# Proving this homomorphism has a nontrivial kernel

Given: $|G|=n$, $H$ is a subgroup of $G$ and $|G/H|=k$, where $n$ does not divide k!.

WTP: The left action map on $G/H$ has a nontrivial kernel.

I have not put the entire problem I am trying to work on here, just the part I am stuck on. I think this should be true, but I am having trouble showing nontriviality. I know if we take an element of G/H then the kernel of the map are all elements a such that $agH=gH$, but I cannot prove that there are any $a$'s that will do this other than the identity. Can anyone help, or tell me if this is not true?

• Work this out with $S_3$ and a subgroup of order $2$. Apr 20, 2015 at 3:17
• The kernel are all elements $a$ such that $agH=gH$ for every $g$, i.e., the intersection of the stabilizers. Apr 20, 2015 at 3:18
• I'm sorry I mistyped, n should not divide k!, obviously it does not divide k,k-1,k-2,...,1. I have modified the question. Apr 20, 2015 at 3:22

Hint: The left action map of $G$ on $G/H$ induces a homomorphism $\phi: G \to S(G/H) \simeq S_k$, given by $a \mapsto \{gH \mapsto agH\}$.
The kernel of the action is the kernel of this homomorphism. If the kernel were trivial, then $\phi$ would be injective, giving an embedding of $G$ as a subgroup of $S_k$. Why can't this happen?
• No idea what you mean by "I think the homomorphism I am looking at is..."; remember that $G/H$ is not necessarily a group. The induced homomorphism I speak of is given by $a \mapsto \{gH \mapsto agH\}$. What can be said about this homomorphism if the kernel is trivial? Apr 20, 2015 at 3:44