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.


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 $G_1$.

  • 7
    $\begingroup$ This looks good. $\endgroup$
    – Gary Moon
    Apr 3 '19 at 2:36
  • $\begingroup$ Essentially correct (just have to type the latex nicely). Just need to always remember this whole trick of using the group homomorphism. $\endgroup$
    – Everiana
    Apr 3 '19 at 2:41
  • 2
    $\begingroup$ must show for any $h \in Ker(\phi)...$; also "Therefore, since $h\in Ker(\phi), $ then $\phi(h)=1$" belongs before $\phi(g)\phi(h)\phi(g^{-1})=\phi(g)\cdot1\cdot\phi(g^{-1})$ $\endgroup$ Apr 3 '19 at 2:45
  • 1
    $\begingroup$ You can render $\ker \phi$ with the LaTeX $\ker \phi$. $\endgroup$ Apr 3 '19 at 2:51
  • 2
    $\begingroup$ 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. $\endgroup$
    – Servaes
    Apr 3 '19 at 3:03

You're right.

An alternative method, following yours right until the end, goes like this:

\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}

The conclusion is, of course, the same.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.