$\mathbb{Z}_2\times\mathbb{Z}_2$ 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 $\langle 1_2,1_2\rangle$.


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 $\mathbb{Z}_2\times\mathbb{Z}_2$?

  • $\begingroup$ Remember that the order of a group is a somewhat distinct concept from the order of an element. The order of a group is the cardinality of that group (in this case, 4---the elements can be explicitly written fairly easily), while the order of an element $a$ is the smallest natural number $n$ such that $a^n$ is the identity (here, the maximal order of an element is 2). $\endgroup$ – Xander Henderson Nov 29 '17 at 18:54
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    $\begingroup$ The order of a group means its size. $\endgroup$ – Ittay Weiss Nov 29 '17 at 18:54
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    $\begingroup$ $(1,1)$ does not generate the element $(0,1)$ $\endgroup$ – Doug M Nov 29 '17 at 18:55
  • $\begingroup$ $\langle 1, 1\rangle$ only generates all of $\Bbb Z_n\times \Bbb Z_m$ if the greatest common factor of $n$ and $m$ is 1. Here it is 2. $\endgroup$ – MJD Nov 29 '17 at 18:57

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 $\mathbb{Z}_4$.)


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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