It is known that the symmetric group $S_3$ of order $1+2+3$ has the class equation $1+2+3$. But a non-abelian group of order $10 = 1+2+3+4$(which is just $D_5$) cannot have the class equation $1+2+3+4$. My question is the following: "are there finite groups $G$ (other than $S_3$) of order $\frac{n(n+1)}{2}(n \in \Bbb N$) with class equation $1+2+3+\dots+n$ "
I've no idea how to find an example of this/disprove this claim. Kindly shed some light on this matter.
Best Answer
Note that one necessary condition for this to occur is that $\binom{n+1}{2} = \frac{(n+1)n}{2}$ must be divisible by $1, 2, \ldots, n-1, n$, since all the terms in the class equation are indices of centralizers, hence are divisors of the order of the group. Since $n$ and $n-1$ are relatively prime, this means that $n(n-1) \mid \frac{(n+1)n}{2}$, so $n(n-1) \leq \frac{(n+1)n}{2}$. But this implies \begin{align*} n-1 \leq \frac{n+1}{2} \implies 2n - 2 \leq n+1 \implies n \leq 3 \, . \end{align*}
Since $n=2$ corresponds to $\mathbb{Z}/3\mathbb{Z}$ which has class equation $1+1+1$, then $S_3$ and the trivial group are the only examples with class equation $1 + \cdots + n$.