# Math Help - G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

1. ## G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

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?

Notation: 1 stands for the trivial subgroup of G.

2. ## Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

Hi enriquestefanini,

well, by the definition of a function $f: G \rightarrow H$, and you have g $\in$ G,
then there is at most one image f(g) = h $\in$ H (there cannot be two distinct elements of H being
the image of g, since this would be in contrary to the definition of a function).

So if you have at least two elements in the image of f, e.g. $im(f) = \{h_1 , h_2\}$, then
you have at least two different elements of G. Then it follows that M cannot be 1 alone, since if it was,
there would be only one element in f(M).

Further, the image f(N) $\subset$ H.

Is that clear to you now?

3. ## Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

Originally Posted by mastermind2007
Hi enriquestefanini,

well, by the definition of a function $f: G \rightarrow H$, and you have g $\in$ G,
then there is at most one image f(g) = h $\in$ H (there cannot be two distinct elements of H being
the image of g, since this would be in contrary to the definition of a function).

So if you have at least two elements in the image of f, e.g. $im(f) = \{h_1 , h_2\}$, then
you have at least two different elements of G. Then it follows that M cannot be 1 alone, since if it was,
there would be only one element in f(M).

Further, the image f(N) $\subset$ H.

Is that clear to you now?
i don't think this alone by itself is enough. surely the pre-image of A has more than one element in it, but we need to show more: that φ-1(A)∩N cannot be trivial.

but it's really deceptively simple: since A is non-trivial, we have a ≠ 1 in A.

since A is contained in φ(N), we have a = φ(n), for some n in N.

this n is in φ-1(A)∩N, and is a non-identity element.

4. ## Re: G, H groups, phi: G -> H epimorphism, N normal in G, A normal in H. Then ...

Thanks very much, guys. It's pretty clear now.