Theorem 11 2

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Statement:

Let $G_{1},G_{2},...,G_{n}$ be groups. For $(a_{1},a_{2},...,a_{n})$ and $(b_{1},b_{2},...,b_{n})$ in $\prod_{i=1} ^{n} G_{i}$, define $(a_{1},a_{2},...,a_{n})(b_{1},b_{2},...,b_{n})$ to be the element $(a_{1}b_{1},a_{2}b_{2},...,a_{n}b_{n})$. Then $\prod_{i=1} ^{n} G_{i}$ is a group, the direct product of the groups $G_{i}$, under the binary operation.

Proof:

Note that since $a_{i} \in G_{i}$, $b_{i} \in G_{i}$, and $G_{i}$ is a group, we have $a_{i}b_{i} \in G_{i}$. Thus the definition of the binary operations on $\prod_{i=1} ^{n} G_{i}$ given in the statement of the theorem makes sense; that is, $\prod_{i=1} ^{n} G_{i}$ is closed under the binary operation.

The associative law in $\prod_{i=1} ^{n} G_{i}$ is thrown back onto the associative law in each component as follows:

(1)
\begin{equation} (a_{1},a_{2},...,a_{n})[(b_{1},b_{2},...,b_{n})(c_{1},c_{2},...,c_{n})] \end{equation}
(2)
\begin{equation} =(a_{1},a_{2},...,a_{n})(b_{1}c_{1},b_{2}c_{2},...,b_{n}c_{n}) \end{equation}
(3)
\begin{equation} =(a_{1}(b_{1}c_{1}),a_{2}(b_{2}c_{2}),...,a_{n}(b_{n}c_{n})) \end{equation}
(4)
\begin{equation} =((a_{1}b_{1})c_{1},(a_{2}b_{2})c_{2},...,(a_{n}b_{n})c_{n}) \end{equation}
(5)
\begin{equation} =(a_{1}b_{1},a_{2}b_{2},...,a_{n}b_{n})(c_{1},c_{2},...,c_{n}) \end{equation}
(6)
\begin{equation} =[(a_{1},a_{2},...,a_{n})(b_{1},b_{2},...,b_{n})](c_{1},c_{2},...,c_{n}) \end{equation}

If $e_{i}$ is the identity element in $G_{i}$, then clearly, with multiplication component-wise, $(e_{1},e_{2},...,e_{n})$ is an identity in $\prod_{i=1} ^{n} G_{i}$. Finally, an inverse of $(a_{1},a_{2},...,a_{n})$ is $(a_{1} ^{-1},a_{2} ^{-1},...,a_{n} ^{-1})$; compute the product component-wise. Hence $\prod_{i=1} ^{n} G_{i}$ is a group.

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