Let \(G\) be a group of order \(240 = 2^4 \cdot 3 \cdot 5\).
How many \(p\)-Sylow subgroups might \(G\) have, for \(p = 2, 3, 5\)?
If \(G\) has a subgroup of order 15, show that it has an element of order 15.
Say \(G\) doesn’t have a subgroup of order 15. Show that the number of \(3\)-Sylows is 10 or 40.
Solution
Part (a).
1. \(n_2 \equiv 1 \pmod 2\) and \(n_2 \mid 15\), so \(n_2 \in \{1, 3, 5, 15\}\).
Sylow’s third theorem; the odd divisors of \(15\) are \(1, 3, 5, 15\).
2. \(n_3 \equiv 1 \pmod 3\) and \(n_3 \mid 80\), so \(n_3 \in \{1, 4, 10, 16, 40\}\).
the divisors of \(80 = 2^4 \cdot 5\) that are \(\equiv 1 \pmod 3\) are \(1, 4, 10, 16, 40\).
3. \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid 48\), so \(n_5 \in \{1, 6, 16\}\).
the divisors of \(48 = 2^4 \cdot 3\) that are \(\equiv 1 \pmod 5\) are \(1, 6, 16\).
Part (b).
1. Let \(H \le G\) be a subgroup of order \(15 = 3 \cdot 5\).
hypothesis.
2. \(H\) has a normal Sylow \(5\)-subgroup.
\(n_5(H) \equiv 1 \pmod 5\) and \(n_5(H) \mid 3\), so \(n_5(H) = 1\).
3. \(H\) has a normal Sylow \(3\)-subgroup.
\(n_3(H) \equiv 1 \pmod 3\) and \(n_3(H) \mid 5\), so \(n_3(H) = 1\).
4. Hence \(H \cong \ZZ/3 \times \ZZ/5 \cong \ZZ/15\).
both Sylow subgroups are normal and intersect trivially, so \(H\) is their direct product, which is cyclic of order \(15\).
5. Therefore \(H\) (and hence \(G\)) has an element of order \(15\).
a cyclic group of order \(15\) has a generator of order \(15\).
6. Q.E.D. for parts (a) and (b).
3 (a) and 5 (b).
Part (c).
1. Let \(P\) be a Sylow \(3\)-subgroup of \(G\).Then\[ |N_G(P)|=\frac{|G|}{n_3}=\frac{240}{n_3}. \]
The conjugation action of \(G\) on its Sylow \(3\)-subgroups is transitive, and the stabilizer of \(P\) is \(N_G(P)\). Orbit–stabilizer therefore gives \(n_3=[G:N_G(P)]\).
2. Under the hypothesis that \(G\) has no subgroup of order \(15\), one has \(5\nmid |N_G(P)|\).
The subgroup \(P\) is normal in \(N_G(P)\) by definition of the normalizer. If \(5\mid |N_G(P)|\), Cauchy’s theorem gives a subgroup \(Q\le N_G(P)\) of order \(5\). Since \(P\trianglelefteq N_G(P)\), the product \(PQ\) is a subgroup of \(N_G(P)\), and \[ |PQ|=\frac{|P||Q|}{|P\cap Q|}=3\cdot5=15, \] because \(P\cap Q=1\). This contradicts the hypothesis.
3. Therefore \(n_3\) is either \(10\) or \(40\).
2. Thus only \(n_3=10\) or \(40\) remain.
4. Q.E.D.
Combine <1>1–<1>3.