Complements of a maximal cyclic subgroup in C_p times C_p need not be unique

Given: The objects and hypotheses in the example.

Let $G$ be a finite abelian $p$-group and let $a\in G$ have maximal element order. Then there is a subgroup $H\le G$ such that $G=\langle a\rangle\oplus H.$ (A maximal-order cyclic subgroup splits off a finite abelian p-group).

Let $G$ be a group and let $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$; in additive notation one writes $G=H\oplus K$. Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$. The following are equivalent: the $N_i$ form an internal direct product of $G$; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).