# Dummit Foote, Group Theory, Section 1.6, Problem 23 [closed]

Dummit Foote, Group Theory, Section 1.6, Problem 23.

Let $$G$$ be a finite group which possesses an automorphism $$\phi$$ such that $$\phi(g) = g$$ if and only if $$g = 1$$. If $$\phi^2$$ is the identity map from $$G$$ to $$G$$, prove that $$G$$ is abelian. (Such an automorphism $$\phi$$ is called fixed point free of order $$2$$.)

Hint: Show that every element of $$G$$ can be written in the form $$x^{-1} \phi(x)$$ and apply $$\phi$$ to such an expression.

I have seen the solution. I don't understand why and how would I think of expressing each element as $$x^{−1}\phi(x)$$? What is the intuition for this? Please help.

• The hint alludes to the fact that we can “rename” our elements by the given formula. Isomorphisms are just the renaming of groups. Hence, we should see if this “renaming” is an isomorphism.
– Nic
Commented Aug 13 at 7:18
• Thank you, but I don't understand why and how would I think of expressing each element as "x−1ϕ(x)"? As in, why out of a different ways of renaming I would come up with this particular renaming procedure?
– S_M
Commented Aug 13 at 7:26
• Besides “because it was in the hint”, I’m not sure how much I can say about the why. As for the how, “expressing each element as…” leads me to think about there being a one-to-one correlation between the $x$ and $x^{-1}\phi (x)$.
– Nic
Commented Aug 13 at 7:35
• Intuition cannot be taught. It is acquired by experience. There are many proofs of mathematical results that involve some clever trick, and of course it s a common reaction to think "that's clever but I would never have thought of that by myself". Commented Aug 13 at 9:19

Note: As slightly less cumbersome notation, I write $$g'$$ for the inverse of $$g \in G$$ rather than $$g^{-1}$$.

A short answer is that you practice with similar problems over time and eventually spot common tricks (approaches used once) or techniques (approaches used at least twice). Below is a longer answer.

We want to take $$x \in G$$ and somehow productively apply $$\phi$$ to it. Suppose we can write $$x = ab$$ for $$a, b \in G$$. Then we would have: $$\phi(ab) = \phi(a)\phi(b)$$. A dead end! In particular, it seems that we should be rigging up our $$x$$ in some way that uses $$\phi$$ of an element; this will be advantageous when we re-apply $$\phi$$ because we have the additional given of $$\phi^2$$ being the identity map on $$G$$.

New plan: Suppose for any $$x \in G$$ we can write $$x = a\phi(b)$$ for some $$a,b \in G$$. Let's apply $$\phi$$ to this:

$$\phi(a\phi(b)) = \phi(a)b$$

I don't see where to go from here; moreover, there are too many letters flying around. This is where the leap (?) perhaps takes place: what if, instead of $$b$$, we could rig this up using $$a'$$? In other words, suppose it were the case that any $$x \in G$$ can be expressed as $$x = a\phi(a')$$. On second thought, mixing $$\phi$$ and the inverse of $$a$$ seems more complicated than necessary; after all, this last idea is equivalent to $$x = g'\phi(g)$$ where we have set $$g=a'$$. At this point, we have the desired setup of $$x = g'\phi(g)$$, which means applying $$\phi$$ will yield:

$$\phi(g'\phi(g)) = \phi(g')\phi^2(g) = [\phi(g)]'g = (g'\phi(g))'$$

where the final equality involves an extra bit of noticing. This has the wonderful property that $$\phi(x) = x'$$, which, if true for every $$x \in G$$, is equivalent to $$G$$ being an abelian group. In particular, for any $$x,y \in G$$:

$$xy = \phi(x') \phi(y') = \phi(x'y') = \phi([yx]') = [yx]'' = yx$$

whence $$G$$ is abelian as desired.

My guess is that D&F include this final equivalence/proof (or assign it as an exercise) earlier in the book.

As a slight postscript, my response here does not actually show that every element can be written in the desired form. And that may be a good thing: Although we used some of the givens (e.g., properties of $$\phi$$ being an automorphism – at least a homomorphism) we never did bring in the finiteness of $$G$$. So, the proof that elements can be expressed in the desired form is likely to include that condition!