# 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)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)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:

- 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:

- 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 $(G, \cdot)$. $\blacksquare$