# $G\:/\:H$, where $G = \mathbb{Z} \oplus \mathbb{Z}$ and $H = \langle(1,3) , (3,1)\rangle$

Let $$G$$ be the additive group $$\mathbb{Z} \oplus \mathbb{Z}$$. Let $$H$$ be the subgroup generated by $$(1,3)$$ and $$(3,1)$$.

a) Determine the order of $$(0,1) + H$$ in $$G\:/\:H$$.

b) Express $$G\:/\:H$$ as a direct sum of cyclic groups.

Here are my thoughts so far:

a) In general, let $$gH \in G\:/\:H$$. Then $$|gH| = n$$ if and only if $$n$$ is the smallest positive integer for which $$g^n\in H$$.

Here, $$H$$ consists of all elements in $$\mathbb Z\oplus\mathbb Z$$ of the form $$m(1,3) + n(3,1)$$, where $$m,n \in \mathbb Z$$. On the other hand, for the element $$gH = (0,1) + H \in G\:/\:H$$, we have $$(gH)^k = (0,k)$$, where $$k$$ is a positive integer. We need to find the smallest $$k$$ such that $$(0,k) = m(1,3) + n(3,1)$$ for integers $$m,n \in \mathbb Z$$. It's not hard to see that $$k$$ must be equal to $$8$$.

Thus, the order of $$(0,1) + H$$ in $$G\:/\:H$$ is 8. Am I correct ?

b) I'm not sure how to begin this part. I've never encountered a quotient group being written as a direct sum of cyclic groups. How can I gain some insight into this?

Thanks!

• Shouldn't it be $k(gH) = (0,k)$, not $(k,1)$? Dec 7, 2019 at 2:19
• Do you have tools available to tell you that $\#(G/H)=8$ (without looking that the specific point $(0,1)$)? Dec 7, 2019 at 2:20
• @GregMartin To observe the index of $H$ in $G$, correct ? Dec 7, 2019 at 3:27

We first define an isomorphism $$\phi: G\rightarrow G$$, sending $$(a, b)$$ to $$(a, b - 3a)$$. This is an isomorphism, because it has an inverse $$\psi$$ sending $$(a, b)$$ to $$(a, b + 3a)$$.

It suffices to solve the problem after applying $$\phi$$.

The group $$\phi(H)$$ is generated by $$\phi(1, 3)$$ and $$\phi(3, 1)$$, namely $$(1, 0)$$ and $$(3, -8)$$. I claim that it is also generated by $$(1, 0)$$ and $$(0, 8)$$. In fact, let $$H'$$ be the subgroup generated by $$(1, 0)$$ and $$(0, 8)$$, we want to show that $$\phi(H) = H'$$.

Since $$(0, 8) = 3(1, 0) - (3, -8)\in H$$, we have $$H'\subseteq \phi(H)$$

Similarly, $$(3, -8) = 3(1, 0) - (0, 8)\in H'$$ shows that $$\phi(H) \subseteq H'$$.

Therefore $$\phi(H) = H'$$.

For a), we want to determine the smallest positive integer $$k$$ such that $$k(0, 1) \in H$$. This is of course the same as the smallest $$k$$ such that $$k\phi(0, 1) = (0, k)\in \phi(H)$$.

Since $$\phi(H)$$ is generated by $$(1, 0)$$ and $$(0, 8)$$, it's obvious that $$k = 8$$ is what we need.

For b), we know that the quotient $$G/H$$ is isomorphic, via $$\phi$$, to the quotient $$G/\phi(H) = (\mathbb Z \oplus \mathbb Z)/(\mathbb Z \oplus 8\mathbb Z) \simeq \mathbb Z/8\mathbb Z$$.

More precisely, we define a group homomorphism $$\tau: G\rightarrow \mathbb Z/8\mathbb Z$$, sending an element $$(a, b)$$ to the image of $$b$$ in $$\mathbb Z/8\mathbb Z$$. It is obviously surjective, because for every $$\beta \in \mathbb Z/8\mathbb Z$$, there is an integer $$b$$ whose image in $$\mathbb Z/8\mathbb Z$$ is $$\beta$$, and we have $$\tau(0, b) = \beta$$ by definition.

What is the kernel of $$\tau$$? It's all the elements $$(a, b)\in G$$ such that $$b$$ is a multiple of $$8$$. But these are exactly those elements that can be written as $$x(1, 0) + y(0, 8)$$ for integers $$x, y$$, i.e. they form exactly the subgroup $$\phi(H)$$.

Hence the homomorphism $$\tau$$ induces an isomorphism $$G/\phi(H)\simeq \mathbb Z/8\mathbb Z$$.

Since $$G$$ is generated by $$x=(1,0)$$ and $$y=(0,1)$$, then $$G/H$$ is generated by their images $$\bar{x}$$ and $$\bar{y}$$. Now you can use the following observations:

• in $$G/H$$, $$3\bar{x}+\bar{y}=0$$, so $$\bar{y}=-3\bar{x}$$ is in the group generated by $$\bar{x}$$, which means that $$G/H$$ is actually generated by $$\bar{x}$$.

• we also have $$\bar{x}+3\bar{y}=0$$, so $$9\bar{x}=3(3\bar{x})=3(-\bar{y})=-3\bar{y}=\bar{x}$$, which tells you that $$8\bar{x}=0$$.