# Finite abelian groups - direct sum of cyclic subgroup

Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be written as the direct sum $G=\langle g \rangle \oplus H$ for some $H \leq G$ (subgroup of $G$). For a proof see this for example (page 2).

My question: Do we need that $G$ is a $p$-group or does it also work for arbitrary finite abelian groups?

I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example.

The result follows for arbitrary finite abelian groups from the $$p$$-group case.
Remember that a finite abelian group $$G$$ is the direct sum of its $$p$$-parts, $$G = G(p_1)\oplus \cdots\oplus G(p_n),$$ where $$p_1,\ldots,p_n$$ are the distinct primes that divide $$|G|$$, and $$G(q) = \{ a\in G\mid q^ma = 0 \text{ for some }m\geq 0\},\qquad q\text{ a prime.}$$
If $$a\in G$$ is of maximal order, then we can write $$a=a_1+a_2+\cdots+a_n$$, where $$a_i\in G(p_i)$$. Since $$a$$ is of maximal order in $$G$$, then $$a_i$$ is of maximal order in $$G(p_i)$$. By the $$p$$-group case, we can write $$G(p_i) = \langle a_i\rangle\oplus H_i$$ with $$H_i\leq G(p_i)$$. Then $$H_1+\cdots+H_n$$ is a subgroup of $$G$$, it is the internal direct sum of the $$H_i$$, and since $$G(p_i) =\langle a_i\rangle\oplus H_i$$, then \begin{align*} G &= G(p_1)\oplus \cdots \oplus G(p_n)\\ &= (\langle a_1\rangle\oplus H_1) \oplus \cdots \oplus (\langle a_n\rangle \oplus H_n)\\ &= (\langle a_1\rangle\oplus\cdots \oplus\langle a_n\rangle) \oplus (H_1\oplus\cdots\oplus H_n). \end{align*} To finish off, note that $$\langle a_1\rangle\oplus\cdots\oplus \langle a_n\rangle = \langle a\rangle$$ (e.g., by the Chinese Remainder Theorem).