Question

List all subgroups of ${\mathbb{Z}}_{12}/H$, where$H=\{0,6\}$ .

Short answer

The subgroups of ${\mathbb{Z}}_{12}/H$ are $\left\{0\right\},{\mathbb{Z}}_2,{\mathbb{Z}}_3$ and ${\mathbb{Z}}_6$.

Step-by-step solution

Referring to Lagrange’s Theorem

If K is a subgroup ofa finite group G. Then the order of K divides the order of G. In particular, $\left|G\right|=\left|K\right|[G:K]$.

Finding all subgroups of  ℤ12/H

The order of ${\mathbb{Z}}_{12}/H$ can be given as:

$\begin{array}{c}O({\mathbb{Z}}_{12}/H)=\left|{\mathbb{Z}}_{12}\right|/\left|H\right|\\ =\frac{12}{2}\\ =6\end{array}$

Since the order of ${\mathbb{Z}}_{12}/H$ is 6 and according toLagrange’s Theorem, all the subgroups of ${\mathbb{Z}}_{12}/H$ must have order as multiples of 6.

Hence, the subgroups are$\left\{0\right\},{\mathbb{Z}}_2,{\mathbb{Z}}_3$ and ${\mathbb{Z}}_6$.