# If H is a subgroup of G then show that H and xHx^-1 are isomorphic? [closed]

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

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 ProvostOct 11 at 15:45

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• Do you mean $xHx^{-1}=\{xhx^{-1}\mid h\in H\}$? – cansomeonehelpmeout Oct 11 at 10:45
• 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 $$xHx^{-1}$$?