$xHx^{-1}= \{xHx^{-1} | h \in H \}$

I know how to show that $xHx^{-1}$ is a subgroup but don't know how to show isomorphic.

closed as off-topic by José Carlos Santos, Christopher, Alex Francisco, amWhy, Alex Provost Oct 11 at 15:45

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  • 1
    Do you mean $xHx^{-1}=\{xhx^{-1}\mid h\in H\}$? – cansomeonehelpmeout Oct 11 at 10:45
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    You need a mapping $f:H \to xHx^{-1}$ and you need to show that it is one-to-one and onto. Gee, what should $f(h)$ be set equal to? – steven gregory Oct 11 at 10:49

To show that $H\cong xHx^{-1}$ you need to construct a homomorphism $$\phi:H\rightarrow xHx^{-1}\tag{1}$$ and then check that it the following holds:

  • $\phi(hh')=\phi(h)\phi(h')$, that is, $\phi$ is a homomorphism.
  • $\phi(h)=\phi(h')\Rightarrow h=h'$, that is, $\phi$ is injective.
  • Given $h'\in xHx^{-1}$, find $h\in H$ such that $\phi(h)=h'$.

What do you think $\phi(h)$ should look like to map to $xHx^{-1}$?

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