Show that a finite group with certain automorphism is abelian

Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)

With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.

Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50

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If you are copying a problem from a book, or if this is homework... it would be nice of you to add that information. In any case, it is often not appreciated by the community here to be told what to do: ask a question instead (where are you stuck? for example) –  Mariano Suárez-Alvarez May 26 '11 at 4:13
It is considered bad form (and poor manners) to simply quote a problem and post it in the imperative, as if you were assigning homework to the group. You should tell us where you encountered this problem (self-study? assignment? a course? which course?), and what you've tried or where you are confused. This will increase the chances of getting an answer that will benefit you the most. –  Arturo Magidin May 26 '11 at 4:14
ok, thank you @Arturo and @Mariano –  leo May 26 '11 at 4:18
This is a problem from herstein which says if $T^{2}=I$, then $G$ is abelian. –  user9413 May 26 '11 at 8:46
@leo: The question after this which is there in herstein is my all time Favorite problem. –  user9413 May 26 '11 at 8:49

Metahint (hint for the hint): Show that if $x^{-1}f(x) = y^{-1}f(y)$, then $x=y$. Conclude that the map (not necessarily a group homomorphism, at this point it's just a set-theoretic function) $x\mapsto x^{-1}f(x)$ from $G$ to itself is one-to-one.

Note that the two things you don't use once you have the hint are the property that $f(z)=z$ implies $z=e$, and the condition that $G$ is finite. I wonder if that will play a role in establishing the hint...?

Additional notes. The conditions that $G$ be finite and that $f$ have no fixed points are both necessary for the conclusion to follow.

An example of an infinite nonabelian group $G$, an isomorphism $f\colon G\to G$ with no fixed points, and with $f\circ f = \mathrm{id}_G$ is given by letting $G$ be the free group of rank $2$, freely generated by $x$ and $y$, and $f\colon G\to G$ be the map that swaps $x$ and $y$. It has no fixed points, since a reduced word that begins with $x$ or $x^{-1}$ has image that begins with $y$ or $y^{-1}$, and viceversa; composing the map with itself gives the identity; but $G$ is not abelian. Even though the (set theoretic) map $w\mapsto w^{-1}f(w)$ is still one-to-one, it is no longer onto: the image contains only words where the sum of the exponents is equal to $0$, so for example $x$ cannot be written as $w^{-1}f(w)$ for any $w\in G$.

For an example with $G$ finite but not abelian if $f$ has fixed points, simply take any nonabelian group with a noncentral element $g$ of order $2$ (any nonabelian simple group will do), and $f$ to be conjugation by $g$.

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Coincidentally, $x\mapsto x^{-1}f(x)=x^{-2}$ is a homomorphism since $G$ is abelian of odd order (though this is of course unnecessary for the argument). –  Zev Chonoles May 26 '11 at 4:37
@Zev: Once you know that the group is abelian, yes. But I did not want the OP to try to show it's a group homomorphism at the point of establishing the hint. –  Arturo Magidin May 26 '11 at 4:39
Indeed, trying to show that directly would be a big pain. I just felt the warning might be better phrased as "not necessarily a group homomorphism" or just "the set-theoretic function". –  Zev Chonoles May 26 '11 at 4:46
@Zev: good point! –  Arturo Magidin May 26 '11 at 4:49

For proving the second part, that is $f \circ f = I$ implies $G$ is abelian you can proceed as follows:

Consider $g \in G$. By the above $\mathsf{Hint}$ every $g \in G$ can be written as $g = x^{-1} f(x)$ for some $x \in G$.

We then have \begin{align*} g = x^{-1} f(x) \Longrightarrow f(g) &= f(x^{-1}f(x))= f(x^{-1}) \cdot f(f(x)) \\ &=(f(x))^{-1} \cdot x = \bigl[x^{-1} f(x)\bigr]^{-1}=g^{-1} \end{align*}

Now we prove that if $f: G \to G$ defined by $f(g) = g^{-1}$ is a Automorphism if $G$ is abelian.

To see this note that $f(ab)=(ab)^{-1} = b^{-1} \cdot a^{-1}$. But $$f(ab) = f(a) \cdot f(b) \Rightarrow f(ab) = a^{-1}b^{-1}$$ From above we have $$b^{-1}a^{-1} = a^{-1}b^{-1} \Rightarrow (ab)^{-1} = (ba)^{-1} \Rightarrow ab=ba$$ Hence $G$ is abelian.

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