Determine the possible class equation for a group of order 21?
Until now I have found the following:
$1+3+3+7+7$
$1+1+1+3+3+3+9$
$1+1+1+1+1+1+1+7+7$
$1+1+1+1+1+1+1+1+1+3+3+3+3$
$1+1+1+\cdots +1 \ (21 \ \text{times})$
Is there any way to eliminate the choices from this equation? More importantly, how would we know that this is a complete list (Until now my attempt has just been guess and check after I found possible occurences of 1's) Is there any easier way to determine the class equation?
$\endgroup$ 44 Answers
$\begingroup$Hint: the number of 1's is the order of $Z(G)$, the center of $G$. Also, $|G/Z(G)|$ can not be a prime number. And of course $|Z(G)|$ divides the order of $G$. This leaves you with the first and last possibilities. In the last case the group is abelian.
$\endgroup$ 11 $\begingroup$Given that |G|=21.Obviously,second,third and forth can be not possible class equations . Because from 1+1+1+1+1+1+1+7+7 ,we see that |Z(G)|=7 and |G/Z(G)|=3, which is prime.So G is abelien.Then we must have G=Z(G),which implies that |G|=|Z(G)| but |Z(G)|=7. Again from 1+1+1+3+3+3+9,we see that |Z(G)|=3 and |G/Z(G)|=7, which is prime.So G is abelian.So we must have G=Z(G) and |G|=|Z(G)| but |Z(G)|=3.finally from 1+1+1+1+1+1+1+1+1+3+3+3+3,we see that |Z(G)|=9, also we know Z(G) is a subgroup of G and |G| is divisible by |Z(G)| but |Z(G)|=9 ,|G| is not divisible by |Z(G)|
$\endgroup$ $\begingroup$Second and third are not possible to be the class equation of a group of order 21,due to lack of the divisions of 21 by 9 and the total centre 9 respectively. but last one is surely possible due to the presence of a Abelian group of of order 21 and others have to be checked by the possible conjugacy class of the elements of 21
$\endgroup$ 1 $\begingroup$I think $G$ is abelian. So $|Z(G)|=21$. Hence class equation is all of $1$'s.
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