# $\ker(\phi)$ is a normal subgroup.

Let $$G_1$$ and $$G_2$$ be groups and suppose $$\phi: G_1\mapsto G_2$$ is a homomorphism. Then $$\ker (\phi)\unlhd G_1$$.

Need some feedback and help proving this. I am still new to proofs, but here's my attempt.

Proof.

We want to show $$\ker (\phi)$$ is normal, therefore, we must show that for any $$h\in \ker (\phi)$$ and $$g\in G_1$$, then $$ghg^{-1}\in \ker (\phi)$$. Since $$h\in \ker (\phi)$$, then $$\phi(h)=1$$. Thus,

\begin{align} \phi(ghg^{-1})&=\phi(g)\phi(h)\phi(g^{-1})\\ &=\phi(g)\cdot 1\cdot\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1})\\ &=\phi(g\cdot g^{-1})\\ &=\phi(1)\\ &=1. \end{align}

Hence, $$ghg^{-1}\in \ker (\phi)$$, which implies $$\ker (\phi)$$ is a normal subgroup in $$G_1$$.

• This looks good. Apr 3 '19 at 2:36
• Essentially correct (just have to type the latex nicely). Just need to always remember this whole trick of using the group homomorphism. Apr 3 '19 at 2:41
• must show for any $h \in Ker(\phi)...$; also "Therefore, since $h\in Ker(\phi),$ then $\phi(h)=1$" belongs before $\phi(g)\phi(h)\phi(g^{-1})=\phi(g)\cdot1\cdot\phi(g^{-1})$ Apr 3 '19 at 2:45
• You can render $\ker \phi$ with the LaTeX $\ker \phi$. Apr 3 '19 at 2:51
• It would not hurt to emphasize that the last step holds because $$\phi(g)\phi(g^{-1})=\phi(g)\phi(g)^{-1}=1 \qquad\text{ or }\qquad \phi(g)\phi(g^{-1})=\phi(gg^{-1})=\phi(1)=1,$$ whichever you prefer. Apr 3 '19 at 3:03

\begin{align} \phi(ghg^{-1})&=\phi(g)\phi(h)\phi(g^{-1}) \\ &=\phi(g)\cdot 1\cdot \phi(g^{-1})\\ &=\phi(g)\phi(g)^{-1}\\ &=1. \end{align}