I am still practicing for a qual exam in algebra (freebie qual I can take the week before grad school starts). I have learned a lot about good proof writing just from the previous two post but I'm sure there will always be helpful feedback I can receive. I'm especially interested in learning to write cleaner and more clever proofs. I checked the qual prep page for my university and they actually give points for more interesting ways of proving a statement so I am also interested in methods which are a little beyond the level of math in the question. Here's my attempt at the proof...
$\bf{Theorem}$: If $\phi : G \rightarrow G$ given by $\phi(g) = g^2$ is an automorphism of a group $G$, then $G$ must be abelian
$\mathbf{Proof}$:
Note that $\phi(fg) = \phi(f)\phi(g) = f^2g^2$ and $\phi(fg) = fgfg$
Applying the proper group operations on $fgfg = f^2g^2$ we get $gf = fg$
Since $f$ and $g$ are arbitrary in $G$ we have that $G$ is abelian and we are done.
My immediate self-critique is that if we are trying to prove that $G$ has a property the proof should start with elements of $G$ and not a map on elements of $G$ for the sake of clarity. But I may be overthinking or seeing it from my perspective and the perspective of a professor grading my qual.
This proof is basically correct, but lacks an introduction and some details, in my opinion. Here is how I would do it:
Consider any $g,g'\in G$ and transform $gg'$ by $\phi$, which is an automorphism by hypothesis – which means that $$\phi(gg')=(gg')^2= gg'\,gg'=\phi(g)\phi(g')=gg\,g'g'.$$ By the (left and right) cancellation rule in a group, we instantly deduce that for any $g, g'\in G$, $$g'g=gg' .$$