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Calculus and Beyond Homework Help
F is an isomorphism from G onto itself,...., show f(x) = x^-1
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[QUOTE="fishturtle1, post: 6288811, member: 606256"] [B]Homework Statement:[/B] 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}##] [B]Relevant Equations:[/B] . 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? [/QUOTE]
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F is an isomorphism from G onto itself,...., show f(x) = x^-1
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