# Generators of a product of finite abelian groups

Let $n_1,...,n_r$ be positive integers. Consider the group $$G={\bf Z}/n_1 {\bf Z} \times \cdots\times {\bf Z}/n_r {\bf Z}$$ When does a given element $(k_1,\cdots,k_r)$ generate $G$?

Obviously $G$ must be cyclic. For instance if the $n_i$ are pairwise coprime and, for each $i$, $(n_i,k_i)=1$, then $(k_1,\cdots,k_r)$ generates $G$.

Equally obviously, this condition is too strong. Can one weaken it? For instance is it enough to take $(n_1,\cdots,n_r)=1$?

Edit: many good answers, but I must only accept one.

No, you need the $n_{i}$ to be pairwise coprime. If for instance $\gcd(n_{1}, n_{2}) > 1$, then the subgroup $\mathbf{Z}/n_{1} \mathbf{Z} \times \mathbf{Z}/n_{2} \mathbf{Z}$ is not cyclic, so $G$ itself is not cyclic.

The reason is that $\mathbf{Z}/n_{1} \mathbf{Z} \times \mathbf{Z}/n_{2} \mathbf{Z}$ has order $n_{1} n_{2}$, but exponent the lcm of $n_{1}, n_{2}$, which is $$\frac{n_{1} n_{2}}{\gcd(n_{1}, n_{2})} < n_{1} n_{2}.$$

• Yes thank you! And I believe the same kind of argument works to show that the assumption on the k_i is also necessary. – Lepanais Oct 7 '14 at 20:32

There is no weaker condition.

The chinese remainder theorem states that $\mathbb{Z}_n \simeq \mathbb{Z}_{j} \times \mathbb{Z}_k$ if and only if $j$ and $k$ are coprime and $n = jk$. Furthermore, the group $G = \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \cdots \times \mathbb{Z}_{n_r}$ is cyclic if and only if it is isomorphic to some $\mathbb{Z}_n$. Thus, if $G$ is cyclic, $\prod_{i=1}^r n_i = n$ and $(n_i,n_j) = 1$ for $i \neq j$.

Furthermore, if $G \simeq \mathbb{Z}_n$, then the set of generators of $G$ is precisely the set that you described. Suppose that $S$ is the set of all $r$ tuples $(k_1,\dots,k_r) \in G$ with $(k_i,n_i) = 1$. Let $\mathbf{k} = (k_1,\dots,k_r) \in S$. Then $(o(k_i),o(k_j)) = (n_i,n_j) = 1$, so $$o(k_1,\dots,k_r) = o(k_1)\cdots o(k_r) = n_1 \cdots n_r = n.$$ Therefore $\mathbf{k}$ is a generator of $G$. The size of $S$ is equal to $\phi(n_1)\cdots \phi(n_r) = \phi(n)$ which is the number of generators of $\mathbb{Z}_n \simeq G$, thus $S$ is the set of all generators of $G$.

You can find the order of the element $(k_1, \ldots, k_r)$, it will be the LCM of the order of the $k_i$'s. But $\text{ord}(k_i) \mid n_i$ and so $$\text{ord}(k_1, \ldots, k_r) \mid \text{LCM}(n_1, \ldots, n_r)$$

This equals $n_1\ldots n_r$ iff $n_i$, $n_j$ relatively prime for $i \ne j$ and $\text{ord}\,k_i = n_i$ for all $i$.

Well any cyclic group is isomorphic to $\mathbb{Z}/{nZ}$, so If $G={\bf Z}/n_1 {\bf Z} \times \cdots\times {\bf Z}/n_r {\bf Z}$ is cyclic then $G \cong \mathbb{Z}/{nZ}$.

So ${\bf Z}/n_1 {\bf Z} \times \cdots\times {\bf Z}/n_r {\bf Z}$ is cyclic iff $(n_i,n_j)=1$, which is a clear consequence of this corollary from J.A.Gallian...