Properties of a set of all isomorphisms $ f: G \to G $

Question

I'm kinda stuck with this task.

Let $G$ be a group and $ S $ the set of all isomorphisms $ f: G \to G$. I first want to show that $ (S, \circ) $ is also a group.

I believe I've shown that all the properties of a group is fulfilled with $ (S, \circ) $:

i) Assume that $ x \in $ and $ f_1,f_2 \in S$. Then $f_2(x) = y \in G$, since $ f_1 $ is an ismorphism. $ f_2(y) = z \in G $ since $ f_2 $ is an ismorphism. Then $ f_1(f_2(x)) = f_1(y) = z = f_1 \circ f_2(x),$ hence $f_1, f_2 \in S \longrightarrow f_1 \circ f_2 \in S.$

ii) $id$ is an isomorphism $ \longrightarrow id \in S$.

iii) $ f $ is an isomorphism $ \longrightarrow \exists f^{-1}$, one can show that $ f^{-1} $ is also an isomorphism, $ \longrightarrow f^{-1} \in S$.

Now, assume that $ | S | = 1$, then I want to show that $ G $ is abelian and each element has an order of 1 or 2.

Im kinda lost with that $ | S | = 1 $. If $ f \in S $ then $f^{-1} \in S $ due to previous result, should not $ | S | = 1 $ imply that $ f = f^{-1} = id$? And how do I move forward from this? Any hints is muy appreciated!

Answer 1

Hint: conjugation by any element of the group is an isomorphism in any group and taking inverse is an isomorphism in any abelian group.

Answer 2

Since $Id_G\in \operatorname{Aut}(G)\subseteq\operatorname{Sym}(G)$, then (subgroup test) you have to prove that from $f,f_1,f_2\in \operatorname{Aut}(G)$ follow that $f_1f_2$ and $f^{-1}$ are operation-preserving. So:

• $(f_1f_2)(gh)=f_1(f_2(gh))=f_1(f_2(g)f_2(h))=f_1(f_2(g))f_1(f_2(h))=(f_1f_2)(g)(f_1f_2)(h)\Longrightarrow f_1f_2\in\operatorname{Aut}(G)$
• $f^{-1}(gh)=f^{-1}(g)f^{-1}(h)\iff f(f^{-1}(gh))=f(f^{-1}(g)f^{-1}(h))\iff gh=gh\Longrightarrow f^{-1}\in\operatorname{Aut}(G)$

Now, $gh=g(hg)g^{-1}=\varphi_g(hg)$ for every $g,h\in G$, where $\varphi_g$ is the automorphism "conjugation by $g$". So, if $\mid\operatorname{Aut}(G)|=1$, then $\varphi_g=Id_G$ for every $g\in G$, whence $gh=hg$ for every $g,h\in G$, and $G$ is abelian.