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)

l c m (r, s)

Proof
Let $l = l c m (r, s)$ . Then, $l = r a = s b$ for some $r, b \in Z$ . Observe that

\begin{aligned} (x, y)^{l} & = (x^{l}, y^{l}) = (x^{r a}, y^{s b}) = (e_{1}, e_{2}) \end{aligned}

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{aligned} (x, y)^{l^{'}} & = (x^{l^{'}}, y^{l^{'}}) = (e_{1}, e_{2}) \end{aligned}

But since $| x | = r$ and $| y | = s$ , then $r | l^{'}$ and $s | l^{'}$ . But this means that $l c m (r, s) = l | l^{'}$ . Since $l^{'} | l$ and $l | l^{'}$ , then $l = l^{'}$ as we wanted to show.

Example 2

Proposition

Z_{m} \times Z_{n} ≅ Z_{m n}

if and only if

g c d (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
$⟶:$ Suppose $Z_{m} \times Z_{n} ≅ Z_{m n}$ . Suppose for the sake of contradiction that $g c d (m, n) = d > 1$ . Now, take an arbitrary element $(a, b) \in Z_{m} \times Z_{n}$ . Add $(a, b) + . . . + (a, b)$ $\frac{m n}{d}$ times. We will see that

\begin{aligned} (a \frac{m n}{d}, b \frac{m n}{d}) & = (a \frac{n}{d} m, b \frac{m}{d} n) \\ = ([0]_{m}, [0]_{n}) \end{aligned}

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 $(a, b) \leq \frac{m n}{d} < m n$ . So the order of any element $(a, b) \in Z_{m} \times Z_{n}$ is strictly less than $m n$ . This means that no element of $Z_{m} \times Z_{n}$ can be a generator of the group. So $Z_{m} \times Z_{n}$ is not cyclic. This is a contradiction because $Z_{m} \times Z_{n}$ isomorphic to $Z_{m n}$ and so it is cyclic.

$⟵:$ Now suppose that $g c d (m, n) = 1$ . We want to show that $Z_{m} \times Z_{n}$ is isomorphic to $Z_{m n}$ . Notice that the order of the element 1 has order $m$ in $Z_{m}$ and the order of the element 1 has order $n$ in $Z_{n}$ . This tells us that the order of $(1, 1) = l c m (m, n) \frac{m n}{g c d (m, n)} \frac{m n}{1} = m n$ . So $(1, 1)$ has order $m n$ which means it generates the group. So it must be cyclic. So $Z_{m} \times Z_{n} = ⟨ (1, 1) ⟩$ . We know all cyclic groups that have the same order are isomorphic. So $Z_{m} \times Z_{n} ≅ Z_{m n}$ .

References

The order of an element in a direct group is the lcm of both elements orders