TITLE: $\mathbb{Z}_2\times\mathbb{Z}_2$ has order $4$? How so?
QUESTION [1 upvotes]: $\mathbb{Z}_2\times\mathbb{Z}_2$ has order $4$, the neutral element is $(0_2,0_2)$ and the other elements have order $2$.
  Therefore, $\mathbb{Z}_2\times\mathbb{Z}_2$ is not cyclic,so it is the Klein group.

The elements of $\mathbb{Z_2}$ are $\{0_2,1_2\}$ and the order of $1_2$ is 2.
If we take the direct product $\mathbb{Z}_2\times\mathbb{Z}_2$ then we have a generator of the Group product $\langle 1_2,1_2\rangle$.
Question:
How can the author state $\mathbb{Z}_2\times\mathbb{Z}_2$ has order $4$? Is $\langle 1_2,1_2\rangle$ not a generator of order 2 of $\mathbb{Z}_2\times\mathbb{Z}_2$?

REPLY [3 votes]: The order of a group is the number of elements in it. $\mathbb{Z}_2\times\mathbb{Z}_2$ has four elements: $(0,0), (0,1), (1,0), (1,1)$.
But as you have seen, each element has order $\le 2$. (This, in particular, proves that $\mathbb{Z}_2\times\mathbb{Z}_2$ is not isomorphic to $\mathbb{Z}_4$.)

REPLY [1 votes]: The elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$ are
(0,0), (0,1), (1,0), (1,1)
There are 4 of them, thus the order of the group is 4.
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