# Is $G/N$ cyclic if $N$ is a subgroup generated by $(2,3)$?

I am given $G=\mathbb{Z}_6 \oplus\mathbb{Z}_6$ and I am told to let $N$ be its cyclic subgroup generated by $(2,3)$. Is the group $G/N$ cyclic? Why? And if it is, what is its generator?

So:

Since the order of $\langle 2\rangle$ in $\mathbb{Z}_6$ is $3$ and the order of $\langle 3\rangle$ in $\mathbb{Z}_6$ is $2$, the order of $(2,3)$ in $\mathbb{Z}_6 \oplus\mathbb{Z}_6$ is $6$.

This is where im stuck. I'm not sure how to find out if $\mathbb{Z}_6 \oplus\mathbb{Z}_6/ \langle (2,3)\rangle$ is cyclic. I know a group is cyclic if $G=\langle g\rangle=\{g^{n}\mid n \in \mathbb(Z)\}$. I also know it has to do with relatively prime numbers, but I can't see how to put it together.

• Do you know the classification of group of order 6? – Nitrogen Oct 2 '15 at 17:52
• Try finding a subgroup $H$ of $G$ of order $6$ with $H \cap N = 1$. – Geoff Robinson Oct 2 '15 at 18:50

The quotient is an abelian group of order $6$, which is necessarily cyclic by the fundamental structure theorem for finitely generated abelian groups.

• ahh ok, I havent heard of that theorem...I will look it up. SInce I know its cyclic, how can I go about finding its generator? – The Physics Student Oct 2 '15 at 18:01
• You also can use Cauchy's theorem: there exists in a quotient an element of order $2$ and anelement of order $3$. What is the order of their product? – Bernard Oct 2 '15 at 18:16

You already showed that the quotient group is of order $6$ and as others pointed out that the only abelian group of order $6$ is the cyclic group.

I would like to propose another approach, which also addresses your question on generator. We will show that image under the standard projection of $(1,1)$ generates $G/N$. This is clear because $$(n,n)=(2m,3m)\pmod 6$$ only if $n$ is a multiple of $6$.

There's another (better) way to see this. Think of $\mathbb Z_6\oplus \mathbb Z_6$ as a $6\times 6$ grid. Draw the line through $(2,3)$ and $(0,0)$, this is the group $N$. The line generated by $(1,1)$ should intersect the $N$-line only at $(0,0)$. (Note: A line here should have $6$ points.)

Similar argument as in linear algebra should show that the original $\mathbb Z_6\oplus \mathbb Z_6$ is decomposed into the direct sum $$\langle (2,3)\rangle \oplus \langle (1,1)\rangle.$$

Similarly, one can find all possible $(k,l)$ such that its coset generates $G/N$. The answer is that the matrix $$\left(\begin{array}{cc}2&k\\3&l\end{array}\right)$$ is invertible, i.e. having invertible determinant.

• ahh very good! This makes sense, thanks! – The Physics Student Oct 2 '15 at 23:14

First of all, order of quotient group $\vert G/N \vert$ is given by

$\vert G/N \vert = \vert G\vert/ \vert N \vert$

Hence order of your quotient group will be = 36/6 = 6. As $G = Z_6 × Z_6$ which abelian group. Hence its quotient group is also abelian.

Now as there are only two groups of order 6, one is symmetric group $S_3$ which is non-abelian and other one is $Z_6$ Which is abelian. So your quotient group $G/N = Z_6 × Z_6/<(2, 3)> ≈ Z_6$ So that your quotient group is cyclic group of order 6. Since $Z_6$ is cyclic group of order 6.