The Abelian Group of the Product of Abelian Subgroups

# The Abelian Group of the Product of Abelian Subgroups

 Definition: Let $(G, \cdot)$ be an abelian group and let $(S, \cdot)$ and $(T, \cdot)$ be (abelian) subgroups of $(G, \cdot)$. Then the Product of $S$ and $T$ denoted $ST$ is defined to be $ST = \{ x \cdot y : x \in S, y \in T \}$.

The groups $(G, \cdot)$, $(S, \cdot)$, and $(T, \cdot)$ need not be abelian to define the product $ST$. On this page in particular, $(G, \cdot)$ will be abelian though and inherently $(S, \cdot)$ and $(T, \cdot)$ will also be abelian.

For example, consider the abelian group of integers with respect to standard addition, $(\mathbb{Z}, +)$. We define $3\mathbb{Z}$ and $4 \mathbb{Z}$ as follows:

(1)
\begin{align} \quad 3\mathbb{Z} = \{ 0, \pm 3, \pm 6, \pm 9, ... \} = \{ z \in \mathbb{Z} : z = 3k, k \in \mathbb{Z} \} \end{align}
(2)
\begin{align} \quad 4 \mathbb{Z} = \{ 0 \pm 4, \pm 8, \pm 12, ... \} = \{ z \in \mathbb{Z} : z = 4k, k \in \mathbb{Z} \} \end{align}

In other words, the set $3\mathbb{Z}$ is the set of all integer multiples of $3$ and the set $4 \mathbb{Z}$ is the set of all integer multiples of $4$. It's not hard to prove that $(3\mathbb{Z}, +)$ and $(4\mathbb{Z}, +)$ are abelian subgroups of $(\mathbb{Z}, +)$. Now consider the product of $3\mathbb{Z}$ and $4 \mathbb{Z}$:

(3)
\begin{align} \quad (3\mathbb{Z})(4\mathbb{Z}) = \{ (x + y) \in \mathbb{Z} : x \in 3 \mathbb{Z}, y \in 4 \mathbb{Z} \} = \{ (x + y) \in \mathbb{Z} : x = 3k, y = 4j \: \mathrm{where} \: k, j \in \mathbb{Z} \} \end{align}

We see that $(3\mathbb{Z})(4\mathbb{Z})$ is the set of all sums of multiples of $3$ and multiples of $4$ and it can be shown that $((3\mathbb{Z})(4\mathbb{Z}), +)$ is also a group.

The following theorem will ensure this for us.

 Theorem 1: Let $(G, \cdot)$ be an abelian group and let $(S, \cdot)$ and $(T, \cdot)$ be (abelian) subgroups of $(G, \cdot)$. Then $(ST, \cdot)$ is an abelian subgroup of $(G, \cdot)$.
• Proof: Since $S, T \subseteq G$ we have that $ST \subseteq G$, so we only need to show that $ST$ is closed under $\cdot$ and that for each $a \in ST$ there exists an $a^{-1} \in ST$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$.
• Let $a, b \in ST$. Then $a = x_1 \cdot y_1$ and $b = x_2 \cdot y_2$ for $x_1, x_2 \in S$ and $y_1, y_2 \in T$. Using the fact that $(G, \cdot)$ is an abelian group and inherently the subgroups $(S, \cdot)$ and $(T, \cdot)$ are abelian (so that $\cdot$ is commutative), we have:
(4)
\begin{align} \quad a \cdot b = (x_1 \cdot y_1) \cdot (x_2 \cdot y_2) = \underbrace{(x_1 \cdot x_2)}_{\in S} \cdot \underbrace{(y_1 \cdot y_2)}_{\in T} \in ST \end{align}
• Therefore $ST$ is closed under $\cdot$.
• Now let $a \in ST$. Then $a = x \cdot y$ where $x \in S$ and $y \in T$. Since $x \in S$ we have $x^{-1} \in S$ since $(S, \cdot)$ is a group, and similarly, since $y \in T$ we have that $y^{-1} \in T$ since $(T, \cdot)$ is a group. Therefore:
(5)
\begin{align} \quad a^{-1} = (x \cdot y)^{-1} = \underbrace{y^{-1}}_{\in T} \cdot \underbrace{x^{-1}}_{\in S} \in ST \end{align}
• Therefore for each $a \in ST$ there exists an $a^{-1} \in ST$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$.
• Therefore $(ST, \cdot)$ is an abelian group, and more specifically, $(ST, \cdot)$ is an abelian subgroup of $(G, \cdot)$. $\blacksquare$