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I'm having a difficult time with groups terminology. My note says $C_2 \times C_2$ is not cyclic as the order of the elements have orders $1,2,2,2$. Could anyone please explain this concept?


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Order (group theory). A cyclic group is generated by one element, so at least one element has the same order as the cardinality of the group in finite cases. – Tyler Mar 9 '14 at 22:11

Let $G$ be a group. The order of an element $g \in G$ is the smallest $n \in {\mathbb N}, n \geq 1$ such that $g^n = 1$.

If a group is cyclic, then there is an element $c$ such that $\{1, c, c^2, \dots, \}$ is the whole group. In the case that this set is finite, you have $c^n = 1$ for some $n \geq 1$ (and after that the sequence repeats); the least such $n$ is the order of $c$ and also the order of the group (i.e., its number of elements).

So, a finite group is cyclic if and only if it has an element whose order is equal to the order of the group.

In your case $C_2 \times C_2$ has 4 elements, but none of those elements has order 4, so it's not cyclic.

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The order of an element is the order of the cyclic group generated by the element. I believe it would be more correct to say the period of an element, but to say order i common. In your case $C_2\times C_2=\{e,a,b,ab\}$ with $a^2=b^2=(ab)^2=e$. So the order (more correctly: period) of $a$ is 2 as the cyclic group generated by $s=a$ has two elements. If $C_2\times C_2$ were cyclic it would have an element of period 4.

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The group $C_2 \times C_2$ is the same as $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$ under addition. The order of $(0,0)$ is of course $1$. The order of all other elements is $2$, because:

\begin{align*} (1,1) + (1,1) &= (0,0) \\ (1,0) + (1,0) &= (0,0) \\ (0,1) + (0,1) &= (0,0). \\ \end{align*}

In general, the order of an element $g$ of a group $G$ under operation $*$ is just the smallest number of $g$s you have to multiply together to get the identity $e$. So it is the smallest $n$ such that $\underbrace{g * g * g * \cdots * g}_{n \text{ times}} = e$.

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