# List all the subgroups H of C_(12)

#### AutGuy98

##### New member
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.

#### Euge

##### MHB Global Moderator
Staff member
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 $|G|/|H|$.

#### AutGuy98

##### New member
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.

#### Euge

##### MHB Global Moderator
Staff member
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.

#### AutGuy98

##### New member
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.

#### Euge

##### MHB Global Moderator
Staff member
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 $12/6 = 2$.