Group homomorphism induced by another homomorphism I need to prove a theorem that if $\phi: G\to H$ is some homomorphism, $M\leq H$ $\phi^{-1}(M) := \{g\in G:\phi(g)\in K\}$, $M \trianglelefteq H$, then $\phi$ induces an injective homomorphism $G/\phi^{-1}(M)\to H/M$.
(I've already proved that $\phi^{-1}(M) \leq G$ and $\phi^{-1}(M)\trianglelefteq G$).
I'm willing to prove the theorem stated above, but I don't know what it means when a homomorphism induces another homomorphism. Didn't find anything about this subject in Abstract Algebra by Dummit and Foote (looked in the Glossary as well). I'm wondering if someone could please give me a good recommendation of what to read on this subject (I'm a beginner in Group Theory).
 A: In math, inducing means to create (by way of some other operation). In your case, the fact that you have a homomorphism from $G$ to $H$ gives rise to a homomorphism (you don't know what it is a priori) between $G/\phi^{-1}(M)$ and $H/M$.
I'll give you a hint/primer so that you can better understand exactly what is going on. Induced maps/morphisms/etc can be quite confusing at first since they take a trained eye to spot and understand.
First you have to show that there is some function between $G/\phi^{-1}(M)$ and $H/M$ before you can even show it is a homomorphism. The obvious question is: what is that map? Well $G/\phi^{-1}(M)$ is a set of equivalence classes of the form $g.\phi^{-1}(M)$. Likewise, $H/M$ is a set of equivalence classes of the form $h.M$ which obviously contains elements of the form $\phi(g).M$.
What is a nice map between these two sets? Using $\phi$, the reasonable map would be to send $g.\phi^{-1}(M)$ to $\phi(g).M$ since it seems to be the only way to truly pair things up between these two sets. This is where the "inducing" comes from. $\phi$ initially acted on $G$, but now it can be sort of thought of as acting on $G/\phi^{-1}(M)$.
Before you can prove that the above "function" from $G/\phi^{-1}(M)$ to $H/M$ is a homomorphism, you have to show that it is well-defined, meaning that it is constant on equivalence classes. If $g.\phi^{-1}(M) = g'.\phi^{-1}(M)$, then $f$ had better send them to the same thing in $H/M$ if it is actually a function - otherwise it sends the same element to two different images.
A: Hint:
$\phi^{-1}(M)$ is the kernel of the composition: $G\to H\to H/M$, where the last map is the canonical homomorphism $h\mapsto hM$.
