Quotient group concerning $\mathbb Z / 24 \mathbb Z$

Question

Let $G=\mathbb Z / 24 \mathbb Z$ and let $H$ be cyclic subgroup of $G$ generated by class $16+ 24 \mathbb Z$.

The question is whether $G/H$ is cyclic? I don't even know what elements of $G/H$ are supposed to be, classes containing classes? Maybe the fact that $\mathbb Z / 24 \mathbb Z$ is isomorphic to $\mathbb Z_{24}$ could help by simplifying things.

Also, what would be the order of $G/H$? By Lagrange's theorem, it should be $|G|/|H|=8$, is that correct? But if I identify $\mathbb Z / 24 \mathbb Z$ with $\mathbb Z_{24}$, I get more than $8$ elements in $G/H$.

Any help would be appreciated. Thanks!

Answer 1

If we view $G = \mathbb{Z}/24\mathbb{Z}$, we talk about the quotient group of the additive group $\mathbb{Z}$ and a subgroup $24\mathbb{Z}$. Hence, we "mod out" $24$ and set it equivalent to $0$. If we denote $n \mod 24$ as $\bar{n}$, $$G = \{\bar{0}, \bar{1}, \cdots, \bar{23}\}$$ and has order $24$. Now that we are used to this group, we again take a quotient group with the subgroup $H$, generated by $16 \mod 24$. Hence, $H$ looks like $$H = \{\bar{0}, \bar{8}, \bar{16}\}.$$

Indeed, $\#G/H$ is then 8, and since $H$ is generated by one element, it is by definition cyclic, hence so is $G/H$

Answer 2

$H=\langle 16\rangle =\{[0],[8],16]\}$

Then $|G/H|=8$ and note that quotient group of a cyclic group is cyclic(Just use that $\Bbb Z_{24}/H=\langle [1]+H\rangle$

Answer 3

Hint: Every homomorphic image of a cyclic group is cyclic.

Answer 4

A quotient of a cyclic group is necessarily cyclic since being cyclic neans there is a surjective group homomorphism from $\mathbf Z$ onto the group. So you just have to compose the canonical surjective homomorphism $\;\mathbf Z\twoheadrightarrow G=\mathbf Z/24\mathbf Z $ with the no less canonical $G\twoheadrightarrow G/H$.

This being said, since $1$ has order $24$ in $\mathbf Z/24\mathbf Z$, $16=16\cdot 1$ has order $\dfrac{24}{\gcd(16,24)}=3$, so that $\;|G/H|=\dfrac{24}3=8$.

Finally, observe that $$16(\mathbf Z/24\mathbf Z)=(16\mathbf Z+24\mathbf Z)/24\mathbf Z=8\mathbf Z/24\mathbf Z\simeq\mathbf Z/3\mathbf Z.$$ In particular, the subgroup generated by $16$ is also generated by $8=-16$.