Let $G = (\Bbb Z/18\Bbb Z, +)$ be a cyclic group of order $18$.

Question

I'm currently taking abstract algebra and I'm very lost.

Let $G = (\Bbb Z/18\Bbb Z, +)$ be a cyclic group of order $18$.

(1) Find a subgroup $H$ of $G$ with $|H|= 3.$

(2) What are the elements of $G/H$?

(3) Find a familiar group that is isomorphic to $G/H$.

For one I think I understand that since it is a cyclic group we need a generator so I choose $\langle [6]\rangle$. $[6]+[6]=[12]$ and $[6]+[6]+[6]=[18]=[0]$ so $H=\langle [6]\rangle=\{[0],[6],[12]\}$. Here we see $18$ divided by $6$ is $3$ so $|H| = 3.$

The next part are the elements $G/H$ just the subgroup I wrote down before?

The last question is confusing me the most. In order to be isomorphic to one another the group that I select must have three elements as well, correct? The problem is there is no other subgroup of $G$ that has an order $3$.

Answer 1

$G/H$ has 6 elements since $|G/H| =|G|/|H|=\frac{18}{3}=6$. We we are looking for a group with 6 elements. We say that $x,y \in G$ are in the same equivalence class if $x-y \in H$, ie $x-y=0,6$ or $12$, so:

$0=6=12$

$1 = 7 = 13$

$2=8=14$

$3=9=15$

$4=10=16$

$5=11=17$

Therefore, the elements of $G/H=\{\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{5}\}$.

Answer 2

Here $\lvert G/H\rvert=\lvert G\rvert/\lvert H\rvert=18/3=6$.

Also, $G\cong\langle a\mid a^{18}\rangle,$ so $$\begin{align} G/H&\cong\langle b\mid b^{18/3}\rangle \\ &\cong \langle b\mid b^6\rangle \\ &\cong \Bbb Z_6. \end{align}$$

Answer 3

For the last question, you can use the third isomorphism theorem:

$g=\mathbf Z/18\mathbf Z$, $H==6\mathbf Z/18 \mathbf Z$, so $$G/H=\mathbf Z/18\mathbf Z\Big/6\mathbf Z/18\mathbf Z\simeq\mathbf Z/6\mathbf Z.$$