Group theory proof

Poirot

Banned
Let
G be a group with normal subgroups H1 and H2 with H2 not a subset of H1. Let K = H1 intersect H2.

Show that if G/H1 is simple, then G/H1 is isomorphic to H2/K.

My first thought was to set up a homomorphism with K as the kernel but soon realised that the fact that H2 was not normal is H1 scuppered this tactic. G/H1 being simple implies that H1 is the largest proper normal subgroup but where to go from there?

Poirot

Banned
Just realised that my last sentence is incorrect. G/H1 being simple means there is no normal subgroup A of G which H1 is normal in.

Opalg

MHB Oldtimer
Staff member
The quotient map $\pi:G\to G/H_1$ maps $H_2$ to a normal subgroup of $G/H_1$. This normal subgroup contains more than just the identity element, so by simplicity it must be the whole of $G/H_1$. Now show that the kernel of the homomorphism $\pi|_{H_2}$ is equal to $K$.

Poirot

Banned
The quotient map $\pi:G\to G/H_1$ maps $H_2$ to a normal subgroup of $G/H_1$. This normal subgroup contains more than just the identity element, so by simplicity it must be the whole of $G/H_1$. Now show that the kernel of the homomorphism $\pi|_{H_2}$ is equal to $K$.
Ah, my first ideas were correct.