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We set $G$ as a group of order $55$. Let $H$ be a sub group of order $5$ and such that $N_G(H) = H$. https://en.wikipedia.org/wiki/Centralizer_and_normalizer

Finally let's call $N$ a normal sub group of order $11$.

I'm trying to show that $G$ is isomorphic to a subgroup of $S_{11}$

What I was thinking is that we can use the induced morphism by the action of $G$ on a group of order $11$. My question is the following :

There are basically 3 sets of order $11$ in this problem :

  • $N$
  • the set of the subgroups conjugate of $H$ (I think they are $11$ by a Sylow theorem )
  • $ G \backslash H$

Is it possible to prove the isomorphism for any of those 3 sets ? I have succeded in proving the isomorphism for the last set (meaning $ G \backslash H$.)

Any other method is welcomed.

Thank you !

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  • $\begingroup$ The second is the "traditional" one to use, but any faithful action on a set of order $11$ will work. $\endgroup$ – user3482749 Jan 12 at 11:25
  • $\begingroup$ If the third one is the set of left cosets then note that it should be $G/H$, not $G\backslash H$. The set of right cosets is $H\backslash G$. $\endgroup$ – Mark Jan 12 at 11:43
  • $\begingroup$ @Mark in fact I didn't how to write the left backslash... $\endgroup$ – Marine Galantin Jan 12 at 12:12
  • $\begingroup$ @user3482749 can you be more precise ? Or could you please detail your thoughts ? I'm just starting with group theory. $\endgroup$ – Marine Galantin Jan 12 at 12:13
  • $\begingroup$ Consider a map $\phi\colon G\rightarrow S_{11}$ given by the second action. Show that $\ker (\phi)=1$, and that $\phi$ is a homomorphism. Then $G$ is a subgroup of $S_{11}$. For details see this duplicate. $\endgroup$ – Dietrich Burde Jan 12 at 12:39

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