Let $G$ be a group of order $p^\alpha m$, and denote $\text{Syl}_p(G)$ to be the set of all Sylow $p$-subgroups. Define $n_p = |\text{Syl}_p(G)|$. Sylow's Third Theorem asserts that $n_p \equiv 1\text{ (mod $p$)}$, and $n_p \mid |G|$.
Now we fix $\alpha \in \mathbb{Z}^+$. For any $n_p$ that satisfies the two conditions stated above, can we always construct a group $G$ of order $p^\alpha m$ such that $|\text{Syl}_p(G)| = n_p$?
Thanks in advance.
Best Answer
No. For example, there is no group of order $30$ with $6$ Sylow $5$-subgroups.