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:
- well-defined
- bijective
- 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?
Best Answer
Hint:
Since $\varphi$ maps element to element, it is well-defined.
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.
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.