for a qual exam in algebra (the freebie attempt we get the week before my first semester of grad school). I'm especially interested in learning to write cleaner and more clever proofs or proofs that use different methods besides a direct manipulation of definitions to expand my tool-kit. Does anyone have feed back on my proof of the titular question or perhaps an alternative proof? Here's my attempt at the proof...

$\mathrm{Inn}(G)$ is all maps of $G$ such that $\phi_g$is defined by $\phi_g(x) = gxg^{-1}$

Let $f \in \mathrm{Aut}(G)$, then $f: G \rightarrow G$ is an isomorphism

Consider the set $f$ $\mathrm{Inn}(G)$ $f^{-1}$ = $\{ f(\phi_g(f^{-1})) | \phi_g \in \mathrm{Inn}(G) \}$

$f\circ\phi_g\circ f^{-1}(x) = f(\phi_g(f^{-1}(x))) = f(gf^{-1}(x)g^{-1}) = f(g)xf(g^{-1}) = \phi_{f(g)}(x)$

Since $f(g)$ is an isomorphism and we pull our $g$ from all of $G$, $f$ $\mathrm{Inn}(G)$ $f^{-1} = \mathrm{Inn}(G)$ and we are done.

This question is not a duplicate of Inner automorphisms form a normal subgroup of Aut(G). This question is looking for feedback on a particular attempt at a proof to help OP learn better proof writing through critique on their own writing which is why it is labeled "proof-writing". Also this question ask for alternative proofs. At any rate, it is not a duplicate.

Answer 1

Your proof is perfectly fine, though the writing feels rather clunky - largely because it starts by repeating definitions rather than stating a goal. I think you can clean this up a bit by using a more standard form (not because it's always good to use a standard form - but it works here) and by writing out definitions when you need them, not at the start:

Theorem: The set of inner automorphisms $\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$.

Proof: Let $\phi_g(x)=gxg^{-1}$ be an inner automorphism of $G$ and $f$ be an automorphism of $G$. We may show that $f\phi_g f^{-1} = \phi_{f(g)}$ by applying the left hand side to an arbitrary $x\in G$ and manipulating the result: $$f(\phi_g(f^{-1}(x)))=f(gf^{-1}(x)g^{-1})=f(g)xf(g)^{-1}=\phi_{f(g)}(x).$$ Since $\operatorname{Inn}(G)$ is the set automorphisms of the form $\phi_g$, this shows that $\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$.

This does assume that you have already shown that $\operatorname{Inn}(G)$ is actually a subgroup of $\operatorname{Aut}(G)$, but you can basically use the above form for each of the axioms you would need to show to prove that. It also largely assumes that you have already defined $\operatorname{Inn}(G)$ and $\operatorname{Aut}(G)$, but if you're writing just this proof on an exam, that's pretty much a given, and if you're writing a more substantial piece of text, it would be better to introduce the concepts before trying to prove a theorem about them.