Homomorphism between a group of exponent $m$ and $\mathbb{Z}/m\mathbb{Z}$

Question

Let $G$ be an abelian group of exponent $m$, where $m\in\mathbb{N}$. Is there always a nontrivial group homomorphism between $G$ and $\mathbb{Z}/m\mathbb{Z}$ ?

For example, if we have $G=\mathbb{Z}/m\mathbb{Z}$ the map $f(x)=x$ is one such homomorphism. If we have $G= \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}$, then $(x,y)\mapsto(x+y)\text{ ( mod }m)$ is another such map.

In the above examples, we have a precise description of $G$. What if we don't know what $G$ looks like? Can we construct a nontrivial homomorphism from $G$ to $\mathbb{Z}/m\mathbb{Z}$ ?

Answer 1

If you can use the structure theorems for finite abelian groups, then it is easy because

In the invariant factor decomposition, the largest factor has order equal to the exponent of $G$.

If you cannot use the structure theorems, see Chapter 2 of the book The Theory of Finite Groups: An Introduction, by Kurzweil and Stellmacher. It contains this elementary theorem, which does not depend on the structure theorem of finite abelian groups:

If $G$ is a finite abelian group and $U$ is a cyclic subgroup of maximal order $m$ in $G$, then there $o(y)$ divides $m$ for all $y \in G$.

From this, it follows that $m=\exp G$ because $o(y) \mid m$ for all $y$ implies $\exp G \mid m$, and $U=\langle u \rangle$ implies $m=ord(u) \mid \exp G$.

The next theorem in that book is:

If $G$ is a finite abelian group and $U$ is a cyclic subgroup of maximal order in $G$, then there exists a complement $V$ of $U$ in $G$, that is, $G=UV$ and $U \cap V =1$. In particular, $G \cong U \times V$.

This gives you a surjective group homomorphism $G \to U \cong\mathbb{Z}/m\mathbb{Z}$.

Answer 2

The First Prüfer Theorem: An abelian group of bounded exponent is isomorphic to a direct sum of cyclic groups.

If $G$ is an abelian group of exponent $m\in\mathbb{N}$, then $G$ is a direct sum of cyclic groups according to the theorem above, from which we may assume that $$G\cong\bigoplus_{p}\,\bigoplus_{k=1}^{l(p)}\,\left(\mathbb{Z}/p^k\mathbb{Z}\right)^{\oplus \alpha(p,k)}\,,$$ where $p$ runs over the prime natural numbers dividing $m$, $l(p)$ is the highest exponent of $p$ that divides $m$, and $\alpha(p,k)$ is a cardinal number for each $k=1,2,\ldots,l(p)$. Note that $\alpha\big(p,l(p)\big)>0$ for all primes $p$ dividing $m$ (or $G$ would be of a smaller exponent). Let $$H:=\left(\bigoplus_{p}\,\left(\mathbb{Z}/p^{l(p)}\mathbb{Z}\right)^{\alpha\big(p,l(p)\big)-1}\right)\oplus\left(\bigoplus_{p}\,\bigoplus_{k=1}^{l(p)-1}\,\left(\mathbb{Z}/p^k\mathbb{Z}\right)^{\oplus \alpha(p,k)}\right)\,,$$ where we interpret $\alpha\big(p,l(p)\big)-1$ as $\alpha\big(p,l(p)\big)$ if $\alpha\big(p,l(p)\big)$ is an infinite cardinal number. Then, $$G\cong \left(\bigoplus_{p}\,\left(\mathbb{Z}/p^{l(p)}\mathbb{Z}\right)\right)\oplus H\cong (\mathbb{Z}/m\mathbb{Z})\oplus H\,.$$ Hence, we have a nontrivial group homomorphism $$G\overset{\cong}{\rightarrow} (\mathbb{Z}/m\mathbb{Z})\oplus H\overset{\pi}{\twoheadrightarrow} \mathbb{Z}/m\mathbb{Z}\,,$$ where $\pi$ is the canonical projection.