2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Math Feb 13 at 15:05
  • $\begingroup$ so what are the possibilities for the character table? $\endgroup$ – Math Feb 14 at 11:56
0
$\begingroup$

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…

$\endgroup$
  • $\begingroup$ sorry I don't understand. the characters of $\mathbb{Z}_2$ are known? $\endgroup$ – Math Feb 13 at 15:24
  • 1
    $\begingroup$ @Math Can you answer such questions without knowing the characters of $\mathbb{Z}_2$? $\endgroup$ – egreg Feb 13 at 15:29
  • 1
    $\begingroup$ sorry I misread your answer, I though it said characters of $\mathbb{Z}_2$ are unknown $\endgroup$ – Math Feb 13 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.