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 $3$.
Best Answer
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 $$G/H=\mathbf Z/18\mathbf Z\Big/6\mathbf Z/18\mathbf Z\simeq\mathbf Z/6\mathbf Z.$$