# Group Direct Product of Cyclic Groups

## Theorem

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

Then their group direct product $G \times H$ is cyclic if and only if $g$ and $h$ are coprime, that is, $g \perp h$.

### 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 $n_j$, $n_1 \perp n_j$.

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 the order of $G \times H$ is:

- $\order G \cdot \order H = g h$

Therefore by Order of Cyclic Group equals Order of Generator:

- $\order {\tuple {x, y} } = g h$

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:

- $\order x = g$
- $\order y = h$

Thus we have that:

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

Moreover by Product of GCD and LCM we have that:

- $\lcm \set {g, h} = \dfrac {g h} {\gcd \set {g, h} }$

These two equalities imply that:

- $\gcd \set {g, h} = 1$

That is, $g$ and $h$ 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

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 43 \gamma$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $18$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $13$: Direct products: Proposition $13.1 \ (3)$