I'm just wondering how to show that a quotient group $H = (G/N)$ is cyclic?
Let $G= \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$
Let $N = \left<(2,3)\right>$ , where N is a cyclic subgroup of G
Is it correct to say that $G$ has order $24$ and $N$ has order $2$?
Can I then say that the order the the quotient group $H$ is $24/2 = 12$?
How should I go about showing that H is cyclic?
Best Answer
$G$ has order $24$ because $\mathbb{Z}/4\mathbb{Z}$ has $4$ elements and $\mathbb{Z}/6\mathbb{Z}$ has $6$ elements and $4\cdot 6 = 24$.
In order to see if $N$ has order $2$ you should check what are the elements in $N:\;$ we have the trivial element, of course, and we also have $(2,3)$. Since $(2,3)+(2,3)=(0,0),$ there are no more elements in $N,$ and so it is of order $2$.
Yes, the order of $H=G/N$ is $\frac{|G|}{|N|}$ where $|G|=24$ is the order of $G$ and $|N|=2$ is the order of $N$.
If $H$ is of order $12$ it necessarily takes one of the following forms
$H$ is isomorphic to $\mathbb{Z}/12\mathbb{Z},$ which is cyclic. Or
$H$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
The latter is not cyclic, so you need to show that $H$ can't take that form. In order to do this, show that $(1,0)+N\in H$ is not of order $2$ but is of order $4$. (This is easy because $(2,0)\not\in N$ but $(4,0)=(0,0)\in N$.)