Let $\phi:G\to G'$ be a group homomorphism, and let $N'$ be a normal subgroup of $G'$. Show that $\phi^{-1}[N']$ is a normal subset of $G'$.
My attempt: $\phi^{-1}[N']=\{g\in G:\phi(n)\in N'\}$
$$g' n' (g')^{-1}\in N'$$
$$\Rightarrow\phi^{-1}(g' n' (g')^{-1})=\phi^{-1}(g')\phi^{-1}(n')\phi^{-1}(g'^{-1})$$
Thus $\phi^{-1}[N']$ is normal.
Is this thinking correct? I feel like I am missing something.
Best Answer
Step $1$: Show that $\phi^{-1}[N]\neq \emptyset $
Step $2$: Take $g_1,g_2\in \phi^{-1}[N]$ show that $g_1g_2\in \phi^{-1}[N]$
Step $3$:Take $g\in G,x\in \phi^{-1}[N]$ show that $gxg^{-1} \in \phi^{-1}[N]$