# For a normal subgroup, the factor group is isomorphic to the range mod the image of a surjective homomorphism.

Suppose that $G_{1} \trianglelefteq G$ and $\phi \colon G\rightarrow H$ a surjective homomorphism. I need to prove (or disprove, but somehow I get the feeling that the theorem is true) that $G/G_{1}\cong H/\phi(G_{1})$.

I've been trying to think how the isomorphism theorems bear on this, which doesn't seem obvious since I don't know the kernel of the mapping so the first theorem seems inappropriate and the other two don't seem to say anything directly about homomorphisms. If I try to use them anyway, I end up saying this:

From the first homomorphism theorem $G/\ker\phi \cong \phi(G)$. If I restrict the map to $G_{1}$ then I get $G_{1}/\ker\phi \cong \phi(G_{1})$. If I'm to use the second homomorphism theorem then I have to pick something to be the "main group" and then two subgroups. I could choose $G$ with subgroups $G_{1}$ and $\ker \phi$. Then I'd have $G_{1}/(G_{1}\cap \ker \phi) \cong G_{1}\ker\phi / \ker\phi$. That doesn't seem to help. Alternately I could choose the group $H$ but I don't see more than one natural choice for a subgroup of that.

In the third isomorphism theorem I don't see how I can apply it here at all since I don't think I have any reason to believe that $\ker\phi \subset G_{1}$.

[As a further thought, perhaps the theorem is false: It sounds like it wouldn't be hard to think of an example where $\phi(G_{1})=H$ in which case $H/\phi(G_{1})$ is the trivial group, while it wouldn't necessarily be true that $G/G_{1}$ is. But this seems to go against the hint that I was given in the problem, that I should use the isomorphism theorems.]

This is not true without some more restrictions on $\phi$. Let $G$ be $\Bbb Z^3$, $G_1$ the subgroup generated by $(1,0,0)$, $H=\Bbb Z$, and $\phi$ projection onto the last coordinate.
$\Bbb Z^3/\Bbb Z \not\approx \Bbb Z/(0)$
If you demand that $\phi^{-1}\circ\phi G_1 = G_1$, then you might be on to something.
• Presumably $H$ is $\mathbb{Z}$? Oct 16 '14 at 22:53
• Yeah, with $H$ as $\mathbb{Z}$ you're right. In spite of the example, should I be seeing any relevance to the isomorphism theorems? Oct 16 '14 at 22:56