I want to prove that $|xHx^{-1}|=|H|$ where $H$ is a subgroup of $G$ and $x\in G$. I know how to prove that for each element in $H$, $|xhx^{-1}|=|h|$, but I'm not sure how to extend that to the whole subgroup.
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Note that conjugation is a (self-) group action, so $\varphi_x(g) = x g x^{-1}$ is a homomorphism, so is either injective, so an isomorphism, or not. You need to argue that "or not" doesn't happen. – Eric Towers Dec 15 '15 at 09:28
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You just need to find a bijection between $H$ and $xHx^{-1}$. As drhab suggested the most natural one is: $$ f : H \rightarrow xHx^{-1} \mbox{ with } f(h)=xhx^{-1} \mbox{, where } h \in H $$
It is easy to see $f$ is injective since $ f(h_1) = f(h_2) $ implies $xh_1x^{-1}=xh_2x^{-1}$ and thus $h_1 = h_2$.
It is also easy to see that $f$ is surjective since for any $xhx^{-1} \in xHx^{-1}$ we have $xhx^{-1}=f(h)$ for $ h \in H$.
NNec
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Hint:
For any fixed $x$ the map $H\to xHx^{-1}$ prescribed by $h\mapsto xhx^{-1}$ is invertible (so is bijective).
drhab
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