Thread: Find all the sub groups of a given group

Can someone please explain to me step-by-step how to solve this problem:
How many sub-groups of order 9 does the group (Z₃₎X(Z₃)X(Z₉) 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.

LaTex seems to only be sporadically working, for some reason.