How to describe conjugacy classes for elements of $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$?

Question

I am totally new to mathematical analysis and just learn what group is. In a problem it says to describe conjugacy class for $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. But what is the conjugate class this group?

Answer 1

If you have a group $G$, the conjugacy class of an element $h\in G$ is all the elements of the form $g\cdot h\cdot g^{-1}$ for $g\in G$. In your case, the conjugacy class of $(0,1)$ consists of the elements:

$$ \begin{split} (0,0)+(0,1)+(0,0)&=(0,1) \\ (0,1)+(0,1)+(0,1)&=(0,1) \\ (1,0)+(0,1)+(1,0)&=(0,1) \\ (1,1)+(0,1)+(1,1)&=(0,1) \end{split} $$

Note that there is only one element in the conjugacy class of $(0,1)$ which is $(0,1)$. This is because the group is abelian so in all cases $g\cdot h\cdot g^{-1}=g\cdot g^{-1}\cdot h=h$, which is what Adam Hughes meant by the conjugation being trivial. The conjugacy classes of $\mathbb{Z}_2\times \mathbb{Z}_2$ are therefore the sets consisting of each element individually:

$$ \begin{split} Cl((0,0))&=\{(0,0)\} \\ Cl((0,1))&=\{(0,1)\} \\ Cl((1,0))&=\{(1,0)\} \\ Cl((1,1))&=\{(1,1)\}\end{split} $$

So $\mathbb{Z}_2\times \mathbb{Z}_2$ has four conjugacy classes each consisting of one element.