# The preimage of a normal subgroup under a group homomorphism is normal

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.

• I don't know if this is just happening on my phone, but the \prime command looks terrible. Normally it is sufficient to use an apostrophe. – Matt Samuel Nov 3 '15 at 4:41
• @MattSamuel One can use \prime but it needs to be a superscript, like $g^\prime$. Of course $g'$ is simpler to type, and is the shorthand for the former. – user147263 Nov 3 '15 at 4:42

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 $$gxg^{-1} \in \phi^{-1}[N]$$