Originally Posted by

**Arczi1984** Hi,

**Problem**: $\displaystyle G$ - simple $\displaystyle \rightarrow$ has $\displaystyle 8$ subgroups of order $\displaystyle 7$ and we could embedding it into $\displaystyle S_8$.

**Solution**: First part is easy to show from Sylow Theorem.

Suppose that we have $\displaystyle H_1,...,H_8$ - Sylow $\displaystyle 7$-subgroups and consider following mapping:

$\displaystyle \phi : G \to S(\{H_1,...,H_8\})\cong S_8$

$\displaystyle \phi(g)(H_i)=gH_ig^{-1}$, $\displaystyle g \in G$ and $\displaystyle i=1,...,8$.

$\displaystyle \phi$ is homomorphism (it is easy to show that $\displaystyle \phi(gh)(H_i)=(\phi(g)\circ\phi(h))(H_i)$)

$\displaystyle \mbox{Ker}\phi=\{g\in G : \forall_{i=1,...,8} \phi(g)(H_i)=H_i\}=\{g\in G : \forall_{i=1,...,8} gH_ig^{-1}=H_i\}$ **<--- may I describe kernel such that?** If yes then:

$\displaystyle \mbox{Ker}=\{1\}$ or $\displaystyle \mbox{Ker}=G$ - because $\displaystyle G$ is simple.

If $\displaystyle \mbox{Ker}=G$, then $\displaystyle gH_i=H_ig$, so $\displaystyle H_1,...,H_8$ are simple - bad choice.

So then $\displaystyle \mbox{Ker}=\{1\}$ $\displaystyle \rightarrow$ $\displaystyle \phi$ is monomorphis. **[qed]**

Is this solution good? Thanks for any advices.