This exercise generalizes the previous two. Let .G; / and .H; / be groups. Their direct

Chapter 41, Problem 41.4

(choose chapter or problem)

This exercise generalizes the previous two. Let .G; / and .H; ?/ be groups. Their direct product is a new group .G; / .H; ?/ whose elements are all ordered pairs .g; h/ where g 2 G and h 2 H. The operation for this group (lets use the symbol ) is defined by .g1; h1/ .g2; h2/ D .g1 g2; h1 ? h2/: For example, lets consider .Z 5 ; / .Z3; /. We have Z 5 D f1; 2; 3; 4g and Z3 D f0; 1; 2g and therefore the elements of .Z 5 ; / .Z3; / are f.1; 0/; .1; 1/; .1; 2/; .2; 0/; .2; 1/; .2; 2/; .3; 0/; .3; 1/; .3; 2/; .4; 0/; .4; 1/; .4; 2/g: Copyrigh The operation .g1; h1/.g2; h2/ produces the value .g; h/ where g D g1g2 (operatingin Z5) and h D g1 g2 (operating in Z5). For example,.2; 1/ .3; 2/ D .2 3; 1 2/ D .1; 0/:You should convince yourself that if .G; / and .H; ?/ are groups, then .G; /.H; ?/ is also a group. [Optional: Write a formal proof of this.]We now come to the point of this problem: If .G; / and .H; ?/ are finite cyclicgroups, then sometimes .G; / .H; ?/ is cyclic and sometimes it is not. The questionis, under what condition(s) is the direct product of two finite cyclic groups also cyclic?41.5.