# Thread: Permutation representation argument validity

1. ## Permutation representation argument validity

Hello,
I would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you.

Problem statement: Let G be a group of order 150. Let H be a subgroup of G of order 25. Consider the action of G on G/H by left multiplication: $g*aH=gaH$. Use the permutation representation of the action to show that G is not simple.

My attempt: Let $S_6$ be the group of permutations on G/H. Then, the action of G on G/H defines a homomorphism $f:G \rightarrow S_6$. We know the order of $S_6$ is 720. Since the order of G does not divide 720, and $f(G)$ is a subgroup of $S_6$, f cannot be one-to-one. Thus, $\exists g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implies f(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple.

Any suggestions or corrections?

2. ## Re: Permutation representation argument validity

I would add a reason why the action of $G$ on $G / H$ defines a homomorphism $f: G \to S_6$. Namely, $[G:G/H] = 6$.

3. ## Re: Permutation representation argument validity

Hi,
Unless you have previously discussed the permutation representation of a group G on the left cosets of a subgroup H of G, I would provide more detail on why any g in G induces a permutation (one to one onto mapping) by left multiplication of said cosets. Finally, the symbol G/H is normally reserved for the quotient group (so H must be normal in G), and not the left cosets of H in G. Otherwise your proof is fine.

Your statement follows immediately from the proposition:

Let G be a finite group with a subgroup H of index n. Then G has a normal subgroup N with n dividing |G/N| and |G/N| divides n!.

You might want to try and prove this. Hint: G acts transitively on the left cosets of H.