# Is the quotient group $\mathbb{Z}_4 \times \mathbb{Z}_6 / \langle(2,2)\rangle$ cyclic?

Is the quotient group $$\mathbb{Z}_4 \times \mathbb{Z}_6 /\langle(2,2)\rangle$$ cyclic?

How do I find if quotient group is cyclic?

I see that $$\langle (2,2)\rangle= \{(0,0), (2,2), (0,4), (2,0), (0,2), (2,4)\}$$ and the order of quotient group is $$4$$.

What's the next step I need to take?

• Is the divisor $(2,2)$ or $(2,3)$? Oct 1 '20 at 13:48
• Sorry! I edited it. It's (2,2)
– jun
Oct 1 '20 at 13:50
• See this question for an idea. Oct 1 '20 at 14:07

## 2 Answers

No.

A presentation for $$\Bbb Z_4\times \Bbb Z_6$$ is

$$\langle a,b\mid a^4, b^6, ab=ba\rangle.\tag{1}$$

Here $$a\mapsto ([1]_4,[0]_6)$$ and $$b\mapsto ([0]_4,[1]_6)$$, so $$([2]_4, [2]_6)$$ corresponds to $$a^2b^2$$. The quotient by $$\langle ([2]_4, [2]_6)\rangle$$ amounts, then, to killing $$a^2b^2$$ in $$(1)$$, like so:

\begin{align} \Bbb Z_4\times \Bbb Z_6/\langle ([2]_4, [2]_6)\rangle &\cong\langle a,b\mid a^2b^2, a^4, b^6, ab=ba\rangle\\ &\cong\langle a,b\mid a^2=b^{-2}, a^4, b^6, ab=ba\rangle\\ &\cong\langle a,b\mid a^2=b^{-2}, b^4, b^6, ab=ba\rangle\\ &\cong\langle a,b\mid a^2=b^2, b^2, ab=ba\rangle\\ &\cong\langle a,b\mid a^2, b^2, ab=ba\rangle\\ &\cong\Bbb Z_2\times\Bbb Z_2, \end{align}

which is not cyclic.

It may be worth saying that one can solve all problems of this sort algorithmically: no thought is required.

Think of the group additively.

Write down the matrix of relations: in this case $$\begin{bmatrix} 4 & 0\\ 0 & 6\\ 2 & 2\\ \end{bmatrix}.$$

Using Gaussian reduction (with no divisions) reduce this to Smith Normal Form: in this case

$$\begin{bmatrix} 4 & 0\\ 0 & 6\\ 2 & 2\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 2\\ 4 & 0\\ 0 & 6\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 2\\ 0 & -4\\ 0 & 6\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 2\\ 0 & -4\\ 0 & 2\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 2\\ 0 & 2\\ 0 & -4\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 2\\ 0 & 2\\ 0 & 0\\ \end{bmatrix} \sim \begin{bmatrix} 2 & 0\\ 0 & 2\\ 0 & 0\\ \end{bmatrix}$$

Hence the group is $$\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\simeq\mathbb{Z}_2\oplus\mathbb{Z}_2$$ which is not cyclic.