# H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n!. [duplicate]

Prove that if G is a group and H a subgroup of index n, then G has a normal subgroup K with [G : K] ≤ n!

I'm having trouble proving this because frankly I have no idea where to start. Any tips?

• There is a natural group homomorphism $G\to \text{Perm}(G/H)$, where $\text{Perm}(G/H)$ is the group of permutations on the set of cosets $G/H$. – Batominovski Mar 1 '17 at 6:48
• Related: 1,2. Use core as a buzzword to search for more within our tag group-theory. – Jyrki Lahtonen Mar 1 '17 at 7:00

## 1 Answer

Start by considering the action of $G$ on the set of left cosets of $H$ in $G$. From this we get a homomorphism (the associated permutation representation), whose kernel $K$ is normal and contained in $H$. Can you take it from here?

• I'm assuming you mean the homomorphism from G -> G/N? And why is K normal and in H? – user3491700 Mar 1 '17 at 7:11
• Yeah, thats the same homomorphism I was referring to. The kernel of any homomorphism is always normal, to see this note that a subgroup $N$ is normal iff $gNg^{-1} \subseteq N$ for all $g \in G$, use this to show that the kernel of any homomorphism is normal. You actually don't need to use the fact that $K$ is contained in $H$ for the result you are trying to prove. – Miguel Landeros Mar 1 '17 at 7:20