There is only one possible class equation of G

Question

Assume $G$ is non-abelian group of order 15. Prove that $Z(G) = 1$. Prove that there is at most one possible class equation for $G$.

So to show first thing: I used Lagrange theorem that is, since $Z(G) \leq G \implies |Z(G)| \mid |G| = 15$. The only divisors that works is 1 as others I will get contradictions. I don't know how to approach part 2 of this problem.

Your LaTeX is fine, but just add $$$'s to all your mathematics. — Krijn, Sep 20 '15 at 05:28
Where did you find this question? Any group of order $15$ is abelian, cyclic even. — Ben West, Sep 20 '15 at 06:11
@Dude I know, I'm just curious because non-abelian groups of order $15$ don't exist. — Ben West, Sep 20 '15 at 07:03

score 3 · Answer 1 · edited Sep 20 '15 at 12:37

Recall how a class equation is made of:

It consists of conjugacy classes i.e $|G|=|Z(G)|+\sum_{i=1}^n |cl(a_i)|$ where $cl(a_i)=\{xa_ix^{-1}:x\in G\}$ where $a_i$ are the distinct class representatives.Also $|(cl(a_i)|$ must divide order of the group.

Since $Z(G)$ has order 1 so $15=1+\sum |cl(a_i)|$ so $14 $ must be distributed over $3,5$ ;only possible way is $14=3+3+3+5$(Check).

Hence only possible equation is $15=1+3+3+3+5$

There is only one possible class equation of G

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