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

Question

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?

Answer 1

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…