Let $G=\mathbb Z_4 \times \mathbb Z_2$. Find all $H$ subgroups of $G$ of order 2, so that $G / H$ is cyclic.

Question

Let $G=\mathbb Z_4 \times \mathbb Z_2$. Find all $H$ subgroups of $G$ of order 2, so that $G / H$ is cyclic.

I tried going brute force, and realized there are 3 candidates:

$H_1=\{(0,0), (2,0) \}$ $H_2=\{(0,0), (0,1) \}$ $H_3=\{(0,0), (2,1) \}$

But how do I continue from here?

Answer 1

Elements of $G$ are of the form

$$(i,j)$$

Now, we want all the elements such that

$$(2i,2j)=0$$

So $2i\equiv0\pmod 4$ and $2j\equiv0 \pmod2$

The posibilities, as $i,j$ are in the intervals $[0,4]$ and $[0,2]$, are, $i=0,2$ and $j=0,1$.

So the elements: $(0,1)$, $(2,0)$ and $(2,1)$ generates all the subgroups of order 2.

Answer 2

Note that for each $i \in \{1,2,3\}$, we have $\left|G/H_i\right|=4$. So there are two possibilities: either these are isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ or they are isomorphic to $\mathbb{Z_4}$ (the latter is cyclic and former is not).

Now consider $G/H_1$. It will be $$G/H_1=\{H_1, (1,0)+H_1, (1,1)+H_1, (2,1)+H_1\}$$ Recall that the order of $aH$ is the smallest positive integer $k$ such that $a^k \in H$. Using this you can easily see that none of them has order $4$, hence NOT cyclic. Now you can do the rest similarly.