# The equality condition for cosets of a subgroup.

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.

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 $H/K$.