# 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. Nov 3, 2015 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, 2015 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 $$gxg^{-1} \in \phi^{-1}[N]$$

## Step 1:

Note that $$N'\unlhd G'$$ implies $$\phi(e_G)=e_{G'}=e_{N'}$$. Thus $$e_G\in\phi^{-1}(N')$$, so $$\phi^{-1}(N')\neq \varnothing$$.

## Step 2:

Let $$g,h\in \phi^{-1}(N')$$. Then there exist $$m,n\in N'$$ with $$\phi(g)=m,\phi(h)=n$$. Now

\begin{align} \phi(gh)&=\phi(h)\phi(h)\\ &=mn\\ &\in N' \end{align}

since $$N'$$ is a subgroup of $$G'$$. Therefore, $$gh\in\phi^{-1}(N')$$.

## Step 3:

Let $$g\in G, x\in\phi^{-1}(N').$$ Then there is some $$r\in N'$$ with $$\phi(x)=r$$. We have

\begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)\phi(x)\phi(g)^{-1}\\ &=\phi(g)r\phi(g)^{-1}\\ &=g'rg'^{-1}, \end{align}

where $$g'=\phi(g)\in G'$$, which implies $$g'rg'^{-1}\in N'$$ as $$N'\unlhd G'$$. Hence $$gxg^{-1}\in\phi^{-1}(N')$$.