The Order of an Element in an External Direct Product of Two Groups

Recall from The External Direct Product of Two Groups page that if $(G, *)$ and $(H, +)$ are two groups then the external direct product of these groups is $(G \times H, \cdot)$ where $\cdot$ is the operation defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by:

(1)

\begin{align} \quad (g_1, h_1) \cdot (g_2, h_2) = (g_1 * g_2, h_1 + h_2) \end{align}

We will now look at a nice theorem which tells us the order of each element $(g, h) \in G \times H$.

Theorem 1: Let $(G, *)$ and $(H, +)$ be two groups and let $(G \times H, \cdot)$ be their external direct product. If $g$ has order $\mid g \mid = m$ and $h$ has order $\mid h \mid = n$ then the order of $(g, h)$ is $\mathrm{lcm} (\mid g \mid, \mid h \mid) = \mathrm{lcm} (m, n)$.

• Proof: Let $e_1$ be the identity in $G$ and let $e_2$ be the identity in $H$. Then $e = (e_1, e_2)$ is the identity in $G \times H$. Note that if $L = \mathrm{lcm} (m, n)$ then $L$ is the smallest positive integer such that $m \mid L$ and $n \mid L$ and so there exists $k_1, k_2 \in \mathbb{Z}$ such that $mk_1 = L$ and $nk_2 = L$, so:

(2)

\begin{align} \quad (g, h)^L = (g^L, h^L) = (g^{mk_1}, h^{nk_2}) = ((g^m)^{k_1}, (h^n)^{k_2}) = (e_1^{k_1}, e_2^{k_2}) = (e_1, e_2) = e \end{align}

• Now suppose that there exists a smaller number, $M < L$ such that $(g, h)^M = e$. Then $(g^M, h^M) = (e_1, e_2)$ so $g^M = e_1$ and $h^M = e_2$. So we must have that $m \mid M$ and $n \mid M$. But since $L = \mathrm{lcm} (m, n)$ we see that $M \geq L$ which is a contradiction.

• Therefore the order of $(g, h)$ is $L = \mathrm{lcm} (\mid g \mid, \mid h \mid)$. $\blacksquare$