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?
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.
