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 Sylow $p$-subgroups of G. Let $P \in X$, let N be the normalizer of P and $r = \mathrm{Card}(X)$.

(a) Show that $r = (G:N)$. Deduce that $r$ belongs to the set R of positive divisors of $m$ which are congruent to 1 $(\bmod\ p)$. In particular, if R is $\{1\}$, P is normal in G.

(b) If $a = 1$, show that the number of elements of G of order $p$ is $r(p - 1)$. If further $N = P$, show that the elements of G of order different from $p$ are equal in number to $m$; if further $m$ is a power of a prime number $q$, deduce that a Sylow $q$-subgroup Q of G contains all the elements of G of order different from $p$ and hence is normal in G.

(c) Let H be a group of order 210. Show that H contains a normal subgroup G of order 105 (cf. Exercise 29). Let $p = 3$ (resp. 5, 7). Show that, if G has no normal Sylow $p$-subgroup, the group G contains at least 14 (resp. 84, 90) elements of order $p$. Conclude that H and G have property (ST) (cf. the preceding exercise).

(d) Show that every group of order $n \leq 100$ has property (ST), except possibly if $n = 24, 36, 48, 60, 72, 96$ (same method as in (c)).

(e) Show that the group $\mathfrak{S}_4$, of order 24, does not have property (ST). Construct analogous examples for $n = 48, 60, 72, 96$. (For $n = 36$, see the following exercise.)