Theorem
The Sylow theorems
Statement
Let \(G\) be a finite group with \(|G| = p^n m\), where \(p\) is prime, \(n \geq 0\), and \(p \nmid m\). A subgroup of order \(p^n\) is called a Sylow \(p\)-subgroup of \(G\); write \(\mathrm{Syl}_p(G)\) for the set of all Sylow \(p\)-subgroups and \(n_p = |\mathrm{Syl}_p(G)|\). Then: (Existence) \(\mathrm{Syl}_p(G) \neq \varnothing\), i.e. \(G\) contains a subgroup of order \(p^n\). (Conjugacy) \(G\) acts transitively on \(\mathrm{Syl}_p(G)\) by conjugation; equivalently, any two Sylow \(p\)-subgroups are conjugate in \(G\), and every \(p\)-subgroup of \(G\) is contained in some Sylow \(p\)-subgroup. (Counting) \(n_p \equiv 1 \pmod p\) and \(n_p \mid m\); moreover \(n_p = [G : N_G(P)]\) for any \(P \in \mathrm{Syl}_p(G)\).
Why it matters
Lagrange's theorem tells you which subgroup orders are possible (divisors of \(|G|\)) but says nothing about which are achieved. The Sylow theorems close that gap for the maximal prime-power divisors: they guarantee existence where Lagrange is silent, and they hand you sharp numerical constraints (\(n_p \equiv 1 \pmod p\), \(n_p \mid m\)) that are often strong enough to pin down \(n_p = 1\) outright, forcing a normal subgroup out of thin air.
This is the single most powerful elementary tool for classifying finite groups of a given order and for proving non-simplicity: almost every "show \(G\) is not simple" or "classify groups of order \(pq\)/\(p^2q\)/..." argument in a first course is a Sylow counting argument in disguise.
Hypotheses
\(G\) is a finite group.Infinite groups need not have any finite-index maximal-\(p\)-power subgroup at all in the relevant sense — e.g. \((\mathbb{Q}, +)\) is a \(p\)-divisible group with no subgroup of index a power of anything meaningful; the entire counting argument relies on finite orbits and finite index, so it evaporates without finiteness.
\(p\) is prime.The statement is about the prime factorisation \(|G| = p^n m\) with \(p \nmid m\); for composite \(q\) there is generally no subgroup of order \(q\) even when \(q \mid |G|\) — e.g. \(A_4\) has order \(12\) but no subgroup of order \(6\), so "Sylow for \(q=6\)" fails completely.
\(p^n\) is the exact power of \(p\) dividing \(|G|\) (i.e. \(p \nmid m\)).If you instead ask for a subgroup of order \(p^k\) with \(k \lt n\), existence still holds (by the same Sylow subgroup, or directly by Cauchy plus induction) but it is a different, weaker statement — the sharp counting congruence \(n_p \equiv 1 \pmod p\) and the conjugacy statement are specifically about the maximal subgroups of order \(p^n\); subgroups of order \(p^k\), \(k \lt n\), need not be unique up to conjugacy in the same controlled way and are not what \(n_p\) counts.
Proof
We prove existence via a group action on subsets (the standard action-counting argument), then conjugacy and containment via a second action, then the counting statement by combining the two.
1
\text{Existence. Let } |G| = p^n m,\ p \nmid m. \text{ Let } \Omega \text{ be the set of all subsets of } G \text{ of size } p^n.
Set-up for a counting argument; \(|\Omega| = \binom{p^n m}{p^n}\). A
2
p \nmid \binom{p^n m}{p^n}
Kummer's theorem / direct valuation: writing \(\binom{p^n m}{p^n} = \prod_{i=0}^{p^n - 1} \frac{p^n m - i}{p^n - i}\), pair each factor \(p^n m - i\) with \(p^n - i\); both have the same \(p\)-adic valuation \(v_p(p^n - i)\) since \(p^n \mid p^n m\), so each ratio contributes \(p\)-valuation \(0\), and the product is coprime to \(p\). C
3
G \text{ acts on } \Omega \text{ by left translation: } g \cdot A = gA \text{ for } A \in \Omega.
Left translation is a group action on subsets, since \((gh)A = g(hA)\) and \(1 \cdot A = A\), and translation preserves cardinality so \(gA \in \Omega\). A
4
\Omega = \bigsqcup_j \mathcal{O}_j \ \Rightarrow\ |\Omega| = \sum_j |\mathcal{O}_j|,\quad p \nmid |\Omega| \Rightarrow \exists\, j_0:\ p \nmid |\mathcal{O}_{j_0}|.
The orbit decomposition partitions \(\Omega\); since \(p\) does not divide the total (Step 2), it cannot divide every summand. A
5
\text{Fix } A \in \mathcal{O}_{j_0}\text{ and let } H = \mathrm{Stab}_G(A) = \{g \in G : gA = A\}. \text{ Then } |\mathcal{O}_{j_0}| = [G:H] \text{ (Orbit–Stabiliser).}
Orbit–Stabiliser theorem for a group action. B
6
\text{Fix } a \in A. \text{ The map } H \to A,\ h \mapsto ha \text{ is injective, so } |H| \leq |A| = p^n.
