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

  • #1
fishturtle1
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81
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?
 
Last edited:

Answers and Replies

  • #2
fresh_42
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Consider ##x \longmapsto xf(x)^{-1}##. Can you show that it is bijective?
 
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  • #3
fishturtle1
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81
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}##.
 
  • #4
fishturtle1
392
81
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!
 

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