# Finding Subgroups in Z

• Nov 8th 2008, 10:59 AM
apsis
Finding Subgroups in Z
I'm a little confused by exactly what this question is asking and am thinking I might be over simplifing it.

The question reads "Find all subgrounps of ${Z}_{12}$".

Does this simply mean list
{null, {0}, {1}, ... , {0, 1}, {0, 2}, ..., {0,1,2,3,4,5,6,7,8,9,10, 11}} as all of the subgrounds?

There is no operation defined on the group, so I'm not sure what else to do?

Thanks!
• Nov 8th 2008, 04:03 PM
ThePerfectHacker
Quote:

Originally Posted by apsis
I'm a little confused by exactly what this question is asking and am thinking I might be over simplifing it.

The question reads "Find all subgrounps of ${Z}_{12}$".

Does this simply mean list
{null, {0}, {1}, ... , {0, 1}, {0, 2}, ..., {0,1,2,3,4,5,6,7,8,9,10, 11}} as all of the subgrounds?

There is no operation defined on the group, so I'm not sure what else to do?

Thanks!

This is the set $\{ [0]_{12}, [1]_{12}, ... , [11]_{12} \}$ with binary operation "addition modulo $12$".
• Nov 9th 2008, 08:42 AM
apsis
How do I count the subgroups then? If I say that [0], [1], ... , [11] are the subgroups its not correct, because the closure property is not satisfied, i.e., [1] + [1] is not a member of [1] in ${Z}_{12}$, and so [1] is not a subgroup. So I would have to expand that group to include [2], but then I run into the same issue of [1] + [2]. Eventually I would include all elements and rebuild the group ${Z}_{12}$.

The same argument applies for all elements of the group (except [0]). So does that mean the only proper subgroup is [0]? Or am I thinking of this wrong?

Thanks
• Nov 9th 2008, 08:54 AM
ThePerfectHacker
The subgroup of a cyclic group is cyclic. Therefore, any subgroup must be generated by an element.
Take for example $[2]_{12}$ then the subgroup generated by it is $\{ [0]_{12},[2]_{12},[4]_{12},[6]_{12}, [8]_{12},[10]_{12} \}$.
• Nov 9th 2008, 09:09 AM
apsis
So if I'm thinking of this right then the subgroups are

[0] -> {[0]}
[1] -> ${Z}_{12}$
[2] -> {[0], [2], [4], [6], [8], [10]}
[3] -> {[0], [3], [6], [9]}
[4] -> {[0], [4], [8]}
[6] -> {[0], [6]}

Any other generator just results in the set ${Z}_{12}$, which would be repetitive.
• Nov 9th 2008, 09:14 AM
ThePerfectHacker
Quote:

Originally Posted by apsis
So if I'm thinking of this right then the subgroups are

[0] -> {[0]}
[1] -> ${Z}_{12}$
[2] -> {[0], [2], [4], [6], [8], [10]}
[3] -> {[0], [3], [6], [9]}
[4] -> {[0], [4], [8]}
[6] -> {[0], [6]}

Any other generator just results in the set ${Z}_{12}$, which would be repetitive.

Yes (Clapping)! In fact for each $k|12$, $1\leq k\leq 12$, we can generate a distinct subgroup. Furthermore, $\left < [k_1]_{12} \right> \subseteq \left< [k_2]_{12}\right>$ if and only if $k_1|k_2$.

You want to try doing a similar problem for $\mathbb{Z}_{60}$ by finding all the subgroups using the above result.
• Nov 9th 2008, 09:22 AM
apsis
wicked thanks! those results make it much easier to determine the subgroups so for ${Z}_{60}$ the generators of distinct subgroups are the integers that divide 60.

But for your second result showing the subgroups of subgroups, isn't that backwards? In my list I have <6> $\subseteq$ <3>. Not the other way around as your result would predict?
• Nov 9th 2008, 09:27 AM
ThePerfectHacker
Quote:

Originally Posted by apsis
But for your second result showing the subgroups of subgroups, isn't that backwards? In my list I have <6> $\subseteq$ <3>. Not the other way around as your result would predict?

Yes! Sorry about that. It should say $k_2|k_1$ and not $k_1|k_2$.