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 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:
 .
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?
##(\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?
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