Let $G$ be a group. Show that, for every $a\in G$, the map $\phi_a:G\to G$, defined by $\phi_a(g) := aga^{−1}$ ($g\in G$), is a group automorphism.

Question

Let $G$ be a group. Show that, for every $a \in G$, the map $\phi_a : G \to G$, defined by $\phi_a(g) := aga^{−1}~ (g \in G)$, is a group automorphism.

Don't really understand how exactly to go about this one - do I have to show that $ϕa$ is a homomorphism and then bijective?

Answer 1

do I have to show that $G$ is a homomorphism and then bijective?

I assume that was a typo and you meant "$\phi_a$ is a homomorphism and then bijective". Yes, that's what you need to show.

Proof.

1. $\phi_a$ is a homomorphism: Indeed, let $g,h\in G$ and $e\in G$ be the neutral element. Then $$\phi_a(gh)=agha^{-1}=ageha^{-1}=aga^{-1}aha^{-1}=\phi_a(g)\phi_a(h)$$
2. $\phi_a$ is bijective: It is enough to construct the inverse of $\phi_a$. So consider $\phi_{a^{-1}}$. Then we have $$\phi_a\big(\phi_{a^{-1}}(x)\big)=\phi_a(a^{-1}xa)=aa^{-1}xaa^{-1}=x$$ Analogously $\phi_{a^{-1}}\big(\phi_a(x)\big)=x$. Therefore $\phi_{a^{-1}}$ is the inverse of $\phi_a$ which completes the proof. $\Box$

Answer 2

Yes, that's what you must show. By definition, an automorphism is an isomorphism from a group to itself- it is, therefore, a permutation of its elements. So, for example, $\phi _a$ is an automorphism because: i) it is a homomorphism; ii) it is a permutation of the elements of $G$. There are many ways of proving these things, but, if you're just a begginer, I'd recommend to do it more sistematically.

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I'll give an example of how you could do it sistematically- if you want to answer the question by yourself, there's no need to read this. As an example, to prove it is a homomorphism, just show that, if $b,c\in G$, $$\phi _a(bc)=abca^{-1}=aba^{-1}aca^{-1}=\phi _a(b) \phi _a (c).$$ To prove it is a bijection, show it is injective and surjective. To show the first part, let $\phi _a(b)=\phi _a(c).$ Therefore, $aba^{-1}=aca^{-1}$, so that $b=c$. To show it is surjective, let $y=a^{-1}ba$. Therefore, $\phi_a(y)=b$, so that our funciont is surjective.