Let $G$ be a group and let $H$ be a subgroup of $G$ which has exactly two distinct cosets. Let $C = \{H' \subset G ~|~ H' = gHg^{−1}$ for some $g ∈ G\}$. How many elements does the set $C$ have?

Since $H$ has only two left cosets, it is a normal subgroup. So $gHg^{-1}=H$. Hence $C$ has only one element, namely $H$. Am I right?

  • $\begingroup$ but is $gHg-1$ or $gHg^{-1}$ ? $\endgroup$ – WLOG Dec 27 '13 at 7:51
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    $\begingroup$ your solution is right $\endgroup$ – WLOG Dec 27 '13 at 8:00

Another way is to consider the normalizer of $H$ in $G$, called $N_G(H)$. Since $H\unlhd G$ so we have $$N_G(H)=G$$ Now show this map: $$\phi:\{H^g\mid g\in G\}\to \{N_G(H)g\mid\in G\}\\ \phi(H^g)=N_G(h)g$$ is a group isomorphism and so the numbers of all conjugations of $H$ is equal to $[G:N_H(G)]$ which is $1$ here.

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  • $\begingroup$ @amWhy: Congratulations! I know this makes you well. $15$ golds. Yahoooooo. :-) $\endgroup$ – mrs Dec 30 '13 at 14:08

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