Suppose that $\phi : G \to G'$ is a homomorphism between the groups $G$ and $G'$. Let $N'$ be a normal subgroup of $G'.$ Prove that the inverse image of $N'$ is a normal subgroup of $G$.
How can I prove this using the defintions?
A proof that was given confused me. The proof states:
Consider the homomorphism $\phi': G' \to G'/N'$ then $N'$ is the kernal of $\phi'$ (why?). Now consider the composition of $\phi$ and $\phi'$ then the inverse image $N$ is the kernal of $\phi' \circ \phi$ (again why?).
Best Answer
Alternative proof:
Define $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $. Then $x,y\in N\Rightarrow xy^{-1}\in N$ since $\phi\left(xy^{-1}\right)=\phi\left(x\right)\phi\left(y\right)^{-1}\in N'$. This proves that $N$ is a subgroup of $G$. Let $n\in N$ and $g\in G$. Then $\phi\left(gng^{-1}\right)=\phi\left(g\right)\phi\left(n\right)\phi\left(g\right)^{-1}\in N'$ because $N'$ is normal. This shows that $gng^{-1}\in N$ and proved is now that $N$ is a normal subgroup of $G$.
When $\nu:G'\rightarrow G'/N'$ is the natural map then $\psi=\nu\circ\phi:G\rightarrow G'/N'$. Notice that $N'$ is the kernel of $\nu$ so the set $N=\left\{ x\in G\mid\phi\left(x\right)\in N'\right\} $ is the kernel of $\psi$.
Every kernel is a normal subgroup, so this also proves that $N$ is a normal subgroup.
In your question $\nu$ is denoted as $\phi'$.