On the Orders of Elements in Direct Product Groups

Proposition 1. Let $a$ be an element of order $m$ in a group $G$, and let $b$ be an element of order $n$ in a group $G$. Then the element $(a,b)$ in the direct product $G \times G$ has order $[m,n]$.

Proof: Let the order of $(a,b)$ be $q$. On the one hand,
$$
(a,b)^{[m,n]} = (a^{[m,n]}, b^{[m,n]}) = (e,e).
$$
Hence
$$
q \mid [m,n].
$$

On the other hand, since
$$
(a^q,b^q) = (a,b)^q = (e,e),
$$
we have
$$
a^q = e \quad \text{and} \quad b^q = e.
$$
Thus
$$
m \mid q, \qquad n \mid q,
$$
and therefore
$$
[m,n] \mid q.
$$

Combining the two divisibility relations, we obtain
$$
q = [m,n].
$$

Proposition 2. The group $\mathbb{Z}_5 \times \mathbb{Z}_7$ is cyclic.

Proof: The element $\overline{1} \in \mathbb{Z}_5$ has order $5$, and the element $\overline{1} \in \mathbb{Z}_7$ has order $7$. Therefore the element
$$
(\overline{1},\overline{1}) \in \mathbb{Z}_5 \times \mathbb{Z}_7
$$
has order
$$
[5,7] = 35.
$$
Since
$$
|\mathbb{Z}_5 \times \mathbb{Z}_7| = 5 \cdot 7 = 35,
$$
it follows that $\mathbb{Z}_5 \times \mathbb{Z}_7$ contains an element of order equal to the order of the whole group. Hence $\mathbb{Z}_5 \times \mathbb{Z}_7$ is cyclic.

Proposition 3. The group $\mathbb{Z}_2 \times \mathbb{Z}_3$ is isomorphic to $\mathbb{Z}_6$.

Proof: The groups $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are cyclic of orders $2$ and $3$, respectively, and
$$
(2,3) = 1.
$$
Hence, by Proposition 1, the group $\mathbb{Z}_2 \times \mathbb{Z}_3$ contains an element of order
$$
[2,3] = 6.
$$
Since
$$
|\mathbb{Z}_2 \times \mathbb{Z}_3| = 6,
$$
it follows that $\mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic of order $6$. Therefore
$$
\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6.
$$

Proposition 4. The groups $\mathbb{Z}_2 \times \mathbb{Z}_2$ and $\mathbb{Z}_4$ are not isomorphic.

Proof: Every element of $\mathbb{Z}_2$ has order $1$ or $2$. Therefore every element of $\mathbb{Z}_2 \times \mathbb{Z}_2$ has order $1$ or $2$. In particular, $\mathbb{Z}_2 \times \mathbb{Z}_2$ has no element of order $4$, so it is not cyclic.

On the other hand, $\mathbb{Z}_4$ is cyclic. Hence the two groups cannot be isomorphic.

Proposition 5. The groups $\mathbb{Z}_8 \times \mathbb{Z}_2$ and $\mathbb{Z}_4 \times \mathbb{Z}_4$ are not isomorphic.

Proof: In $\mathbb{Z}_8$, the possible element orders are $1,2,4,8$, while in $\mathbb{Z}_2$, the possible element orders are $1,2$. Hence the direct product $\mathbb{Z}_8 \times \mathbb{Z}_2$ contains elements of order $8$, and the largest possible order of an element is $8$.

By contrast, every element of $\mathbb{Z}_4$ has order $1,2$, or $4$. Therefore every element of $\mathbb{Z}_4 \times \mathbb{Z}_4$ has order $1,2$, or $4$, and in particular there is no element of order $8$.

Thus $\mathbb{Z}_8 \times \mathbb{Z}_2$ and $\mathbb{Z}_4 \times \mathbb{Z}_4$ cannot be isomorphic.

The cover image of this article was taken at Lake Baikal in Russia.