# List all the subgroups H of C_(12)

• MHB
• AutGuy98
In summary, The conversation is about the poster seeking help with a class assignment involving subgroups and indexes. They have previous received help with parts 2(a) and 2(b) but are unsure about how to proceed with part 2(c). The expert explains that every subgroup of a cyclic group is cyclic and for every positive divisor of n, there is a unique subgroup of Zn of order d. Since the positive divisors of 12 are 1, 2, 3, 4, 6, and 12, the subgroups of Z12 are kZ12 where k=1, 2, 3, 4, 6, 12. Using this correspondence, the subgroups of C12
AutGuy98
Hey guys,

Sorry that it's been a decent amount of time since my last posting on here. Just want to say upfront that I am extremely appreciative of all the support that you all have given me over my last three or four posts. Words cannot express it and I am more than grateful for it all. But, in light of that, I actually have some more questions for an exercise set that I have to do for one of my classes and I'm really unsure how to begin doing them. There are four of them and they all require proofs to some degree. Anyway, I was going to make one post and put all four parts of the same question in it (i.e. 2(a),2(b),2(c), and 2(d)), but was unsure whether or not it would be allowed here. So, for those reasons and to play it safe rather than try to do so, here is the third part that I've been having trouble with. Any help here is, once again, greatly appreciated and will leave me forever further in your gratitude.

Question: 2(c): "List all the subgroups H of C12 and compute the index of H in C12 for every choice of H."

Again, I have no idea where to start with this, so any help is extremely gracious and appreciated.

P.S. If possible at all, I'd need help on these by tomorrow at 12:30 E.S.T., so please try to look this over at your earliest conveniences. Thank you all again for your help with everything already.

There is a well-defined, one-to-one correspondence (in fact, an isomorphism) from $\Bbb Z_{12}$ onto $C_{12}$, the mapping being $\phi : \Bbb Z_{12} \xrightarrow{\simeq} C_{12}$ defined by the equation $\phi([k]_{12}) = \omega^k$. Use this correspondence and your knowledge about subgroups of the group of integers modulo $12$ to determine all the subgroups of $C_{12}$. To compute index, use the fact that if $G$ is a finite group and $H$ is a subgroup of $G$, then the index of $H$ in $G$ is $|G|/|H|$.

Euge said:
There is a well-defined, one-to-one correspondence (in fact, an isomorphism) from $\Bbb Z_{12}$ onto $C_{12}$, the mapping being $\phi : \Bbb Z_{12} \xrightarrow{\simeq} C_{12}$ defined by the equation $\phi([k]_{12}) = \omega^k$. Use this correspondence and your knowledge about subgroups of the group of integers modulo $12$ to determine all the subgroups of $C_{12}$. To compute index, use the fact that if $G$ is a finite group and $H$ is a subgroup of $G$, then the index of $H$ in $G$ is $|G|/|H|$.

Hi Euge,
The same situation applies here as was the case with 2(b). I only have a small amount of understanding when it comes to what is going on in my Elem. Abstract class, so if you would be so kind as to please just provide me with the steps to the solution whenever you are able, as I am still heavily pressed for time and honestly have no idea where to go with the advice you gave me, nor do I have that much time to figure it all out. I would really, really appreciate it more than anything right now. Thank you again for your help with 2(a) and 2(b) though, that helped relieve the tension off of me a tremendous amount.

Every subgroup of a cyclic group is cyclic, and for every positive divisor $d$ of $n$, there is a unique subgroup of $\Bbb Z_n$ of order $d$. The positive divisors of $12$ are $1, 2, 3, 4, 6$, and $12$. Thus, the subgroups of $\Bbb Z_{12}$ are $k \Bbb Z_{12}$ where $k = 1, 2, 3, 4, 6, 12$. By the correspondence the subgroups of $C_{12}$ are $\omega^k C_{12}$ for the same $k$-values.

Euge said:
Every subgroup of a cyclic group is cyclic, and for every positive divisor $d$ of $n$, there is a unique subgroup of $\Bbb Z_n$ of order $d$. The positive divisors of $12$ are $1, 2, 3, 4, 6$, and $12$. Thus, the subgroups of $\Bbb Z_{12}$ are $k \Bbb Z_{12}$ where $k = 1, 2, 3, 4, 6, 12$. By the correspondence the subgroups of $C_{12}$ are $\omega^k C_{12}$ for the same $k$-values.

Thank you Euge for all the help. But, if you would be so kind as to provide the information, how exactly do you find the index values for each choice of H from here? Thank you again.

AutGuy98 said:
Thank you Euge for all the help. But, if you would be so kind as to provide the information, how exactly do you find the index values for each choice of H from here? Thank you again.

As I explained in Post #2, to find the index of a subgroup $H$ of a finite group $G$, compute $|G|/|H|$. Since $C_{12}$ has order $12$, the index $(C_{12} : H) = 12/|H|$. For example, if one of your subgroups has $6$ elements in it, then the index of the subgroup is $12/6 = 2$.

## 1. What is a subgroup?

A subgroup is a subset of a larger group that still follows the same group operations and structure as the original group. In other words, it is a smaller group within a larger group.

## 2. How many subgroups are there in C12?

There are 6 subgroups in C12. These include the trivial subgroup (containing only the identity element), the entire group C12, and 4 non-trivial subgroups.

## 3. What are the non-trivial subgroups of C12?

The non-trivial subgroups of C12 are the cyclic subgroups generated by the elements 2, 3, 4, and 6. These subgroups have orders 6, 4, 3, and 2 respectively.

## 4. How can I determine the elements of a subgroup in C12?

To determine the elements of a subgroup in C12, you can use the following formula: H = {gn | n is a positive integer and g is an element of C12}. This means that you take the element g and raise it to different powers (starting from 1) until you reach the identity element or the original element g again. The resulting elements will form the subgroup H.

## 5. Can there be subgroups of C12 with the same order?

Yes, there can be subgroups of C12 with the same order. For example, both the cyclic subgroup generated by 2 and the cyclic subgroup generated by 6 have orders of 6. This is because they contain the same number of elements, even though the elements themselves may be different.

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