Exercise 1.1.28 (Direct product of groups)

Proof. Let $(A,\star )$ and $(B,◇)$ be groups.

The product on $A\times B$ is defined for all $(a,b)\in A\times B$ and for all $(c,d)\in A\times B$ by

Since $(a\star c,b◇d)\in A\times B$, this defines a law on $A\times B$.

(a)

For all $(a_i,b_i)\in A\times B,i=1,2,3$, since $\star$ and $◇$ are associative laws, $$\begin{array}{llll}\hfill (a_1,b_1)[(a_2,b_2)(a_3,b_3)]& =(a_1,b_1)(a_2\star a_3,b_2◇b_3)& \hfill & \\ \hfill & =(a_1\star (a_2\star a_3),b_1◇(b_2◇b_3))& \hfill & \\ \hfill & =((a_1\star a_2)\star a_3),(b_1◇b_2)◇b_3)& \hfill & \\ \hfill & =(a_1\star a_2,b_1◇b_2)(a_3,b_3)& \hfill & \\ \hfill & =[(a_1,b_1)(a_2,b_2)](a_3,b_3).& \hfill & \end{array}$$

(b)

For all $(a,b)\in A\times B$, $$\begin{array}{llll}\hfill (1,1)(a,b)& =(1\star a,1◇b)=(a,b),& \hfill & \\ \hfill (a,b)(1,1)& =(a\star 1,b◇1)=(a,b),& \hfill & \end{array}$$

so $(1,1)$ is the identity of $A\times B$.

(c)

For all $(a,b)\in A\times B$, $$\begin{array}{llll}\hfill (a,b)(a^{-1},b^{-1})& =(a\star a^{-1},b◇b^{-1})=(1,1),& \hfill & \\ \hfill (a^{-1},b^{-1})(a,b)& =(a^{-1}\star a,b^{-1}◇b)=(1,1).& \hfill & \end{array}$$

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

So $H\times G$ is a group for this multiplicative law. □