# Characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$

Follow up question to $$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$, how would I find the characters of $$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$?

The Cayley table for this group:

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

EDIT:

The character table will have the form:

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

But what is a possibility to make the table correct?

• As this group is also abelian, the characters will take values $\pm 1$ however I don't know whether to use $1$ or $-1$ for any given element – Math Feb 13 at 15:05
• so what are the possibilities for the character table? – Math Feb 14 at 11:56

If $$G$$ is an abelian group, a character of $$G$$ is a homomorphism $$\chi\colon G\to\mathbb{C}^*$$ (multiplicative group of nonzero complex numbers).

Suppose $$G_1$$ and $$G_2$$ are abelian groups and let $$f_1\colon G_1\to G_1\oplus G_2$$ and $$f_2\colon G_2\to G_1\oplus G_2$$ be the canonical embeddings.

Then a homomorphism $$\chi\colon G_1\oplus G_2\to\mathbb{C}^*$$ is completely determined by $$\chi\circ f_1$$ and $$\chi\circ f_2$$.

Similarly if you do the direct sum of more than two groups.

As the characters of $$\mathbb{Z}_2$$ are known…

• sorry I don't understand. the characters of $\mathbb{Z}_2$ are known? – Math Feb 13 at 15:24
• @Math Can you answer such questions without knowing the characters of $\mathbb{Z}_2$? – egreg Feb 13 at 15:29
• sorry I misread your answer, I though it said characters of $\mathbb{Z}_2$ are unknown – Math Feb 13 at 15:32