Finding the order of $(1, 1) + \langle(2, 2)\rangle$ in the factor group $\mathbb{Z} \times \mathbb{Z} / \langle (2, 2)\rangle$

Question

I read somewhere that the order of $(1, 1) + \langle(2, 2)\rangle$ in the factor group $\mathbb{Z} \times \mathbb{Z} / \langle (2, 2)\rangle$ is $2$. I am trying to understand how this is true. I have:

$(1, 1) + (-2, -2) = (-1, -1) \neq (0, 0)$

$(1, 1) + (0, 0) = (1, 1)$

$(1, 1) + (2, 2) = (3, 3)$

$(1, 1) + (4, 4) = (5, 5)$

$(1, 1) + (6, 6) = (7, 7) \neq (0, 0)$

How is the order of $(1, 1) + \langle(2, 2)\rangle$ in the factor group $\mathbb{Z} \times \mathbb{Z} / \langle (2, 2)\rangle$ equal to $2$?

Answer 1

Good typesetting, by the way!

The calculations you are doing are finding elements of the coset $(1, 1) + \langle(2, 2)\rangle$; but that's not what you're asking.

The order of the coset $(1, 1) + \langle(2, 2)\rangle$ (that is, its order in the factor group $(\Bbb Z\times \Bbb Z)/\langle(2, 2)\rangle$) is the smallest positive integer $m$ such that $m$ times $(1, 1) + \langle(2, 2)\rangle$ is equal to the identity element of $(\Bbb Z\times \Bbb Z)/\langle(2, 2)\rangle$.

So to answer this question, you'll need to be able to answer these sub-questions:

• What is the identity element of $(\Bbb Z\times \Bbb Z)/\langle(2, 2)\rangle$?
• How can you tell in general whether $(a, b) + \langle(2, 2)\rangle$ is equal to this identity element?
• What is the definition of "$m$ times $(1, 1) + \langle(2, 2)\rangle$"?
• Is $1\times\big( (1, 1) + \langle(2, 2)\rangle \big)$ equal to the identity element of $(\Bbb Z\times \Bbb Z)/\langle(2, 2)\rangle$? Is $2\times\big( (1, 1) + \langle(2, 2)\rangle \big)$ equal to this identity element? Is $3\times\big( (1, 1) + \langle(2, 2)\rangle \big)$ equal to this identity element? ...

Answer 2

The order of an element $a$ of a group is the smallest positive integer $m$

such that $\underbrace{a+a+\cdot+a}_{m\text{ times}}$ is the identity element of the group.

The order of $(1,1)$ in the quotient group $\mathbb Z\times\mathbb Z/\langle(2,2)\rangle$ is $2$,

because $(1,1)\not\in \langle(2,2)\rangle$, but $(1,1)+(1,1)=(2,2)\in\langle(2,2)\rangle$;

$(0,0)$ and $(2,2)$ are both in the same coset of $\langle(2,2)\rangle$,

which is $\langle(2,2)\rangle$, the identity element of the quotient group.

Answer 3

$\langle(2,2)\rangle=\{(2,2)^i,i\in \mathbb{Z}\}=\{(2i,2i),i\in \mathbb{Z}\}$

$(1,1)+\langle(2,2)\rangle=\{(1+2i,1+2i),i\in \mathbb{Z}\}$

\begin{alignat}{1} \bigl((1,1)+\langle(2,2)\rangle\bigr)^2 &= \{(2+2i+2j,2+2i+2j),i,j\in \mathbb{Z}\} \\ &= \{(2(1+i+j),2(1+i+j)),i,j\in \mathbb{Z}\} \\ &= \{(2k,2k),k\in \mathbb{Z}\} \\ &= \langle(2,2)\rangle \\ \tag 1 \end{alignat}