Which is/are the possible class equation of a group $G$ of order $30$?
a) $1+15+2+2+2+2+2+2+2$
b) $1+1+1+6+6+15$
c) $1+1+1+1+1+5+10+10$
d) $1+1+3+10+15$
I could eliminate option (d) , as $|Z(G)|=2$, so, quotient $G/Z(G)$ is of order $15$, a cyclic group. Now this implies $G$ abelian, contradiction.
And option (a) is clearly class equation of $D_{15}$
Now, as I couldn't eliminate the other two options, could I say this that option (b) and option (c) are also right options.
Or is there any other way to say that both options are true or false (but without considering the groups)?
Best Answer
In cases b) and c), $Z(G)$ has order $3$ and $5$, and in either case, by considering $G/Z(G)$, we see that $G$ has a normal subgroup of order $15$, which must be abelian. (In fact all groups of order $30$ have a normal subgroup of order $15$, but you don't need that.)
So conjugacy classes of elements in that normal subgroup have sizes $1$ or $2$. But that is not the case for decompositions b) and c), so they are not possible.