# Minimal normal subgroups.

Hi: Let G be a group and N normal in G such that 1 and N are the only normal subgroups of G that are contained in N. Let $\phi$ be a surjective (onto) homomorphism from G to a group H. Let $A \neq 1$ be a normal subgroup of $H$ that is contained in $N^\phi$ and let $M = A^{\phi^{-1}} \cap N$. Then $M$ is normal in $G$.

OK. But I am said that, then,$M \neq 1$ since $A \neq 1$. I have been trying hard to prove that $M \neq 1$ but I could not. Any suggestion?

### Re: Minimal normal subgroups.

Thanks a lot. Your explanation is clear as day. I did not see that if y belongs to phi(N) there must be x in N such that phi(x) = y. I just saw that $A \subseteq \phi(N)$ does not imply $\phi^{-1}(A) \subseteq N$.

### Re: Minimal normal subgroups.

If there is some y in A different from e' (the neutral element of H), then since A is contained in phi(N) there is some x in N such that phi(x) = y. But x cannot equal e (the neutral element in G), since phi(e) = e'. And that x is in M. So M = phi^{-1}(A) \cap N contains something other than e, and by the minimality of N we have phi^{-1}(A) \cap N = N, same as N \subseteq phi^{-1}(A) and therefore phi(N) \subseteq A and since we had A \subseteq phi(N) we conclude A = phi(N). So we've shown phi(N) is also minimal normal.

So images of minimal normal subgroups under surjective homomorphisms are also minimal normal.

### 1 11 111 1111…

1 11 111 1111… This could be an example to here, the normal.