Assignment:
(1)  Find the order of the element  (1,1) in Z9 x Z12

(1,1)^36 = (36,36) = (0,0) Since  9 divides 36 and 12 divides 36.

36 is the smallest such number then LCM(9,12) = 36.

(2)  How many subgroups of  Z4 x Z25 are there?

Z4 has 3 subgroups and Z25 has 3 subgroups so there

are 3x3 = 9 subgroups of Z4xZ25.  This is the same question

we had earlier when I asked the number of subgroups of Z100.


(3)  Decompose the multiplicative group of the numbers
relatively prime to 21 into an internal direct sum of 
cyclic groups.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
x x   x x     x    x  x     x        x  x     x  x

The order is 12.  The Group must be Z4 x Z3 or Z2 x Z2 x Z3


 2  4  8 16 11 1
 4 16  1
 5  4 20 16 17 1
 8  1
10 16 13  4 19 1
11 16  8  4  2 1 
13  1
16  4  1
17 16 20  4  5 1
19  4 13 16 10 1
20  1

Since no element is of order 4, it must be Z2 x Z2 x Z3

If I use { 1, 2, 4, 8,16,11},I have to combine it with

{1,13} or {1,20}.  I cannot use {1,8} since 8 already

appears.

G = { 1, 2, 4, 8,16,11} x {1,13}

G = { 1, 4,16} x { 1,8} x {1,13}


(4)  List the possible commutative groups of order 36.

            Z4 x Z9

            Z4 x Z3 x Z3

            Z2 x Z2 x Z9

            Z2 x Z2 x Z3 x Z3