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:
- The group direct product $G \times H$ is cyclic
- $n$ 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
- 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 $13.1 \ (3)$