- #1

twoflower

- 368

- 0

Hi all,

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

Given the group [itex](\mathbb{Z}_{12}, +, 0)[/itex], 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:

[tex]

\langle0\rangle = \{0\}

[/tex]

[tex]

\langle1\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle2\rangle = \{0,2,4,6,8,10\}

[/tex]

[tex]

\langle3\rangle = \{0,3,6,9\}

[/tex]

[tex]

\langle4\rangle = \{0,4,8\}

[/tex]

[tex]

\langle5\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle6\rangle = \{0,6\}

[/tex]

[tex]

\langle7\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle8\rangle = \{0,4,8\}

[/tex]

[tex]

\langle9\rangle = \{0,3,6,9\}

[/tex]

[tex]

\langle10\rangle = \{0,2,4,6,8,10\}

[/tex]

[tex]

\langle11\rangle = \mathbb{Z}_{12}

[/tex]

So after removing duplicities, I have 6 subgroups:

[tex]

\mathbb{Z}_{12}

[/tex]

[tex]

\{0\}

[/tex]

[tex]

\{0,6\}

[/tex]

[tex]

\{0,4,8\}

[/tex]

[tex]

\{0,3,6,9\}

[/tex]

[tex]

\{0,2,4,6,8,10\}

[/tex]

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.

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

Given the group [itex](\mathbb{Z}_{12}, +, 0)[/itex], 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:

[tex]

\langle0\rangle = \{0\}

[/tex]

[tex]

\langle1\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle2\rangle = \{0,2,4,6,8,10\}

[/tex]

[tex]

\langle3\rangle = \{0,3,6,9\}

[/tex]

[tex]

\langle4\rangle = \{0,4,8\}

[/tex]

[tex]

\langle5\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle6\rangle = \{0,6\}

[/tex]

[tex]

\langle7\rangle = \mathbb{Z}_{12}

[/tex]

[tex]

\langle8\rangle = \{0,4,8\}

[/tex]

[tex]

\langle9\rangle = \{0,3,6,9\}

[/tex]

[tex]

\langle10\rangle = \{0,2,4,6,8,10\}

[/tex]

[tex]

\langle11\rangle = \mathbb{Z}_{12}

[/tex]

So after removing duplicities, I have 6 subgroups:

[tex]

\mathbb{Z}_{12}

[/tex]

[tex]

\{0\}

[/tex]

[tex]

\{0,6\}

[/tex]

[tex]

\{0,4,8\}

[/tex]

[tex]

\{0,3,6,9\}

[/tex]

[tex]

\{0,2,4,6,8,10\}

[/tex]

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.

Last edited: