Example 1

Proposition
Suppose \((x,y) \in G_1 \times G_2\) and that the order of the elements are \(|x| = r\) and \(|y| = s\). Then the order of \((x,y)\) is \(lcm(r,s)\)


Proof
Let \(l = lcm(r,s)\). Then, \(l = ra = sb\) for some \(r, b \in \mathbf{Z}\). Observe that

$$ \begin{align*} (x,y)^{l} &= (x^l, y^l) = (x^{ra}, y^{sb}) = (e_1, e_2) \end{align*} $$

From this we see that the order of \((x,y)\) must divide \(l\). If we let \(l' = |(x,y)|\), then \(l' \ | \ l\). Now, observe that

$$ \begin{align*} (x,y)^{l'} &= (x^{l'}, y^{l'}) = (e_1, e_2) \end{align*} $$

But since \(|x| = r\) and \(|y| = s\), then \(r \ | \ l'\) and \(s \ | \ l'\). But this means that \(lcm(r,s) = l \ | \ l'\). Since \(l' \ | \ l\) and \(l \ | \ l'\), then \(l = l'\) as we wanted to show.


Example 2

Proposition
\(\mathbf{Z}_m \times \mathbf{Z}_n \cong \mathbf{Z}_{mn}\) if and only if \(gcd(m,n) = 1\)


This says if you take the product of two cyclic groups, then their product is cyclic if and only if \(m\) and \(n\) are relatively prime.

Proof
\(\longrightarrow:\) Suppose \(\mathbf{Z}_m \times \mathbf{Z}_n \cong \mathbf{Z}_{mn}\). Suppose for the sake of contradiction that \(gcd(m,n) = d > 1\). Now, take an arbitrary element \((a,b) \in \mathbf{Z}_m \times \mathbf{Z}_n\). Add \((a,b)+...+(a,b)\) \(\frac{mn}{d}\) times. We will see that

$$ \begin{align*} (a\frac{mn}{d},b\frac{mn}{d}) &= (a\frac{n}{d}m,b\frac{m}{d}n) \\ &= ([0]_m, [0]_n) \end{align*} $$

Notice here that both \(\frac{n}{d}\) and \(\frac{m}{d}\) are whole numbers since \(d\) is a divisor of both \(m\) and \(n\). So the first entry is a multiple of \(m\) while the second entry is a multiple of \(n\). This implies that the order of \((a,b) \leq \frac{mn}{d} < mn\). So the order of any element \((a,b) \in \mathbf{Z}_m \times \mathbf{Z}_n\) is strictly less than \(mn\). This means that no element of \(\mathbf{Z}_m \times \mathbf{Z}_n\) can be a generator of the group. So \(\mathbf{Z}_m \times \mathbf{Z}_n\) is not cyclic. This is a contradiction because \(\mathbf{Z}_m \times \mathbf{Z}_n\) isomorphic to \(\mathbf{Z}_{mn}\) and so it is cyclic.

\(\longleftarrow:\) Now suppose that \(gcd(m,n) = 1\). We want to show that \(\mathbf{Z}_m \times \mathbf{Z}_n\) is isomorphic to \(\mathbf{Z}_{mn}\). Notice that the order of the element 1 has order \(m\) in \(\mathbf{Z}_m\) and the order of the element 1 has order \(n\) in \(\mathbf{Z}_n\). This tells us that the order of \((1,1) = lcm(m,n) \frac{mn}{gcd(m,n)} \frac{mn}{1} = mn\). So \((1,1)\) has order \(mn\) which means it generates the group. So it must be cyclic. So \(\mathbf{Z}_m \times \mathbf{Z}_n = \langle (1,1) \rangle\). We know all cyclic groups that have the same order are isomorphic. So \(\mathbf{Z}_m \times \mathbf{Z}_n \cong \mathbf{Z}_{mn}\).


References