Finding subgroups of $G=\displaystyle\normalsize{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}}\LARGE_{/}\large_{\langle(1,0)\rangle}$

I'm doing this exercise:

Find all the subgroups of $G=\displaystyle\normalsize{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}}\LARGE_{/}\large_{\langle(1,0)\rangle}$

This is my try:

First, we see that $$\Large\mid\displaystyle\normalsize{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}}\LARGE_{/}\large_{\langle(1,0)\rangle}\Large\mid\normalsize=\displaystyle\frac{2\cdot 4}{2}=4,$$ which, by Lagrange's Theorem, means that the subgroups of the group $G$ must have order $1$, $2$, or $4$.

We classify the objects on $G$ and we end up with this table: \begin{array}{|c|c|c|}\hline Order & Elements \\\hline 1 & \{(0,0)\} \\\hline 2 & (0,2), (1,2), (1,0) \\\hline 4 & (1,1), (1,3), (0,1), (0,3) \\\hline 8 & / \\\hline \end{array}

Now we start classifying subgroups:

Subgroups generated by $1$ element:

$\large·$ $\langle (0,0)\rangle=\{(0,0)\}$

$\large·$ $\langle (0,2)\rangle=\{(0,0),(0,2)\}$

$\large·$ $\langle (1,2)\rangle=\{(0,0),(1,2)\}$

$\large·$ $\langle (1,0)\rangle=\{(0,0),(1,0)\}$

$\large·$ $\langle (0,1)\rangle=\langle (0,3)\rangle=\{(0,0),(0,1),(0,2),(0,3)\}$

$\large·$ $\langle (1,1)\rangle=\langle (1,3)\rangle=\{(0,0),(1,1),(1,3),(0,2)\}$

Now I have to find the subgroups generated by $2$ elements, but I don't know if I'm doing correctly the exercise, if there's another way to do this kind of exercises or if I made a mistake at any point, because we don't have any example like this on our class notes.

Anybody could say me how to proceed in order to find the subgroups generated by $2$ elements? Thank you.

• It seems you are not considering the quotient group, you're classifying the subgroups of $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$ instead. – Daniel Apr 5 '15 at 16:23

It looks like you're classifying subgroups of $\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z$, but remember there's a quotient. For example $(1, 0)$ represents the identity element in your group, but you have it listed as having order $2$.
The first thing you should do is consider what that quotient does. Show that there is an isomorphism $(\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z)/\langle(1, 0)\rangle \simeq \mathbb Z/4\mathbb Z$, classify the subgroups of $\mathbb Z/4\mathbb Z$, and then use the isomorphism to transfer this information over to $(\mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z)/\langle(1, 0)\rangle$.
• Hi @Jim, why can I say that my group is isomorphic to $\mathbb{Z}/4\mathbb{Z}$? – Relure Apr 5 '15 at 16:29
• Morally you should think this because the quotient kills $(1, 0)$ and that element generates the $\mathbb Z/2\mathbb Z$ factor so only the $\mathbb Z/4\mathbb Z$ factor is left. As far as how to prove that this is the case, you'll have to figure out a homomorphism between $G$ and $\mathbb Z/4\mathbb Z$ and then you'll have to check that this homomorphism is a bijection. – Jim Apr 5 '15 at 17:11