4
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From the Cayley table:

\begin{align*} \begin{array}{c | c c c c } & (0,0) & (0,1) & (1,0) & (1,1)\\ \hline (0,0) & (0,0) & (0,1) & (1,0) & (1,1)\\ (0,1) & (0,1) & (0,0) & (1,1) & (1,0)\\ (1,0) & (1,0) & (1,1) & (0,0) & (0,1)\\ (1,1) & (1,1) & (1,0) & (0,1) & (0,0)\\ \end{array} \end{align*}

How would I construct the characters of this group, $G =\mathbb{Z}_2 \oplus \mathbb{Z}_2$?

EDIT: Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian, all characters are one-dimensional so they take on values $\pm 1$. So we have the same character table as the Klein-4 group:

\begin{align*} \begin{array}{c | c c c c } & (0,0) & (0,1) & (1,0) & (1,1)\\ \hline \chi_{(0,0)} & 1 & 1 & 1 & 1\\ \chi_{(0,1)} & 1 & 1 & -1 & -1\\ \chi_{(1,0)} & 1 & -1 & 1 & -1\\ \chi_{(1,1)} & 1 & -1 & -1 & 1\\ \end{array} \end{align*}

If this is correct, I can put it as a solution rather than an edit however feel free to critique my attempt.

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  • $\begingroup$ Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is isomorphic to the Klein-4 group, does this imply the character table will be the same? $\endgroup$ – Math Feb 13 at 12:06
  • $\begingroup$ Yes isomorphic groups have isomorphic character tables. Only possibly the rows and columns would be permuted if you had put the group elements and irreducible characters in a different order. $\endgroup$ – Ben Feb 13 at 12:25
  • $\begingroup$ @Ben just to check, is my orderings correct? $\endgroup$ – Math Feb 13 at 12:35
  • $\begingroup$ I don’t know your definition of the Klein four group, but the character table you wrote seems to just be the character table for $Z/2\oplus Z/2$, so you don’t have to worry about an isomorphism. $\endgroup$ – Ben Feb 13 at 12:43
3
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Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian, all characters are one-dimensional so they take on values $\pm 1$. So we have the same character table as the Klein-4 group:

\begin{align*} \begin{array}{c | c c c c } & (0,0) & (0,1) & (1,0) & (1,1)\\ \hline \chi_{(0,0)} & 1 & 1 & 1 & 1\\ \chi_{(0,1)} & 1 & 1 & -1 & -1\\ \chi_{(1,0)} & 1 & -1 & 1 & -1\\ \chi_{(1,1)} & 1 & -1 & -1 & 1\\ \end{array} \end{align*}

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