Let $T$ be a simple group. Can this equality hold for $l>1$?
$$|T|^l = 2q^{q^2}\left(\frac{q^2-1}{2}\right)^{q^2},$$ where $q$ is a prime at least $5$.
I think that the power of $2$ makes it look impossible, but I don't know how to show it.
For $q=5$, we get $|T|^l=2^{51}3^{25}5^{25}$. Then $2^{51}$ divides $|T|^l$ which means $l$ is either $3$ or $17$, right? If $l=3$ or $17$ then the powers of $3$ and $5$ give some contradiction, not sure how to formulate it. Anyway, any ideas for all the primes?
Observe that $q =4k+1$ or $q =4k+3$. Then, necessarily $\frac{q^2-1}{2} = 2x$ for some $x \in \mathbb{N}$. Moreover, $2x =2^nm$ for some $m,n \in \mathbb{N} $ where $gcd(2,m)=1$ and $n\geq 1$. Then $\lvert T\rvert ^l = 2^{nq^2+1}q^{q^2}m^{q^2}$. Since $q$ is a prime number, necessarily $l=q,q^2$. But observe that $q,q^2\nmid nq^2+1$. Hence, this equality cannot hold for any such $q,l$ pair.