Group theory : order of groups For $(g, h)$ in the product group $G \times H$, prove that $\text{ord}(g, h) = \text{lcm}(\text{ord}\ g, \text{ord}\ h)$,
I was also wondering how would you interpret this when $g$ and $h$ have infinite orders?
 A: The precise statement is

If $g \in G$ and $h \in H$ have finite order, then so does $(g,h) \in G \times H$ and $\operatorname{ord}(g, h) = \operatorname{lcm}(\operatorname{ord} (g), \operatorname{ord} (h))$.

You can extend this to elements of infinite order if you extend the definition of $\operatorname{lcm}$ to $\operatorname{lcm}(n,\infty)=\infty$. With this definition, the statement above holds for elements of infinite order.
Alternatively, we can state this as

If $g \in G$ and $h \in H$, then $(g,h) \in G \times H$ has finite order iff $g$ and $h$ have finite order. In this case, $\operatorname{ord}(g, h) = \operatorname{lcm}(\operatorname{ord} (g), \operatorname{ord} (h))$.

A: Say $g\in G$ and $h \in H$ have finite order. Then, there exist $k,l$ so that $g^k = e = h^l$. Then, $(g,h)^{kl} = ((g^k)^l,(h^l)^k) = (e,e)$, so $(g,h)$ has finite order. 
By definition of order of an element, $\text{ord}(g,h)$ is the smallest number $n$ with $(g,h)^n = (g^n,h^n) = (e,e)$. Then $g^n=e=h^n$, so $n$ must be a multiple of $\text{ord}(g)$ and a multiple of $\text{ord}(h)$. Then, $n$ is the smallest number which is a multiple of $\text{ord}(g)$ and $\text{ord}(h)$, ie the least common multiple of $\text{ord}(g)$ and $\text{ord}(h)$.
A: For the finite case: 
Given $G$ and $H$ groups, consider $G \times H$, the product group of $G$ and $H$. Then given $(g,h)$ $\in G \times H$ we seek a least positive $k$ such that $(g,h)$$^k$ = $(1,1)$. 
But $(g,h)$$^k$ is defined as powers of each component i.e: 
$$ (g^k,h^k) $$
If we set $k$ = $lcm(|g|,|h|)$, then $k$ certainly sends the above value to $(1,1)$ as both components are sent to a power that divides their individual order. This is the least such value as let $s \lt k$.But $(g^s,h^s) = (1,1) $ if and only if |g| divides |s| AND |h| divides |s|. To see why this is true see the below ( I only do $1$ direction, the other is even quicker). 
Suppose |g| = $w$ and $w$ does not divide $s$, and let $g^s$ = $1$. Then $s \gt$ $w$ since $w$ is the least positive integer power of $g$ that sends $g$ to $1$. So then note that $g^s$ = $g^a$ where $a$ = $s$ mod $w$ since if $s$ = $a$ + $bw$, then $g^{a + bw}$ = $(g^a)(g^{bw})$ = $(g^a)(1)$. But then we have $g^a$ = $g^w$ for $a$ $\lt$ $w$ which contradicts $w$ being the least such positive power of $g$ such that $g^w$ = 1. 
Hence $s$ must divide $w$ = |g|. Similarly $s$ must divide |h|.   
