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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?

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  • $\begingroup$ Note that the size of a conjugacy class is the index of a centralizer, so it must be a factor of the group order. $\endgroup$ Nov 3, 2013 at 0:01
  • $\begingroup$ @GerryMyerson: But doesn't all the equations above satisfy your condition? $\endgroup$
    – user104221
    Nov 3, 2013 at 0:31
  • $\begingroup$ All except the one with the 9 in it. Nicky's answer will be more useful to you, but this is one additional check that may be handy sometimes. $\endgroup$ Nov 3, 2013 at 0:43
  • $\begingroup$ 2,3,4 are definitely not the possible options. $\endgroup$ Sep 18, 2017 at 13:57

4 Answers 4

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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.

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  • $\begingroup$ Why would we consider that $G$ is non-abelian? There is no information given. So, wouldn't there be two possibilities namely first and last? Also, if $|G/Z(G)|$ is prime then it means that $G$ is abelian which should be fine too? Could you please say if I am on the correct path? $\endgroup$
    – user104221
    Nov 3, 2013 at 0:26
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    $\begingroup$ Nicky didn't say $G$ is not abelian, just that you can pick out the abelian group(s) from the class equation. And if $G$ is abelian, then the center is the whole group, so that quotient is 1. $\endgroup$ Nov 3, 2013 at 0:45
  • $\begingroup$ @GerryMyerson: I see, so we can have abelian groups. So, following that chain of thought 2 and 3 will also be valid. Right? This is in regards to my other concern of $G$ being abelian $\endgroup$
    – user104221
    Nov 3, 2013 at 0:51
  • $\begingroup$ I'm not sure I understand your question, but if a group is abelian, then its class equation is guaranteed to be $1+1+\cdots+1$. For all $x$ and $g$, $g^{-1}xg=x$ in an abelian group. $\endgroup$ Nov 3, 2013 at 2:58
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    $\begingroup$ @user104221 - no, 2 and 3 are not correct since in that case you have $|G/Z(G)|$ equals 3 or 7 which cannot be the case as I pointed out. By the way there is a nice theorem of Burnside that says that in general if $G$ has odd order then $|G| \equiv s$ mod 16. Here $s$ denotes the number of conjugacy classes of $G$. That gives you $s$=5, if $G$ is non-abelian. $\endgroup$ Nov 3, 2013 at 10:42
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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)|

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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

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  • $\begingroup$ Welcome to this site! The question you have just answered is rather old and has perfectly good answers, to which your post does not seem to add much. More value to this site would be added by answering unanswered questions, or posting answers with significantly new information. $\endgroup$ Jun 24, 2015 at 9:05
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I think $G$ is abelian. So $|Z(G)|=21$. Hence class equation is all of $1$'s.

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  • $\begingroup$ There is an abelian group of order 21. But there is also a nonabelian group of order 21. $\endgroup$ Nov 4, 2013 at 12:26
  • $\begingroup$ Can you give me some example of non-abelian group of oder 21?@GerryMyerson $\endgroup$
    – abelian
    Nov 4, 2013 at 12:59
  • $\begingroup$ Have a look at math.stackexchange.com/questions/117500/… $\endgroup$ Nov 4, 2013 at 23:32

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