# Compute $Z_{4}\times Z_{8}/\langle(1,2)\rangle$

I would like to know a systematic approach of calculating this. My instructor tried a few things on the board and suddenly he decided to take the order of:

$$(2,0)+\langle(1,2)\rangle,\qquad (3,0)+\langle(1,2)\rangle, \qquad \text{and}\qquad (0,1)+\langle(1,2)\rangle$$

and poof it's isomorphic to $Z_8$. Now this doesn't help students like me at all. Can someone perhaps detail what I should be looking for in order to tackle this problem. If there's an alternative way, great. I am aware that every one of these problems can be different in nature so there's no algorithm to tackle it universally.

So far, I see that the subgroup has four elements and hence the order of the factor group would be $8$. So our factor group could be isomorphic to $Z_2\times Z_4$, $Z_8$ or $Z_2\times Z_2\times Z_2$. I believe the first and the third options are not cyclic so if we cannot find a generator it would be one of the two. From here I don't know how to carry on. Please help.

• I almost thought you wanted us to write out every element in the group. – IAmNoOne Nov 3 '14 at 3:04

You can easily check that $\{(1,0),(0,1)\}$ generates $Z_4\times Z_8$, so $\{(1,0)+\langle(1,2)\rangle,(0,1)+\langle(1,2)\rangle\}$ also generates $Z_4\times Z_8/\langle(1,2)\rangle$, but $$(1,0)+\langle(1,2)\rangle = (0,-2)+\langle(1,2)\rangle$$ so $(0,1)+\langle(1,2)\rangle$ generates $Z_4\times Z_8/\langle(1,2)\rangle$; that is, $Z_4\times Z_8/\langle(1,2)\rangle$ is a cyclic group.

• I think this is the clearest answer. – guest Nov 3 '14 at 3:09
• Sorry but how did you get the (1,0)+<(1,2)>=(0,-2)+<(1,2)>? and does this exclude (1,0)? it would be great if you can explain a bit more. – cambelot Nov 3 '14 at 3:13
• @camox You may know that (if not, you can check that) $a+\langle(1,2)\rangle=a-(1,2)+\langle(1,2)\rangle$. and $$(0,-2)+\langle (1,2)\rangle = -[(0,1)+\langle (1,2)\rangle+(0,1)+\langle (1,2)\rangle]$$. – Hanul Jeon Nov 3 '14 at 3:18
• If you don't know how to prove a+<(1,2)>=a-(1,2)+<(1,2)>, enumerate an elements of a+<(1,2)>. It is possible since a+<(1,2)> is an equivalent class over $Z_4\times Z_8$. – Hanul Jeon Nov 3 '14 at 3:20

The way I see that $\mathbb{Z}_4\times \mathbb{Z}_8/\langle(1,2)\rangle$ is isomorphic to $\mathbb{Z}_8$ is first by writing this in a slightly different form. Let $x$ be a generator of the $\mathbb{Z}_4$ component and $y$ of $\mathbb{Z}_8$. Then in the quotient, $xy^2\equiv 1$, or, in other words, $x\equiv y^{-2}$. So, we may replace any $x$ with $y^{-2}$ (and the group $\langle y^{-2}\rangle$ is itself isomorphic to $\mathbb{Z}_4$), so it is like we are embedding the first component into the second, and so the first component can be disregarded.

You're on the right track. To expand on your instructor's insight a bit, the distributions of the orders of the elements are different for the three candidate isomorphism types you've identified:

$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$: all elements have order $1$ or $2$

$\mathbb{Z}_4 \times \mathbb{Z}_2$: all elements have order $1$, $2$, or $4$

$\mathbb{Z}_8$: all elements have order $1$, $2$, $4$, or $8$

So, your quotient is isomorphic to $\mathbb{Z}_8$ iff you can find an element of order $8$. In fact, since the only other possible orders of elements for the candidate isomorphism types are $1$, $2$, $4$, and any such element $g$ satisfies $0 = 4g = 2(2g)$, to show that the group is $\mathbb{Z}_8$, it suffices to find an element such that doubling it twice does not give the identity.

In this case, taking $g = (0, 1) + \langle (1, 2) \rangle$ gives $4g = (0, 4) + \langle(1, 2)\rangle \neq 0$, so it must have order $8$, and so the quotient is isomorphic to $\mathbb{Z}_8$ as desired.

• This is the most straightfoward way to prove/explain this. – Pubbie Nov 4 '14 at 19:04

This is the same thing as taking the quotient of $\mathbf{Z} \times \mathbf{Z}$ by the subgroup generated by $(4,0)$, $(0,8)$, and $(1,2)$.

You can use linear algebra to simplify the presentation: make the three vectors rows of a matrix, and then elementary integer row operations are just finding equivalent generating sets, and elementary integer column operations act as a change of basis on $\mathbf{Z} \times \mathbf{Z}$.

By using these, you can always transform an integer matrix to one whose only entries are along the main diagonal. At that point, it's easy to see what the structure of the quotient group is.

(it may help to think "Euclidean algorithm")

If you've not done integer linear algebra before, the elementary integer row operations are adding a multiple of one row to another, multiplying a row by $-1$, and swapping two rows. Multiplying by any other constant is not elementary, because it's inverse isn't given by an integer matrix.

It turns out that any invertible, square integer matrix is a product of the elementary integer row operations, just as in the usual case of linear algebra over a field.

$$f: \mathbb{Z}_4\times \mathbb{Z}_8 \rightarrow \mathbb{Z}_4\times \mathbb{Z}_2,\ f(x,y)= (2x-y,y)$$

Here $(t,2t)$ is kernel. It is a group homomorphism.