Exercise 38, § 6

Let G be a finite group of order $n = p^a m$, where $p$ is a prime number not dividing $m$, and let X be the set of the Sylow $p$-subgroups of G. Let P $\in$ X, let N be the normalizer of P and let $r = \mathrm{Card}(X)$. a) Now prove that $r = (G : N)$. Deduce that $r$ belongs to the set R of the positive divisors of $m$ which are congruent to 1 (mod $p$). In particular, if R is reduced to 1, P is distinguished in G. b) If $a = 1$, now prove that the number of elements of G of order $p$ is $r(p - 1)$. If moreover N = P, now prove that the elements of G of order different from $p$ are equal in number to $m$; if moreover $m$ is a power of a prime number $q$, deduce that a $q$-Sylow subgroup Q of G contains all the elements of G of order different from $p$, hence is distinguished in G. c) Let H be a group of order 210. Now prove that H contains a distinguished subgroup G of order $n = 105$ (cf. I, p. 139, exerc. 29). Let $p = 3$ (resp. 5, 7). Now prove that, if G has no distinguished $p$-Sylow subgroup, the group G contains at least 14 (resp. 84, 90) elements of order $p$. Conclude that H and G possess the property (ST) (cf. preceding exercise). d) Now prove that every group of order $n \leq 100$ possesses the property (ST), unless perhaps if $n = 24, 36, 48, 60, 72$ or 96 (same methods as in c)). e) Now prove that the group $\mathfrak{S}_4$, of order 24, does not possess the property (ST). Construct analogous examples for $n = 48, 60, 72, 96$. (For $n = 36$, see the following exercise.)