State the Sylow theorems.
Show that in a group of order 105, either the 5-Sylow or the 7-Sylow subgroup is normal.
Show that a group of order 105 has a normal subgroup of order 35.
Show that in a group of order 105, both the 5-Sylow and the 7-Sylow subgroups are normal.
Solution
1. The Sylow theorems have the following form.
For a finite group \(H\) of order \(p^a m\), with \(p\) prime and \(p\nmid m\), there is a subgroup of order \(p^a\), called a Sylow \(p\)-subgroup. Every \(p\)-subgroup of \(H\) is contained in a Sylow \(p\)-subgroup, and all Sylow \(p\)-subgroups are conjugate. Their number \(n_p\) divides \(m\) and satisfies \(n_p\equiv1\pmod p\) [@DF04]. In particular, a Sylow subgroup is normal if and only if it is the unique Sylow subgroup for its prime: conjugacy proves one implication, and conjugation preserves subgroup order for the other.
For the remaining parts, let \(|G|=105=3\cdot5\cdot7\).
2. At least one of the Sylow \(5\)- and \(7\)-subgroups is normal.
The divisibility and congruence conditions in step 1 give \[ n_5\in\{1,21\},\qquad n_7\in\{1,15\}. \] Indeed, the divisors of \(21\) are \(1,3,7,21\), of which only \(1\) and \(21\) are \(1\) modulo \(5\); the divisors of \(15\) are \(1,3,5,15\), of which only \(1\) and \(15\) are \(1\) modulo \(7\).
Distinct subgroups of prime order intersect only in the identity: any nonidentity element of such a subgroup generates it. A subgroup of order \(5\) and one of order \(7\) also intersect trivially, since the order of an element in their intersection divides both primes. Thus \(n_5=21\) and \(n_7=15\) would give \[ 21(5-1)+15(7-1)=174 \] distinct nonidentity elements in a group having only \(104\) of them. This is impossible. Hence \(n_5=1\) or \(n_7=1\).
3. There is a normal subgroup \(N\lhd G\) of order \(35\).
Choose a normal Sylow subgroup \(P\) supplied by step 2, and write \(|P|=p\) and \(\{p,q\}=\{5,7\}\). Then \(|G/P|=3q\). The number of Sylow \(q\)-subgroups of \(G/P\) divides \(3\) and is \(1\) modulo \(q\). Since \(q\geq5\), this number is \(1\). Let \(\overline Q\) be that unique, and therefore normal, subgroup of order \(q\).
For the quotient map \(\rho:G\to G/P\), set \(N=\rho^{-1}(\overline Q)\). If \(g\in G\) and \(n\in N\), normality of \(\overline Q\) gives \(\rho(gng^{-1})\in\overline Q\), so \(N\lhd G\). The restriction \(N\to\overline Q\) is surjective with kernel \(P\); its fibers are the cosets of \(P\). Hence \(|N|=pq=35\).
4. Both Sylow subgroups are normal in \(G\).
Inside \(N\), the number of Sylow \(5\)-subgroups divides \(7\) and is \(1\) modulo \(5\), so it is \(1\). The number of Sylow \(7\)-subgroups divides \(5\) and is \(1\) modulo \(7\), so it too is \(1\). Denote these unique subgroups by \(P_5\) and \(P_7\).
For \(g\in G\), normality of \(N\) implies \(gP_pg^{-1}\subseteq N\) for \(p=5,7\). This conjugate still has order \(p\), so uniqueness inside \(N\) gives \(gP_pg^{-1}=P_p\). Therefore \(P_5\) and \(P_7\) are normal in \(G\). They are Sylow subgroups of \(G\) because \(105\) contains each of \(5\) and \(7\) to the first power. By conjugacy, they are also the unique Sylow subgroups of \(G\) for their respective primes.