# Thread: Finding Subgroups in Z

1. ## 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!

2. 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 $12$".

3. 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

4. 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 $\{ [0]_{12},[2]_{12},[4]_{12},[6]_{12}, [8]_{12},[10]_{12} \}$.

5. 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.

6. 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 ! 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.

7. 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?

8. 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 $k_1|k_2$.