Can someone please explain to me step-by-step how to solve this problem:
How many sub-groups of order 9 does the group (Z_{3)}X(Z_{3})X(Z_{9}) has?
First the number of cyclic subgroups of order 9: If H=<(a,b,c)> has order 9, then c has order 9. So with an appropriate "power", we may take c=1. Further 2(a,b,c) is in H. So there are really just 2 choices each for a and b, giving a total of 4 cyclic subgroups of order 9.
Next if H is not cyclic of order 9, then it is an order 9 subgroup (subspace) of the 3 dimensional vector space over the field with 3 elements. The total number of sets of 2 independent vectors is (3^3-1)(3^3-3)=26*24. Each subgroup has (3^2-1)(3^2-3)=48 sets of 2 independent vectors. Now different subgroups can not have a set of 2 independent vectors in common. So if n is the number of non-cyclic subgroups of order 9, 48n=26*24 or n=13.
Thus finally there are 4+13=17 subgroups of order 9.
alternatively, each distinct subgroup of order 9 contains 6 elements of order 9. since if c is any element of order 9 (for which we have 6 choices), (a,b,c) is of order 9, we have 54 elements of order 9. thus we must have 9 cyclic subgroups (not 4).
i will list them:
<(0,0,1)> = {(0,0,0),(0,0,1),(0,0,2),(0,0,3),(0,0,4),(0,0,5),( 0,0,6),(0,0,7),(0,0,8)}
<(0,1,1)> = {(0,0,0),(0,1,1),(0,2,2),(0,0,3),(0,1,4),(0,2,5),( 0,0,6),(0,1,7),(0,2,8)}
<(1,0,1)> = {(0,0,0),(1,0,1),(2,0,2),(0,0,3),(1,0,4),(2,0,5),( 0,0,6),(1,0,7),(2,0,8)}
<(1,1,1)> = {(0,0,0),(1,1,1),(2,2,2),(0,0,3),(1,1,4),(2,2,5),( 0,0,6),(1,1,7),(2,0,8)}
<(0,1,2)> = {(0,0,0),(0,1,2),(0,2,4),(0,0,6),(0,1,8),(0,2,1),( 0,0,3),(0,1,5),(0,2,7)}
<(1,0,2)> = {(0,0,0),(1,0,2),(2,0,4),(0,0,6),(1,0,8),(2,0,1),( 0,0,3),(1,0,5),(2,0,7)}
<(1,1,2)> = {(0,0,0),(1,1,2),(2,2,4),(0,0,6),(1,1,8),(2,2,1),( 0,0,3),(1,1,5),(2,2,7)}
<(1,2,1)> = {(0,0,0),(1,2,1),(2,1,2),(0,0,3),(1,2,4),(2,1,5),( 0,0,6),(1,2,7),(2,1,8)}
<(2,1,1)> = {(0,0,0),(2,1,1),(1,2,2),(0,0,3),(2,1,4),(1,2,5),( 0,0,6),(2,1,7),(1,2,8)} <--all 54 elements of order 9 accounted for.
now it's clear that any non-cyclic subgroup must be generated by two distinct elements of order 3 which give rise to two distinct subgroups of order 3. since we have 54 elements of order 9, and 1 element of order 1, we have:
81 - 54 - 1 = 26 elements of order 3. each element x of order 3 comes paired with its inverse in <x> = {e,x,x^{-1}}. so we must have 13 distinct subgroups of order 3.
so any non-cyclic subgroup is determined by a choice of 2 distinct subgroups of order 3, and we have 13 choose 2 = 13!/(2!11!) = 78 different ways to choose two such subgroups.
if we call the first subgroup H = <a>, and the second subgroup K = <b>, then:
<a,b> = <a,a+b> = <a+b,b> = <a+b,a-b> = <a-b,b> = <a,a-b>, so we have a six-fold duplication when we count such subgroups. hence we have at most 13 such subgroups. a complete list:
<(1,0,0),(0,0,3)>
<(0,1,0),(0,0,3)>
<(1,1,0),(0,0,3)>
<(1,2,0),(0,0,3)>
<(1,0,0),(0,1,0)>
<(1,0,0),(0,1,3)>
<(1,0,0),(0,2,3)>
<(0,1,0),(1,1,3)>
<(0,1,0),(2,0,3)>
<(1,1,0),(0,1,3)>
<(1,1,0),(0,2,3)>
<(1,0,3),(0,1,3)>
<(1,1,3),(1,2,0)> by which we see there are exactly 13 such subgroups.
so the total number is 9+13 = 22 subgroups of order 9.
oops, my bad. I miscounted the number of cyclic subgroups of order 9. There are exactly 9 of these. As before, if (a,b,c) is a generator for a cyclic group of order 9, by taking a suitable "power", we may assume c=1. Now the elements (a,b,1) all generate cyclic subgroups of order 9. Now if (a,b,1) is in <(c,d,1)>, then k(c,d,1)=(a,b,1). So k is 1 mod 9 and then k is 1 mod 3; i.e. a=c and b=d. So any choices for a, b yield different subgroups. So there are 9 cyclic subgroups of order 9. Thus as Deveno says there are 9 + 13 = 22 subgroups of order 9.
The only advantage of counting this way is that it immediately generalizes:
Could someone direct me to information on how to write Tex in these postings? Wrapping TEX tags around selected text seems to work for simple expressions only.