Find the order of $\displaystyle Z \oplus Z | <(2,2)>$

Now, the elements in $\displaystyle Z \oplus Z$ are (0,0) and (1,1), so the order of it is 2. but I'm confuse about the second one, what is the order of that?

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- Oct 7th 2007, 08:35 PMtttcomraderFactor group of an external direct product
Find the order of $\displaystyle Z \oplus Z | <(2,2)>$

Now, the elements in $\displaystyle Z \oplus Z$ are (0,0) and (1,1), so the order of it is 2. but I'm confuse about the second one, what is the order of that? - Oct 8th 2007, 09:14 AMThePerfectHacker
Here $\displaystyle \mathbb{Z}\oplus \mathbb{Z} = \{... ,(-2,-2),(-1,-1),(0,0),(1,1),(2,2),...\}$

Write out $\displaystyle \left<(2,2)\right>$ which means $\displaystyle \{...,(-4,-4),(-2,-2),(0,0),(2,2),(4,4),...\}$.

Now notice that,

$\displaystyle \left< (2,2) \right> \mbox{ and }(1,1) + \left< (2,2)\right>$.

Form a complete list of co-sets, i.e. the even ones and the odd ones.

Thus, this factor group is of order 2.