1
$\begingroup$

I am trying to go through the proof for the third isomorphism theorem for groups. Given $H,K$ are normal in $G$ and $K$ is a subgroup of $H$, a map $\phi:G/K \to G/H$ is defined by $\phi(Kg)=Hg$ and I would like to show that it is injective, meaning that $\forall g_1,g_2 \in G$, $Hg_1=Hg_2$ implies $Kg_1=Kg_2$.

Since $Hg_1=Hg_2$, we have $g_1g_2^{-1} \in H$ but how can i show that $g_1g_2^{-1} \in K$?

Any help would be greatly appreciated.

$\endgroup$
1
$\begingroup$

You can't prove the map is injective, because generally it isn't.

For instance, consider $G=\mathbb{Z}$, $K=\{0\}$ and $H=2\mathbb{Z}$.

What you have to prove is that the map is well defined, that is, if $Kg_1=Kg_2$, then $Hg_1=Hg_2$. This amounts to observing that $g_1g_2^{-1}\in K$ implies $g_1g_2^{-1}\in H$.

The map is injective only in the trivial case $H=K$: indeed its kernel is precisely $H/K$.

$\endgroup$
1
  • $\begingroup$ Thank you so much. It now makes sense why I couldnt prove it. Appreciate your counter example. $\endgroup$ – James Jun 27 '17 at 14:15

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.