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Hello,

If you could please help me with the following question,

ques. 22.3- Construct the Cayley table for the quotient group Z12/the subgroup generated by [4].

What I don't understand is how to get the elements in that group.

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Here is question 22.7 that anonymous wrote on December 3. I am also having problems with it.

ques. 22.7- If m and n are positive integers and m|n, then the group generated by n is a normal subgroup of the group generated by m (in Z). What is the order of the quotient group- group generated by m/group generated by n?

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First you find the subgroup generated by 4 (call it N). In Z_{12} we have: [4]+[4]=8 and [4]+[4]+[4]=[0]. So, N={[0],[4],[8]}. The set N plays the role of the identity in Z_{12}/N (and you have one of the elements). Since Z_{12} has 12 elements and N has 3, the quotient gruop has 12/3=4 elements (one of which is N). The other 3 elements of the quotient group are the other right cosets. For example, N+[1]={[0]+[1], [4]+[1], [8]+[1]}={[1], [5], [9]} (we have found one more element; two to go). Then N+[2]={[0]+[2], [4]+[2], [8]+[2]}={[2], [6], [10]}; and finaly, the fourth element must be the rest of Z_{12} (since the cosets cover all of that group); indeed N+[3]={[0]+[3], [4]+[3], [8]+[3]}={[3], [7], [11]}. Now that you have all of the elements, you may want to try the Cayley table.

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22.7 Call H the group generated by n, and K the group generated by m. The group generated by n consist of all the multiples of n, and since m|n, all the elements of that group are also multiples of m, and hence H is a subgroup of K. It is certainly normal since all the groups you see in that question are commutative. Finally, it is easy to guess what |K/H| is: take some specific values for n and m and write down all of the right cosets of H in K (for example, take m=3, n=6). I'll leave that part to you.