# Group Theory, group of order 55

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 !

• The second is the "traditional" one to use, but any faithful action on a set of order $11$ will work. – user3482749 Jan 12 at 11:25
• 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 $H\backslash G$. – Mark Jan 12 at 11:43
• @Mark in fact I didn't how to write the left backslash... – Marine Galantin Jan 12 at 12:12
• @user3482749 can you be more precise ? Or could you please detail your thoughts ? I'm just starting with group theory. – Marine Galantin Jan 12 at 12:13
• 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. – Dietrich Burde Jan 12 at 12:39