For cyclic groups $G, H$, find a necessary and sufficient condition when $H$ is a homomorphic image of $G$

Question

I believe the condition to be 'iff there is a normal subgroup of $G$, $N$ such that $H \cong G/N$

I have found an answer that says to use:

a cyclic group is determined by its cardinality, we can give a straighter answer:

1. Each cyclic group is a homomorphic image of $\Bbb Z$, the infinite cyclic group.
2. If $n|m$, we have a surjective homomorphism $\Bbb Z_m\to\Bbb Z_n$.
3. If $\Bbb Z_n\cong\Bbb Z_m/N$ for any (normal) subgroup $N$ of $\Bbb Z_m$, we have $n|m$.

but I could use some help filling in the details.

Thanks

Answer 1

The condition you state is tautological. Let's see some other facts.

• If $G$ is an infinite cyclic group, then every cyclic group is a homomorphic image of $G$.

• If $H$ is infinite cyclic, then $G$ must be infinite cyclic as well.

• If both $H$ and $G$ are finite, then the necessary and sufficient condition is that $|H|$ is a divisor of $|G|$.

In the finite case the condition is necessary by the homomorphism theorem and Lagrange's theorem.

Why is the condition also sufficient?

In order to prove sufficiency, it is not restrictive to assume $G=\mathbb{Z}/n\mathbb{Z}$. Let $m=|H|$ be a divisor of $n=|G|$. Then $G$ has a unique subgroup $N$ of order $n/m$ and $G/N$ is a cyclic group with $m$ elements.

Answer 2

It's easier to see "how to get there" when you already know the answer. But let's see what happens when we have no idea of what we're going to find out.

So we'll start with a cyclic group $G = \langle x\rangle$.

If we have a homomorphism $\phi: G \to L$, we know that $\phi(G) = \{\phi(x)^k: k \in \Bbb Z_0^+\}$.

In short, $\phi(G) = \langle \phi(x)\rangle$, which is cyclic (note that this depends in an essential way on $\phi$ being a homomorphism, so that $\phi(x^{k+l} = \phi(x^kx^l) = \phi(x^k)\phi(x^l) = (\phi(x))^k(\phi(x))^l = (\phi(x)^{k+l}$.

Clearly, then, $\phi(G)$ has order $|\phi(x)|$.

If $G \cong \Bbb Z$ (that is, it is infinite cyclic), we can have two outcomes:

$\phi(x)$ has finite order, or infinite order. Let's look at the second one, first. In this case $x \mapsto \phi(x)$ is an isomorphism of $G$ with $\phi(G)$ (why?).

In the first case when $\phi(x)$ has finite order, by the Fundamental Isomorphism Theorem, we have:

$\phi(G) \cong G/\text{ker }\phi$.

If $m = |\phi(x)|$, then $\text{ker }\phi = \langle x^m\rangle$ (I leave the details of this to you). Thus $\phi(G) = G/mG \cong \Bbb Z/m\Bbb Z$.

So we can find cyclic groups of any order as homomorphic images of any infinite cyclic group (just pick any non-negative integer $m$, and form $G/mG$).

When $|G| = n$, where $n$ is a non-negative integer, it's a bit more interesting.

As before, we have $\phi(G) = G/\text{ker }\phi$ (and $\phi(G)$ is necessarily finite, because homomorphisms are functions, and functions preserve finiteness of sets). It is also immediately clear that $\phi(G)$ will be cyclic.

Again, as before, we have $|\text{ker }\phi| = |\phi(x)| = m$.

Since $|\phi(G)| = |G/\text{ker }\phi| = |G|/|\text{ker }\phi| = n/m$ is a (positive) integer, it follows that it is necessary that $m\mid n$.

On the other hand, suppose $H = \langle y\rangle$ with $|y| = m$, where $m\mid n$.

Define $\phi: G \to H$, by $\phi(x^k) = y^k$. It is not immediately clear this is a well-defined function, for if $x^k = x^l$, it is not clear that $y^k = y^l$.

However, we know that $x^k = x^l$ only when $k \equiv l \text{ mod }n$, that is $k - l = nt$ for some integer $t$. Since $m \mid n$, that is, $n = dm$ for some positive integer $d$, we have $k - l = nt = m(dt)$, so that $k \equiv l\text{ mod }m$, and thus $y^k = y^l$.

The verification that $\phi$ thus defined, is a homomorphism is straight-forward, and left to you. It should also not be too difficult to show that $\phi$ is surjective, and thus $H = \phi(G)$, which shows sufficiency.