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: $\displaystyle g*aH=gaH$. Use the permutation representation of the action to show that G is not simple.

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

Any suggestions or corrections?