Product Groups

Groups can be combined to form new product group or decomposed to simpler groups for analysis.

Theorem. Given $G_1$ and $G_2$ are groups, the product $G_1 \times G_2$ is a group under operation

\[(a_1, a_2) \ast (b_1, b_2) = (a_1 \ast_1 b_1, a_2 \ast_2 b_2)\]

where $a_1, b_1 \in G_1$ and $a_2, b_2 \in G_2$.

Proof.

0. As $c_1 = a_1 \ast_1 b_1 \in G_1$ and $c_2 = a_2 \ast_2 b_2 \in G_2$, $(c_1, c_2) \in G_1 \times G_2$ and is closed.

1. $(e_1, e_2)$ is the identity element

2. $(a^{-1}, b^{-1})$ is the inverse.

3. $G_1 \times G_2$ is associative as

\[\begin{align*} ((a_1, a_2) \ast (b_1, b_2)) \ast (c_1, c_2) &= ((a_1 \ast_1 b_1) \ast_1 c_1, (a_2 \ast_2 b_2) \ast_2 c_2) \\ &= (a_1 \ast_1 (b_1 \ast_1 c_1), a_2 \ast_2 (b_2 \ast_2 c_2)) \\ &= (a_1, a_2) \ast ((b_1, b_2) \ast (c_1, c_2)) \end{align*}\]

Hence, $G_1 \times G_2$ is a group.

Direct Product Theorem

In order to check if a group is direct product of two groups, we have the following conditions:

Theorem. [Direct Product Theorem] Let $H_1, H_2 \le G$ such that

  1. $H_1 \cap H_2 = \set{e}$

  2. $\forall a_1 \in H_1$ and $\forall a_2 \in H_2$, $a_1a_2 = a_2a_1$

  3. $\forall a \in G$, $\exists a_1 \in H_1$ and $\exists a_2 \in H_2$, $a = a_1a_2$

Then $G \cong H_1 \times H_2$.

Proof.

Let $f: H_1 \times H_2 \to G$ defined by $f((a_1, a_2)) = a_1a_2$. $G \cong H_1 \times H_2$ if $f$ is an isomorphism.

By (2),

\[f((a_1, a_2)(b_1, b_2)) = f((a_1b_1, a_2b_2)) = a_1b_1a_2b_2 = a_1a_2b_1b_2 = f((a_1, a_2))f((b_1, b_2))\]

and hence $f$ is a homomorphism.

By (3), $f$ is surjective.

When $f((a_1, a_2)) = a_1a_2 = e$, $a_2 = a_1^{-1}$. By (1), it is only possible if $a_1 = a_2 = e$. Hence, $\ker f = \set{(e, e)}$ and $f$ is injective.

Therefore, $f$ is a bijection and $G \cong H_1 \times H_2$.

References