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I've been having some trouble with the following type of question:

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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.

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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

  1. 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 $\varphi(x)=y$).

  2. 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?

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  • $\begingroup$ Thanks, I will have a think about d), groups is not a strong point! $\endgroup$ – Mike Miller May 5 '14 at 15:06
  • $\begingroup$ Not making much progress. Have shown that $\varphi$ is a bijection, so the identity map together with $\varphi$ must mean that $|Aut(G)| \geq 2$? But that doesn't use part (c) so must be off. $\endgroup$ – Mike Miller May 5 '14 at 15:23
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    $\begingroup$ My original answer wasn't completely right. $\varphi$ is always a bijection, but for it to be an automorphism it also needs to be a group homomorphism. $\endgroup$ – user134824 May 5 '14 at 15:27
  • $\begingroup$ Well as $G$ is abelian, $\varphi$ is a homomorphism from (c) and a bijection from (b), then $\varphi$ is an automorphism. Then this together with the trival map shows $|Aut(G)| \geq 2$? $\endgroup$ – Mike Miller May 5 '14 at 15:33
  • $\begingroup$ Oh I see, and then have to check that the function is not equal to the identity map, which can only be the case is $c \leq 2$. $\endgroup$ – Mike Miller May 5 '14 at 15:39
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$S_G$ 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?

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  • $\begingroup$ I did know this, but still not a clue I'm afraid. $\endgroup$ – Mike Miller May 5 '14 at 14:48
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    $\begingroup$ So how do you prove that a function is bijective? You'll need to show that it is injective and surjective. Alternatively, you can find an explicit inverse function. Neither is very hard in the context of this question, but the latter is very easy to spot. $\endgroup$ – fixedp May 5 '14 at 14:50
  • $\begingroup$ Oh I see, so as $\varphi o \varphi = id_G$, it's bijective and $\varphi \in S_G$? $\endgroup$ – Mike Miller May 5 '14 at 14:57
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    $\begingroup$ Yes, that is correct. For part d) refer to user134824's answer. $\endgroup$ – fixedp May 5 '14 at 15:02

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