# 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

Notation: We take $$a^{b} = b a b^{-1}$$. We use the left action convention.

First we use Sylow theory to establish that there are $$11$$ Sylow-$$5$$ subgroups.

By the Sylow theorems we know that the number of Sylow $$5$$ theorems divides $$11$$ and is congruent to $$1$$ mod 5 and so the possibilities are $$1$$ or $$11$$ Sylow-$$5$$ subgroups. If we had a unique Sylow $$5$$ subgroup then it would be normal. However we are told in the question that there is a Sylow $$5$$ subgroup called $$H$$ which is not normal in $$G$$ (because $$N_{G}(H)=H$$). Thus we conclude there are $$11$$ Sylow $$5$$ subgroups.

Next we show that we have a homomorphism from $$G$$ to $$S_{11}$$.

Possibly too much detail is included in this section and if you are happy with the material in your course on Group Actions you might just want to skip it.

Let $$T$$ be the set of Sylow $$5$$ subgroups. For $$g \in G$$ we define $$\sigma_{g} : T \rightarrow T$$ to be the map given by $$t \mapsto t^{g}$$. By Sylow theory we know that conjugating a Sylow-$$5$$ subgroup gives a Sylow $$5$$ subgroup. Also this map has inverse $$\sigma_{g^{-1}}$$ and so is a bijection. Hence $$\sigma_{g} \in Sym(T)$$.

Next define the map $$\phi : G \rightarrow Sym(T)$$ given by $$g \mapsto \sigma_{g}$$. This map is a homomorphism. Indeed $$\phi(gh)(t)=\sigma_{gh}(t)=t^{gh}= ghth^{-1}g^{-1}=(\sigma_{g} \circ \sigma_{h})(t)=(\phi(g) \circ \phi(h))(t)$$.

Next we want to show that this homomorphism is injective ( the kernel of $$\phi$$ is trivial ).

Notice that if $$g \in Ker \phi$$ then $$\sigma_{g} = Id$$, the identity map. Recall the self normalising group $$H$$ given in the question. We would then have that $$\sigma_{g} (H) = Id(H) = H$$. But $$\sigma_{g}(H)=H^{g}$$, so $$H=H^{g}$$ and thus $$Ker \phi \leq N_{G}(H)=H$$, a cyclic group of order $$5$$ . Hence either $$Ker \phi$$ is $$H$$ or is trivial. By the isomorphism theorems we know that $$Ker \phi$$ is a normal subgroup, and as $$H$$ is not normal we must have that $$Ker \phi$$ is trivial.

Now by the Isomorphism theorems we can get

$$G \cong \frac{G}{\{1\}} \cong \frac{G}{Ker \phi} \cong Im \phi \leq Sym(T)$$

Finally we note that as $$T$$ contains $$11$$ things, we can label these things $$1 - 11$$, and it is clear $$Sym(T) \cong S_{10}$$.