Find all subgroups of the given group

[twoflower]

Hi all,

I've been practising some algebra excercises and don't know how to solve this one:

Given the group (\mathbb{Z}_{12}, +, 0), find all its subgroups. How many elements will have these subgroups?

The only idea which came to my mind is to see the subgroups generated by each individual element of the given group. So I got:

<br /> \langle0\rangle = \{0\}<br />

<br /> \langle1\rangle = \mathbb{Z}_{12}<br />

<br /> \langle2\rangle = \{0,2,4,6,8,10\}<br />

<br /> \langle3\rangle = \{0,3,6,9\}<br />

<br /> \langle4\rangle = \{0,4,8\}<br />

<br /> \langle5\rangle = \mathbb{Z}_{12}<br />

<br /> \langle6\rangle = \{0,6\}<br />

<br /> \langle7\rangle = \mathbb{Z}_{12}<br />

<br /> \langle8\rangle = \{0,4,8\}<br />

<br /> \langle9\rangle = \{0,3,6,9\}<br />

<br /> \langle10\rangle = \{0,2,4,6,8,10\}<br />

<br /> \langle11\rangle = \mathbb{Z}_{12}<br />


So after removing duplicities, I have 6 subgroups:

<br /> \mathbb{Z}_{12}<br />

<br /> \{0\}<br />

<br /> \{0,6\}<br />

<br /> \{0,4,8\}<br />

<br /> \{0,3,6,9\}<br />

<br /> \{0,2,4,6,8,10\}<br />


But since my approach was rather non-rigorous, I wonder if these are all subgroups of the original subgroup. Is there a chance (when using this approach) that I miss some subgroup?

It seems that I can take care only of elements of the original group, which have some common divisor with module (12) greater than 1, because only they seem to generate some interesting subgroups, other elements (5,7,9,10,11) generate always the whole group. Is this also a rule?

I just would like to know some more general approach to find all subgroups of given group, because if the group gets more complicated than this example, this doesn't seem to be comfortable way to do that.

Thank you for any advice.

Best regards.

 

[matt grime]

There really is no better method than simply working them out by hand like this. Sorry if you wanted a clever way of doing it. In general there isn't.

 

[twoflower]

There really is no better method than simply working them out by hand like this. Sorry if you wanted a clever way of doing it. In general there isn't.

Thank you matt. Is it true that I'll get all subgroups doing it this way? Can't I miss any? And if so, may I restrict myself to generating subgroups from elements which have GCD with module of the group > 1?

 

[matt grime]

You have classified the cyclic subgroups. It is now up to you to try to decide if there are non-cyclic subgroups.

 

[mathwonk]

if you know the subgroups of Z, you might look at the surjection from Z to Z/n and use the fact that the inverse image of a subgroup is a subgroup.