I would add a reason why the action of on defines a homomorphism . Namely, .
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: . Use the permutation representation of the action to show that G is not simple.
My attempt: Let be the group of permutations on G/H. Then, the action of G on G/H defines a homomorphism . We know the order of is 720. Since the order of G does not divide 720, and is a subgroup of , f cannot be one-to-one. Thus, distinct in such that . Thus, . Since is a normal subgroup of , we have found a normal subgroup of . Also, since is non-trivial, then is a proper normal subgroup of $G.$ Hence is not simple.
Any suggestions or corrections?
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.