Exercise : Let G finite group, $\alpha$ automorphism of G This morning during an exam, it was given to me the following exercise that i was unable to solve, and i would like to know at least how should be solved.
Maybe it was my bad, but it was the first time that i saw an Automorphism and i didn't really much more that if an isomorphism $f : G \longrightarrow G$
Let $G$ be a finite group and $\alpha$ an Automorphism of $G$ such that $\alpha(x)=x$ if and only if $x =e_G$
$(a)$ Prove that for every $g \in G$ exists $x \in G$ such that $g=x^{-1}\alpha(x)$.
$(b)$ Prove that if it's true that $\alpha(\alpha(x))=x$ for every $x \in G$, then $\alpha(g)= g^{-1}$ for every $g \in G$.
Your help would be amazing, 
Thank you anyway.
 A: Consider $f: G \to G$ defined by $f(x) = x^{-1}\alpha(x)$. Part a) asks to show that $f$ is surjective. Since $|G|$ is finite, it suffices to show that $f$ is injective. So suppose that $f(x) = f(y)$. Then 
$$x^{-1}\alpha(x) = y^{-1}\alpha(y),$$
so multiplying on the left by $y$ and on the right by $\alpha(x)^{-1} = \alpha(x^{-1})$ shows
$$yx^{-1} = \alpha(y)\alpha(x)^{-1} = \alpha(yx^{-1}).$$
But since $\alpha(z) = z$ if and only if $z$ is equal to the identity, this implies that $yx^{-1}$ is the identity, i.e. $x=y$. 
To solve part b), observe that if $g \in G$, then $g = x^{-1}\alpha(x)$ for some $x \in G$ by part a). Then 
$$\alpha(g) = \alpha(x)^{-1} \alpha(\alpha(x)) = \alpha(x)^{-1}x = (x^{-1}\alpha(x))^{-1} = g^{-1},$$
as desired.
A: For part $a$ suppose $x^{-1}\alpha (x)=y^ {-1}\alpha(y)$ and conclude $xy^{-1}=\alpha(xy^{-1})$. Hence $x=y$, since the group is finite this means $x^{-1}\alpha(x)$ covers the group.
For part $b$ notice $\alpha(x^{-1}\alpha(x))=\alpha(x^{-1})x=(x^ {-1}\alpha(x))^{-1}$.
And since $x^{-1}\alpha(x)$ covers the whole group we are done.
A: Let $a=\alpha$ be the given automorphism.
(a) Let us consider the appropiate map (of sets), $G\to G$, $x\to f(x):= x^{-1}a(x)$. Let us show the injectivity of this map. Start with $x,y\in G$, and assume $f(x)=f(y)$. Then we can successively rewrite equivalently:
$$
\begin{aligned}
f(x)&=f(y)\ ,\\
x^{-1}a(x) &= y^{-1}a(y)\ ,\\
yx^{-1}a(x) &= a(y)\ ,\\
yx^{-1} &= a(y)\, a(x)^{-1}\ ,\\
yx^{-1} &= a(y\, x^{-1})\ ,
\end{aligned}
$$
so the element $yx^{-1}$, invariated by $a$ is the neutral element. This implies $x=y$. 
We have thus shown the injectivity.
Since $G$ is finite, the injectivity of the self-map $f$ implies its surjectivity. It is exactly what we have to show in (a). Moreover the $x$ is unique.
(b) Fix some $g$ in $G$. We find an $x$ as in (a). Then:
$$
\begin{aligned}
a(g)
&=
a(x^{-1}a(x))\\
&=
a(x^{-1})\, a(a(x))\\
&=
a(x^{-1})\, x\\
&=
a(x)^{-1}\, x\\
&=
(x^{-1}a(x))^{-1}\\
&=g^{-1}\ .
\end{aligned}
$$
Note: In this case, this also implies that the group $G$ is commutative.
