F is an isomorphism from G onto itself,...., show f(x) = x^-1

[fishturtle1]

Homework Statement

i) Prove that a group is abelian iff $f : G \rightarrow G$ defined as $f(a) = a^{-1}$ is a homomorphism.

ii) Let $f : G \rightarrow G$ be an isomorphism from a finite group $G$ to itself. If $f$ has no nontrivial fixed points (i.e. $f(x) = x \Rightarrow x = e$) and if $f \circ f$ is the identity function, then for all $x \in G$, $f(x) = x^{-1}$, and $G$ is abelian. [Hint: Show for $g \in G$, there exists $x \in G$ such that $g = xf(x)^{-1}$]

Relevant Equations

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i) Proof: Let $a, b \in G$

$(\Rightarrow)$ If $G$ is abelian, then

$\begin{align*} f(a)f(b) &= a^{-1}b^{-1} \\ &= b^{-1}a^{-1} \\ &= (ab)^{-1} \\ &= f(ab) \\ \end{align*}$
So $f$ is a homomorphism.

$(\Leftarrow)$ If $f$ is a homomorphism, then

$\begin{align*} f(a^{-1})f(b^{-1}) &= f(a^{-1}b^{-1}) \\ ab &= (a^{-1}b^{-1})^{-1} \\ ab &= ba \\ \end{align*}$

So $G$ is abelian. []

For ii) I'm stuck. I tried to show the hint first: Let $g \in G$. Then there is $x \in G$ such that $f(x) = g$. We have

$$g = f(x) = f(xxx^{-1}) = f(x)f(x)f(x)^{-1} = f(x)^2f(x)^{-1}$$

and I'm trying to show $g = xf(x)^{-1}$. I know I'm not using the fact that there's no nontrivial fixed points, but I'm not sure how. How to proceed?

 

[fresh_42]

Consider $x \longmapsto xf(x)^{-1}$. Can you show that it is bijective?

 

[fishturtle1]

Consider $x \longmapsto xf(x)^{-1}$. Can you show that it is bijective?

I think so. We know $G$ is finite and we're mapping from $G$ to $G$ so its enough to show the map $x \mapsto xf(x)^{-1}$ is one to one. Let $x, y \in G$ and observe,

$\begin{align*} xf(x)^{-1} &= yf(y)^{-1} \\ y^{-1}x &= f(y)^{-1}f(x) \\ y^{-1}x &= f(y^{-1}x) \\ y^{-1}x &= e \\ x &= y \\ \end{align*}$

This shows the map is one to one and we can conclude its a bijection. This gives us the hint, that for any $g \in G$ there is $x \in G$ such that $g = xf(x)^{-1}$.

 

[fishturtle1]

Ok and to finish the problem:

ii) Proof: Let $g \in G$. We've shown there exists $x \in G$ such that $g = xf(x)^{-1}$. Then $f(g) = f(xf(x)^{-1}) = f(x)f(f(x^{-1})) = f(x)x^{-1}$. But that means $f(g)g = e = gf(g)$ i.e. $f(g) = g^{-1}$. Now by i), this shows $G$ is abelian. []

Thank you for your help on this!