Group Direct Product of Cyclic Groups

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Theorem

Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively.

Then:

The group direct product $G \times H$ is cyclic

if and only if:

$n$ and $m$ are coprime, that is:
$n \perp m$


Corollary

Let $n_1, n_2, \ldots, n_s$ be a finite sequence of integers, all greater than $1$, such that for any pair of them $n_i$ and $n_j$, $n_i \perp n_j$.

Let $G_i$ be a cyclic group of order $n_i$ for each $i: 1 \le i \le s$.

Then $G_1 \times G_2 \times \cdots \times G_s$ is cyclic of order $n_1 n_2 \ldots n_s$.


Proof

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.


Necessary condition

Suppose:

$(1): \quad \order G = n, G = \gen x$
$(2): \quad \order H = m, H = \gen y$
$(3): \quad m \perp n$


Then:

\(\ds \order {\tuple {x, y} }\) \(=\) \(\ds k\)
\(\ds \leadsto \ \ \) \(\ds \tuple {x, y}^k\) \(=\) \(\ds e_{G \times H} = \tuple {x^k, y^k}\) Powers of Elements in Group Direct Product
\(\ds \leadsto \ \ \) \(\ds x^k\) \(=\) \(\ds e_G, y^k = e_H\)
\(\ds \leadsto \ \ \) \(\ds n\) \(\divides\) \(\ds k, m \divides k\)
\(\ds \leadsto \ \ \) \(\ds \lcm \set {n, m}\) \(\divides\) \(\ds k\)
\(\ds \leadsto \ \ \) \(\ds n m\) \(\divides\) \(\ds k\) as $n \perp m$


But then:

$\tuple {x, y}^{n m} = e_{G \times H} = \tuple {x^{n m}, y^{n m} }$

Thus:

$k \divides n m$

So:

$\order {\tuple {x, y} } = n m \implies \gen {\tuple {x, y} } = G \times H$

$\Box$


Sufficient condition

Suppose that $G \times H$ is cyclic.

Let $\tuple {x, y}$ be a generator of $G \times H$.

By Cardinality of Cartesian Product of Finite Sets the order of $G \times H$ is:

$\order G \cdot \order H = n m$

Therefore by Order of Cyclic Group equals Order of Generator:

$\order {\tuple {x, y} } = n m$

On the other hand, by Order of Group Element in Group Direct Product we have:

$\order {\tuple {x, y} } = \lcm \set {\tuple {\order x, \order y} }$


Next we claim that $x$ generates $G$.

Let $x' \in G$.

Then:

$\tuple {x', e_H} \in G \times H$

so there exists $k \in \N$ such that:

$\tuple {x, y}^k = \tuple {x^k, y^k} = \tuple {x', e_H}$

and therefore $x^k = x'$.

Thus the powers of $x$ generate the whole group $G$.

In the same way, it is seen that $y$ generates $H$.


Therefore by Order of Cyclic Group equals Order of Generator:

$\order x = n$
$\order y = m$

Thus we have that:

$n m = \order {\tuple {x, y} } = \lcm \set {n, m}$

Moreover by Product of GCD and LCM we have that:

$\lcm \set {n, m} = \dfrac {n m} {\gcd \set {n, m} }$

These two equalities imply that:

$\gcd \set {n, m} = 1$

That is, $n$ and $m$ are coprime.

$\blacksquare$


Examples

$C_2 \times C_2$ is not Cyclic

Let $C_2$ denote the cyclic group of order $2$.

The group direct product $C_2 \times C_2$ is not a cyclic group.


$C_2 \times C_3$ is Cyclic

Let $C_2$ denote the cyclic group of order $2$.

The group direct product $C_2 \times C_3$ is a cyclic group.


Sources

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