Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{Z}_{6}$.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.