Groups past exam question help I've been having some trouble with the following type of question:

I think I can do part (a) and part (c). 
(a) 
The subset is the set of all elements with self-inverses: $E =\lbrace e,rot_{\pi},ref_0,ref_{\pi /4}, ref_{pi /2}, ref_{3\pi /2} \rbrace $
(c) 
First in the $\rightarrow$ direction. Assume $\varphi \in Hom(G,G)$ and let $x,y \in G$, then $$\varphi(xy)=(xy)^{-1}$$ $$\implies \varphi(x) \varphi(y)=y^{-1} x^{-1}$$ $$\implies x^{-1} y^{-1} =y^{-1} x^{-1},$$ hence abelian. 
In the $\leftarrow$ direction. Assume $G$ is abelian. Let $x,y \in G$, then $$\varphi(xy)=(xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = \varphi(x)\varphi(y).$$ Hence $\varphi \in Hom(G,G)$. 
For (b) I don't really have a clue what I'm trying to show, or the technique to do it. Many thanks. 
 A: $S_G$ is the group of permutations of elements of $G$ (so the set of bijective functions $f:G\to G$). Think you can take it from here?
A: For part (b), you need to show that $\varphi:G\to G$ is a bijection. Two common methods to show that a function is a bijection include


*

*Showing the function is injective (if $\varphi(x)=\varphi(x')$ then $x=x'$) and surjective (if $y\in G$ then there is an $x\in G$ such that $\varphi(x)=y$).

*Showing that the function has an inverse (in both directions): that is, showing that there is a function $\psi:G\to G$ such that $\varphi\circ\psi=1_G$ and $\psi\circ\varphi=1_G$. (By $1_G$ I mean the identity map on $G$.)
Either method will work for your particular problem.
For part (d), the question poser meant to write "using parts (b) and (c)". To show that $|\operatorname{Aut}(G)|\geq2$, all you must do is find a single nontrivial automorphism of $G$. By "nontrivial" I mean "not the identity map" and by "automorphism" I mean "a group homomorphism from $G$ to $G$ which is a bijection". Think about the earlier parts of the problem. Did any bijections appear? Were they nontrivial?
