What's the fastest way to determine all the subgroups of the additive group $\mathbb{Z}_{24}$

Question

Question is as in title.

I know that all of the subgroups of $\mathbb{Z}_{24}$ (under addition) must be cyclic, and I could find them by finding the generating groups for each element of $\mathbb{Z}_{24}$ - but surely there is a quicker way?

Would appreciate any help,

Jack

Answer 1

In a cyclic group we know that we have only one group of order $n$, for each n which divides the order of the group. $24 = 3\cdot2^3$ so we know that we have $8$ subgroups.

All the subgroups of a cyclic group are cyclic, so all the subgroups are isomorphic to: $\mathbb{Z_2},\mathbb{Z_3},\mathbb{Z_4},\mathbb{Z_6},\mathbb{Z_8},\mathbb{Z_{12}},\mathbb{Z_{24}}$.

If you want to know explicitely the elements of the subgroups you must find a generator for the subgroup:

Let's consider $\mathbb{Z_2}$: the order of the group is $2$ so the generator is $\frac{24}{2}=12.$ So the subgroup is ${\{0,12\}}$.

We do the same for the others:

$\mathbb{Z_3}$: Generator = $8$ so the subgroup is $\{0,8,16\}$

$\mathbb{Z_4}$: Generator = $6$ so the subgroup is $\{0,6,12,18\}$

$\mathbb{Z_6}$: Generator = $4$ so the subgroup is $\{0,4,8,12,16,20\}$

$\mathbb{Z_8}$: Generator = $3$ so the subgroup is $\{0,3,6,9,12,15,18,21\}$

$\mathbb{Z_{12}}$: Generator = $2$ so the subgroup is $\{0,2,4,6,8,10,12,14,16,18,20,22\}$

Answer 2

What is important has already been said in comments and other answers: for each $d \mid n$, there is a single subgroup of order $d$ in $\Bbb Z_n$, and it is isomorphic to $\Bbb Z_d$. It can be explicitly described as

$$\left\{ \widehat {\frac {kn} d} \Bigg| 0 \le k \le d-1 \right\} .$$