How to prove that this is a group isomorphism?

Question

The conditions are the following. Let $(G,\circ)$ be a group, and $a$ an element of $G$.
Prove that this is a group isomorphism:

$$\phi : G \to G : \quad\phi(x) = a^{-1}\circ x\circ a$$

I know that I have to prove it is:

1. well-defined
2. bijective
3. homomorphic

The main question I have is can I switch the e.g. $a$ and $x$? So the function would be $\phi(x)= a^{-1} \circ a \circ x$, then it would be pretty easy, but since $G$ is not abelian I am not quite sure how to proof the 3 axioms for the isomorphism.

Can anyone help me?

Answer 1

Hint:

1. Since $\varphi$ maps element to element, it is well-defined.

2. Consider $\phi:G\rightarrow G$ defined by $\phi(x)=a\circ x\circ a^{-1}$ for all $x\in G$. Try to show that $\phi\varphi=\text{id}_G=\varphi\phi$. This shows that $\varphi$ is bijective.

3. Let $x,y\in G$. Check that $$\varphi(x\circ y)=a^{-1}\circ (x\circ y)\circ a=(a^{-1}\circ x\circ a)\circ (a^{-1}\circ y\circ a)=\varphi(x)\circ\varphi(y).$$ This shows that $\varphi$ is a homomorphism.

Answer 2

1) $$x=y\Rightarrow a^{-1}x=a^{-1}y \Rightarrow a^{-1}xa=a^{-1}ya \Rightarrow \varphi(x)=\varphi(y)$$ 2) The implications above are actually equivalences(because of the cancellability in a group), so the function is injective. Let's proof it's surjective: $g\in G$. Since $g=a^{-1}(aga^{-1})a=\varphi(aga^{-1})$, the function is surjective.

3) $$\varphi(xy)=a^{-1}xya=a^{-1}x1ya=a^{-1}x(aa^{-1})ya=(a^{-1}xa)(a^{-1}ya)=\varphi(x)\varphi(y) $$

Answer 3

For the surjectivity consider arbitrary $g\in G$ then we know that $a\circ g \circ a^{-1}\in G $ because $a\in G$ and $G$ is a group but so $\phi(a\circ g \circ a^{-1})=a^{-1}\circ a\circ g\circ a^{-1}\circ a=g$. Because $g$ was arbitrary the map is surjective. Now for the injectivity assume $\phi(g)=\phi(h)$, i.e. $a^{-1}\circ g \circ a=a^{-1}\circ h \circ a$ which is equivalent to $a\circ a^{-1}\circ g \circ a\circ a^{-1}=a\circ a^{-1}\circ g \circ a\circ a^{-1}$, i.e. $g=h$ since we are in a group, so the right and left translation maps are bijective. Now for the homomorphism we have $\phi(g\circ h)=a^{-1}\circ g \circ h\circ a=a^{-1}\circ g\circ a\circ a^{-1} \circ h\circ a=\phi(g)\circ\phi(h)$ so we are done.