# Showing a subgroup of $\mathbb{Z}\times\mathbb{Z}$ is cyclic.

In the group $$G = \mathbb{Z}\times\mathbb{Z}$$, consider the subgroup $$H$$ generated by $$(-5,1)$$ and $$(1,-5)$$. I want to show that $$G/H$$ is cyclic and find the standard cyclic group it is isomorphic to.

I haven't much group theory experience, but understand that $$G$$ is a group. Firstly what is meant by $$H$$ being generated by the mentioned elements of $$G$$? I know it's the intersection of all subgroups that contain those two particular elements, but can it be thought of as all the multiples and linear combinations of the two?

And I am also confused about the rest of the question.

Edit: I think confusion lies with the definition of 'generated by'. I understand its the intersection of all these subgroups that contain the set of elements (or generators) but is there a more useful equivalent definition.

• NB: It's "intersection", not "interception".
– Shaun
Commented Jan 10, 2020 at 15:25
• $5(-5,1)+(1,-5)=(-24,0); (-5,1)+5(1,-5)=(0,-24)$ Commented Jan 10, 2020 at 15:33

As you seem to have perceived, this is as much a linear algebra question as a group theory question, although you have to be careful and do your linear algebra over $$\mathbb Z$$ instead of the usual $$\mathbb R$$. That means you can only use integers where you are used to using arbitrary real numbers.

The group $$H$$ is generated by two integer vectors $$\vec v = (-5,1)$$ and $$\vec w = (1,-5)$$. Since this is an abelian group, then yes, you can say that $$H$$ is the group of all integer linear combinations of $$\vec v$$ and $$\vec w$$.

Now let's put those two vectors into the rows of a matrix: $$M = \begin{pmatrix} -5 & 1 \\ 1 & -5 \end{pmatrix}$$ It follows that the row space of $$M$$ over $$\mathbb Z$$ is $$H$$, i.e. the set of all integer linear combinations of the rows of $$M$$ is $$H$$.

Now use your linear algebra skills to simplify the matrix $$M$$ by doing row operations which do not affect the row space over $$\mathbb Z$$. For example, add $$5$$ times row 2 to row 1 to get $$\begin{pmatrix} 0 & -24 \\ 1 & -5 \end{pmatrix}$$ then switch rows 1 and 2 to get $$\begin{pmatrix} 1 & -5 \\ 0 & -24 \end{pmatrix}$$ and then multiply row $$2$$ by $$-1$$ to get $$\begin{pmatrix} 1 & -5 \\ 0 & 24 \end{pmatrix}$$ You can also do column operations over $$\mathbb Z$$, which have the effect of changing the given basis for $$G$$, but of course that does not affect the isomorphism type of the quotient group $$G/H$$. So, adding $$5$$ times column $$1$$ to column $$2$$ you get $$\begin{pmatrix} 1 & 0 \\ 0 & -24 \end{pmatrix}$$ So now we know that $$G / H \approx (\mathbb Z \oplus \mathbb Z) / (\mathbb Z \oplus 24\mathbb Z) \approx (\mathbb Z / 1 \mathbb Z) \oplus (\mathbb Z / 24\mathbb Z) \approx \mathbb Z / 24\mathbb Z$$ so the quotient is isomorphic to the cyclic group of order $$24$$.

• Note that the column operations actually change the row space. I don't think it is entirely trivial that it doesn't change the quotient. (Well, it does change the quotient, but isomorphically.) It is the row operations that only change the basis of the row space. Commented Jan 10, 2020 at 15:15
• From a group theoretic perspective, I think it helps to understand that a column operation changes the isomorphism between $G$ and $\mathbb Z \oplus \mathbb Z$ (i.e. it changes the basis, as said in my answer), but it does not change $G$ nor $H$. Commented Jan 10, 2020 at 15:38

If $$H$$ is generated by $$h_1$$ and $$h_2$$ then $$H=\{ah_1+bh_2\}$$ where $$a$$ and $$b$$ are integers. If you think of $$G$$ as the set of points in the plane with integer co-ordinates then $$H$$ is the lattice of points with co-ordinates $$(-5a+b, a-5b)$$ where $$a$$ and $$b$$ are integers.

The elements of $$G/H$$ correspond to the co-sets of $$H$$ within $$G$$. Since the determinant of

$$\begin{pmatrix} -5 & 1 \\ 1 & -5 \end{pmatrix}$$

is $$24$$, the area of the parallelogram bounded by $$(0,0)$$, $$(-5,1)$$ and $$(1,-5)$$ is $$24$$ so there are $$24$$ such co-sets.

Since $$G$$ is abelian, $$G/H$$ must also be abelian, so $$G/H$$ is an abelian group of order $$24$$. To show that $$G/H$$ is isomorphic to $$C_{24}$$ and not to some other abelian group with order $$24$$ (such as $$C_{12} \times C_2$$) we must find an element of $$G/H$$ that has order $$24$$. The co-set that contains the point $$(0,1)$$ is a candidate for this, since

$$-5a+b=0 \Rightarrow b=5a \Rightarrow a-5b = -24a$$

so if $$k(0,1) \in H$$ then $$k$$ must be a multiple of $$24$$, so the order of the $$(0,1)$$ co-set within $$G/H$$ is $$24$$.