# Computing Quotient Groups $\mathbb{Z}_4 \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$, $\mathbb{Z} \times \mathbb{Z}_{6}/ \langle (1, 2) \rangle$

Let $G/H = \mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$. I know that $|G/H|$ = 4, so $G/H \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{4}$. Since $G/H$ has an element of order 4, namely $(0, 1) + \langle (2, 4) \rangle$, $G/H \simeq \mathbb{Z}_{4}$. Is my reasoning correct?

Also why is $\mathbb{Z} \times \mathbb{Z}_{6}/ \langle (1, 2) \rangle \simeq \mathbb{Z}_{6}$ and not $\mathbb{Z}$?

Edit

So for the first question, $\mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$ becomes $\mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 0), (0, 2) \rangle \simeq \mathbb{Z}_{10}/ \langle 2 \rangle \times \mathbb{Z}_{4} / \langle 2 \rangle \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}$.

Could you elaborate on how you approached my second question? I still do not follow.

2nd Edit

I see how $\mathbb{Z} \times \mathbb{Z}_{6} / \langle(3,0)\rangle \simeq \mathbb{Z}_{3} \times \mathbb{Z}_{6}$. And we want to do that because it is simple and $(3,0) \in \langle (1, 2) \rangle$. Then I think your next step is to compute $\mathbb{Z}_{3} \times \mathbb{Z}_{6} / \langle (1,2 )\rangle$. But then $|\mathbb{Z}_{3} \times \mathbb{Z}_{6} / \langle (1,2 )\rangle|$ = 6. So the resulting quotient must be $\mathbb{Z}_{6}$.

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Actually, $(0,1)+\langle(2,4)\rangle$ is not of order $4$. Note that $$\langle (2,4)\rangle = \{ (2,4), (0,8), (2,2), (0,6), (2,0), (0,4), (2,8), (0,2), (2,6), (0,0)\}$$ so $$\bigl( (0,1)+\langle (2,4)\rangle\bigr) + \bigl( (0,1)+\langle (2,4)\rangle\bigr) = (0,2)+\langle(2,4)\rangle = (0,0)+\langle(2,4)\rangle.$$

In fact, note that $\langle(2,4)\rangle = \langle (2,0), (0,2)\rangle$, which should make the isomorphism type of the quotient very clear.

For the second, notice that $(3,0)\in\langle(1,2)\rangle$. So you can first mod out by $(3,0)$ as a first approximation; we have $\mathbb{Z}\times\mathbb{Z}_6/\langle(3,0)\rangle \cong \mathbb{Z}_3\times\mathbb{Z}_6$. Now you want to quotient out this by the subgroup generated by (the image of) $(1,2)$. $$\langle(1,2)\rangle = \{ (1,2), (2,4), (0,0)\}.$$ So $(0,1)+\langle (1,2)\rangle$ is of order $6$, which shows that the quotient is cyclic of order $6$.

Intuitively, taking the quotient modulo $\langle(1,2)\rangle$ "identifies" the $1$ in $\mathbb{Z}$ with the $2$ in $\mathbb{Z}_6$; that means that the $1$ in $\mathbb{Z}$ is of order $3$ in the quotient (as we saw), and that twice $(0,1)$ is the same as $(1,0)$. So from $(0,1)$ has order $6$ in the quotient; since $(1,0)$ and $(0,1)$ generate $\mathbb{Z}\times\mathbb{Z}_6$, knowing their images tells you exactly what happens to the whole group.

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Could you take a look at my edit? –  Student Dec 13 '11 at 4:45
First part is correct. As to how I approached the second question, first I noticed that $(3,0)$ was in the subgroup $\langle(1,2)\rangle$, and I quotiented out by that first, because that's something very easy to do. Then we are essentially in a case similar to the one in the first part, with a finite group and a cyclic subgroup, and you can approach it in the same way you attempted to approach the first one. It also tells us that the quotient will have order $6$ (because by quotienting out by $\langle(3,0)\rangle$ you get something of order $18$ (cont) –  Arturo Magidin Dec 13 '11 at 4:53
@Jon: (cont) and then you take the quotient by a subgroup of order $3$, so it will result in a group of order $6$. There is only one abelian group of order $6$ (the cyclic group), so then you just notice you have a generator. –  Arturo Magidin Dec 13 '11 at 4:54
Would you mind taking a look at my second edit? Thanks : ) –  Student Dec 13 '11 at 5:09
@Jon: $\mathbb{Z}_2\times\mathbb{Z}_3$ is isomorphic to $\mathbb{Z}_6$. There is only one abelian group of order $6$. As soon as you know that the quotient is of order $6$ (and of course abelian), you know it must be $\mathbb{Z}_6$. –  Arturo Magidin Dec 13 '11 at 5:15