3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ but is $gHg-1$ or $gHg^{-1}$ ? $\endgroup$ – WLOG Dec 27 '13 at 7:51
  • 2
    $\begingroup$ your solution is right $\endgroup$ – WLOG Dec 27 '13 at 8:00
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.