# Kernel is a normal subgroup of the preimage

I am curious if my logic is correct. I am a little stuck on part 2 and 3.

Let $G$ be a group and $H\unlhd G$. Define $\varphi:G\rightarrow G/H$. Let $K\leqslant G/H$. I have already shown that $\varphi^{-1}(K)\leqslant G$.

My end goal is to show $\ker(\varphi)\unlhd \varphi^{-1}(K)$.

Here is my proof:

1. We want to show that $\ker(\varphi)\subseteq \varphi^{-1}(K)$. Let $q\in \ker(\varphi)=\{q\in G|\varphi(q)=1\}$. Notice $\varphi(q)=1\in K$. Thus $\ker(\varphi)\subseteq \varphi^{-1}(K)$.
2. We will show that $\ker(\varphi)\leqslant \varphi^{-1}(K)$. Notice $1\in \ker(\varphi)$, therefore it is nonempty. Let $x,y\in \ker(\varphi)$. Then $\varphi(x)=1$ and $\varphi(y)=1$. Consider \begin{eqnarray} \varphi(xy^{-1})&=&\varphi(x)\varphi(y^{-1})\\ &=& \varphi(y^{-1}) \end{eqnarray} ????????
3. We will show that $\ker(\varphi)\unlhd \varphi^{-1}(K)$. Want to show that $\forall r\in \varphi^{-1}(K)$ that $r\ker(\varphi)r^{-1}=\ker(\varphi)$.
• if $\varphi(x) = 1$ then $1 = \varphi(1) = \varphi(x^{-1}x) = \varphi(x) \varphi(x^{-1}) = \varphi(x^{-1})$ so $x^{-1} \in ker(\varphi)$. If also $\varphi(y) = 1$ then $1 = \varphi(y) = \varphi(x) \varphi(y) = \varphi(xy)$ so $xy \in ker(\varphi)$ Sep 19, 2016 at 4:06
• @user1952009 Thank you! I ended up using a variation of this approach to show part 2. $G$ is finite (didn't say in question) so I just showed that multiplication was closed under it. I have just got in a habit of the one-step subgroup test. The $xx^{-1}$ is a very handy and clever trick. Appreciate you reminding me of it. Sep 19, 2016 at 5:00

Part 2: $\varphi(y^{-1})=\varphi(y))^{-1}=1$, since $\varphi(y)\varphi(y^{-1})=\varphi(yy^{-1})=\varphi(1)=1$.
Part 3: For any $x\in\ker\varphi$, $\varphi(rxr^{-1})=\varphi(r)\varphi(x)\varphi(r^{-1})=\varphi(r)1\varphi(r^{-1})=1$. So, $rxr^{-1}\in \ker\varphi$.