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Given a group $G$ with $a\in G$ fixed, define $\phi: G \to G$ by $\phi(x) = axa^{-1}, x \in G$, I am wondering when $\phi$ is an isomorphism. I think that it is always an isomorphism because it is one to one, and the order of the domain ($|G|$) and the order of the range are equal. Am I wrong in thinking this?

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    $\begingroup$ Your argument doesn't work if $G$ is infinite. Conjugation is always an isomorphism because it always has an inverse. $\endgroup$ Dec 3, 2012 at 21:46

2 Answers 2

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Given a group $G$ with $a\in G$ fixed, define $\phi: G \to G$ by $\phi(x) = axa^{-1}, x \in G$, prove that $\phi$ is always an isomorphism.

You wrote:

  • I think that it is always an isomorphism because it is one to one, and the order of the domain ($|G|$) and the order of the range are equal. Am I wrong in thinking this?

That gives you that $\phi$ is bijective if $G$ is finite. Otherwise, you can prove that $\phi$ is bijective by establishing that $\phi^{-1}$ exists.

To prove that it is always an isomorphism: you also need to show that $\phi$ is a homomorphism: that is, that $\phi$ satisfies the following property: $$\phi(xy) = \phi(x)\phi(y), \forall x, y \in G.$$ The proof is fairly straightforward:

Let $x, y$ be any elements in $G$. Then:

$$\phi(x)\phi(y) = (axa^{-1})(aya^{-1}) = ax(a^{-1}a)ya^{-1} = a(xy)a^{-1} = \phi(xy)$$

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    $\begingroup$ It doesn't give you that $\phi$ is bijective if $G$ is infinite. $\endgroup$ Dec 3, 2012 at 21:47
  • $\begingroup$ Thanks for the helpful answer and comments! $\endgroup$
    – Moderat
    Dec 4, 2012 at 0:31
  • $\begingroup$ Your very welcome, Josh! $\endgroup$
    – amWhy
    Dec 4, 2012 at 0:37
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== $\,\phi\,$ is $\,1-1\,$: $\,\phi(x)=axa^{-1}=1\Longrightarrow x=a^{-1}a=1\Longrightarrow \ker\phi=\{1\}\,$

== $\,\phi\,$ is onto: for all $\,y\in G\,\,,\,\phi(a^{-1}ya)=a(a^{-1}ya)a^{-1}=y\,$

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    $\begingroup$ I always thought that $1-1=0$, and if $\phi$ is $1-1$ then $\phi$ is the trivial homomorphism. This means it is injective only if $G$ is the trivial group. Oh wait! You meant $1$-$1$, like one-to-one. :-) $\endgroup$
    – Asaf Karagila
    Dec 3, 2012 at 22:11

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