Showing that $G$ is a group under an alternative operation.

Solution 1:

More generally, given a bijection $f\colon X \to G$ from a set $X$ to a group $G$, one can pullback the operation of $G$ to make $X$ into a group by defining $x\star y = f^{-1}(f(x)\cdot f(y))$, where $\cdot$ is the operation of $G$.

There is not much really to prove here: whenever you want to operate with elements of $X$, use $f$ to rename them as elements of $G$, do the operation, and rename the result back to $X$. In other words, you're forcing $X$ to be isomorphic to $G$.

In your case, $X=G$ and $f(x)=xc^{-1}$. The only interesting issue is proving that $f$ is a bijection, which follows readily from the group axioms.

Solution 2:

You must show that $G$ has a $*$-identity, that $*$ is associative, and that every element of $G$ has a $*$-inverse. More explicitly, you must show the following:

(1) There is some $e'\in G$ such that for any $x\in G$ we have $x*e'=x=e'*x$.

(2) For any $x,y,z\in G$, we have $x*(y*z)=(x*y)*z$.

(3) For any $x\in G$ there is some $x'\in G$ such that $x*x'=e'=x'*x$.