defining group homomorphisms in terms of generators

Question

Somebody who was more knowledgeable than I am was helping me with a problem (determining the possible homomorphisms) and kept using the following trick:

If we want to define a homomorphism $f$ between two cyclic groups, say $Z_3=<a>$ and $Z_{12}=<b>$ (these are cyclic groups written multiplicatively), it's sufficient to map the generator $a$ so that $|f(a)|$ divides $|a|$. (In this case, since $a$ has order $3$, there are three possible homomorphism (each determined by where $a$ is sent):

1. $a\mapsto b^0$, the identity
2. $a\mapsto b^4$, since $|b^4|=3$
3. $a\mapsto b^8$, since $|b^8|=3$

Why does this work? Any hints? Or where can I find this statement in a textbook (as a theorem or exercise)? What about the more general statement when the two groups are finitely generated?

Answer 1

There is something very general and somewhat deep behind this:

• $\Bbb Z$ has the universal property that a homomorphism $\Bbb Z\to H$ is "the same" as picking an element $h\in H$ and letting $1\mapsto h$.
• If $G$, $H$ are groups and $N\lhd G$, then a homomrphism $G/N\to H$ is "the same" as a homomorphism $G\to H$ with $N$ contained in its kernel.
• Assume $G$ is generated by a set $S$. Then a homomorphims $G\to H$ is trivial iff all elements of $S$ map to the neutral element.

So with this in mind, a homomorphism $\Bbb Z/n\Bbb Z\to \Bbb Z/m\Bbb Z$ requires us to find a homomorphism $\Bbb Z\to \Bbb Z/m\Bbb Z$ (i.e., pick $f(1)\in\Bbb Z/m\Bbb Z$) such that $n\mapsto 0\in\Bbb Z/m\Bbb Z$ ...