2
$\begingroup$

Can someone please verify these? I'm quite unsure about my answer to the Quaternion Group.

Find the class equation for the following groups:

(a) The Quaternion group

(b) $D_5$

(c) $D_6$

(d) The Klein four group

(a) $1+1+2+2+2$

(b) $1+2+2+5$

(c) $1+1+3+3+2+2$

(d) $1+1+1+1$

$\endgroup$
3
  • 3
    $\begingroup$ They are correct. Not sure this is a useful question, though. $\endgroup$ Aug 21, 2015 at 5:31
  • $\begingroup$ if you mean by the class equation the size of each conjugacy class, yes. Usually though we put together all of the things in the center, which means all the ones that are 1 are put together, so a would be 2+2+2+2, D would be 4, and C would 2+3+3+2+2 $\endgroup$
    – Alan
    Aug 21, 2015 at 6:02
  • $\begingroup$ Can you show how you got these answers, which are by the way correct? $\endgroup$ Aug 21, 2015 at 6:25

1 Answer 1

5
$\begingroup$

Class equations: $|G|=|Z(G)|+\sum[G:c(a)]$, where $a \notin Z(G)$ and $|cl(a)|=[G:c(a)]$

$Q_8=\{1,-1,i,-i,j,-j,k,-k\}$

$Z(Q_8)=\{1,-1\}$. $c(i)=\{1,-1,i,-i\}$,$|c(i)|=|c(j)|=|c(k)|=4$. Therefore $|cl(i)|=|cl(j)|=|cl(k)|=2$

$|Q_8|=2+2+2+2$.

$D_5=\{e,r,r^2,r^3,r^4,s,rs,r^2s,r^3s,r^4s\}$, $r^5=s^2=e,rs=sr^4$ . $Z(D_5)=\{e\},cl(r)=\{r,r^4\},cl(r^2)=\{r^3,r^2\},cl(s)=\{rs,r^2s,r^3s,r^4s,s\}$

$|D_5|=1+2+2+5$.

$D_6=\{e,r,r^2,r^3,r^4,r^5,s,rs,r^2s,r^3s,r^4s,r^5s\}$, $r^6=s^2=e,rs=sr^5$

$Z(D_6)=\{e,r^3\},cl(r)=\{r,r^5\},cl(r^2)=\{r^2,r^4\},cl(s)=\{s,r^2s,r^4s\},cl(rs)=\{rs,r^3s,r^5s\}$

$|D_6|=2+2+2+3+3$

$V_4=\{e,a,b,c\},a^2=b^2=c^2=1$ (since $V_4$ is abelian group)

$|V_4|=4(Z(V_4))$

$\endgroup$

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.