**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.15**

Solution:

(1) Elements in a direct product are uniquely represented, so well-definedness is not an issue. Suppose $\prod a_i$, $\prod b_i$, and $\prod c_i$ are in $G$. Then \begin{align*}\left[(\prod a_i) (\prod b_i)\right] (\prod c_i) =&\ (\prod a_ib_i)(\prod c_i) = \prod (a_ib_i)c_i\\ =&\ \prod a_i(b_ic_i) = (\prod a_i)(\prod b_ic_i)\\ =&\ (\prod a_i)\left[(\prod b_i)(\prod c_i)\right],\end{align*} so that the product in $G$ is associative.

(2)We have $$(\prod 1_i)(\prod a_i) = \prod 1_ia_i = \prod a_i;$$ likewise $(\prod a_i)(\prod 1_i) = \prod a_i$. Thus $1 = \prod 1_i$ is an identity in $G$ under componentwise multiplication.

(3) We have $$(\prod a_i)(\prod a_i^{-1}) = \prod a_ia_i^{-1} = \prod 1_i;$$ likewise $(\prod a_i^{-1})(\prod a_i) = \prod 1_i$. Thus $(\prod a_i)^{-1} = (\prod a_i)$.