Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $T(x) = x$ for $x \in G$ if and only if $x = e.$ Prove that every $g \in G$ can be represented as $g = x^{-1}T(x)$ for some $x \in G$.
I am having trouble understanding what I have to prove and how to get started? What exactly is $g = x^{-1}T(x)$ telling me? I don't see the connection between "$T(x) = x$ for $x \in G$ if and only if $x = e.$" with "can be represented as $g = x^{-1}T(x)$."
Let's be concrete about what $T$ is. It is an automorphism of $G,$ meaning that it is a bijective function $T:G\to G$ which is a homomorphism. A homomorphism satisfies $T(ab) = T(a)T(b)$ for all $a,b\in G.$ Try to prove that homomorphisms must also have the property that $T(e)=e.$ Another way of wording the property $T(e)=e$ is the following: $T(x)=x$ if $x=e.$ Now your question statement tells us that $x=e$ is the only element of $G$ which has $T(x)=x.$ Call this property (*).
Using the above property, you have to prove that for any element $g$ of $G,$ you can find some $x\in G$ such that $g= x^{-1} T(x).$ A helpful way to rethink this statement: You want to show that the function $F:G\to G \ : \ x\mapsto x^{-1}T(x)$ is surjective. This is where the conditions in the question start to come into play. A map from a finite set to itself is surjective if and only if it is injective. We are given that $G$ is finite, so we just need to show that $F$ is injective. That is, if $F(x)=F(y)$ then $x=y.$ Try to show this, keeping (*) in mind.