If \(h_1 a = h_2 a\) then \(h_1 = h_2\) by cancellation in \(G\); the map lands in \(A\) because \(hA = A\) for \(h \in H\). B
7
p \nmid [G:H] = \frac{p^n m}{|H|}\ \text{ and } |H| \leq p^n \ \Rightarrow\ p^n \mid |H| \ \Rightarrow\ |H| = p^n.
From Step 4, \(p \nmid [G:H]\); writing \([G:H] = p^n m / |H|\), if \(|H| \lt p^n\) then a factor of \(p\) would survive in \([G:H]\) unless it is cancelled — precisely, \(|H|\) must absorb the full \(p^n\), else \([G:H]\) retains a positive power of \(p\), contradicting \(p \nmid [G:H]\); combined with \(|H| \leq p^n\) this forces equality. C
8
\therefore H \leq G \text{ is a subgroup with } |H| = p^n, \text{ i.e. } H \in \mathrm{Syl}_p(G) \neq \varnothing.
\(H\) is a stabiliser, hence a genuine subgroup of \(G\), of the required order. This proves existence. B
9
\text{Conjugacy and containment. Fix } P \in \mathrm{Syl}_p(G) \text{ (exists by Step 8). Let } Q \leq G \text{ be any } p\text{-subgroup.}
Set-up for the second action; we show \(Q\) is contained in a conjugate of \(P\). A
10
\text{Let } X = G/P = \{gP : g \in G\},\ |X| = [G:P] = m,\ p \nmid m. \text{ Let } Q \text{ act on } X \text{ by left translation.}
The coset space \(X\) carries a natural \(G\)-action by left translation, which restricts to an action of the subgroup \(Q\). B
11
X = \bigsqcup_i \mathcal{O}_i^Q,\quad |\mathcal{O}_i^Q| \text{ divides } |Q|, \text{ a power of } p.
Orbit sizes for a \(Q\)-action divide \(|Q|\) by Orbit–Stabiliser; since \(|Q|\) is a power of \(p\), each orbit has size \(1\) or a positive power of \(p\). A
12
p \nmid |X| = m \ \Rightarrow\ \text{not every orbit has size divisible by } p \ \Rightarrow\ \exists\, \text{a fixed point } gP \in X\ (|\mathcal{O}|=1).
If every orbit had size a multiple of \(p\), their sum \(|X|\) would be divisible by \(p\), contradicting \(p \nmid m\); so some orbit has size \(p^0 = 1\), i.e. a fixed point. B
13
gP \text{ fixed by } Q \ \Rightarrow\ \forall q \in Q:\ qgP = gP \ \Rightarrow\ g^{-1}qg \in P \ \Rightarrow\ Q \leq gPg^{-1}.
Unwinding "\(qgP=gP\)" gives \(g^{-1}qg \in P\) for every \(q \in Q\), i.e. \(g^{-1}Qg \leq P\), equivalently \(Q \leq gPg^{-1}\). B
14
\therefore \text{ every } p\text{-subgroup } Q \text{ lies in some conjugate } gPg^{-1} \text{ of } P.
This is the containment statement; taking \(Q\) itself to be a Sylow \(p\)-subgroup gives conjugacy, proved next. B
15
\text{If } Q \in \mathrm{Syl}_p(G) \text{ too, then } |Q| = p^n = |gPg^{-1}| \text{ and } Q \leq gPg^{-1} \ \Rightarrow\ Q = gPg^{-1}.
A subgroup contained in another subgroup of the same finite order equals it; \(|gPg^{-1}| = |P| = p^n\) since conjugation is an automorphism of \(G\). This proves conjugacy: all Sylow \(p\)-subgroups are conjugate. A
16
\text{Counting, part 1. } G \text{ acts transitively on } \mathrm{Syl}_p(G) \text{ by conjugation (Step 15)} \ \Rightarrow\ n_p = |\mathrm{Syl}_p(G)| = [G:N_G(P)].
Orbit–Stabiliser applied to the conjugation action on \(\mathrm{Syl}_p(G)\): the orbit of \(P\) is all of \(\mathrm{Syl}_p(G)\) by transitivity, and the stabiliser of \(P\) under conjugation is by definition \(N_G(P) = \{g \in G : gPg^{-1}=P\}\). B
17
P \leq N_G(P) \leq G \ \Rightarrow\ n_p = [G:N_G(P)] \ \bigm|\ [G:P] = m.
\(P\) normalises itself, so \(P \leq N_G(P)\); by the tower law \([G:P] = [G:N_G(P)]\,[N_G(P):P]\), so \([G:N_G(P)]\) divides \([G:P]=m\). This gives \(n_p \mid m\). A
18
\text{Counting, part 2. Let } P \text{ act on } \mathrm{Syl}_p(G) \text{ by conjugation. Then } n_p \equiv |\mathrm{Fix}_P(\mathrm{Syl}_p(G))| \pmod p.
As in Step 11–12, every \(P\)-orbit on the finite set \(\mathrm{Syl}_p(G)\) has size dividing \(|P|=p^n\), hence size \(1\) or a multiple of \(p\); summing orbit sizes, the total is congruent mod \(p\) to the number of size-\(1\) orbits (fixed points). C
19
Q \in \mathrm{Syl}_p(G) \text{ fixed by } P\text{-conjugation} \ \iff\ P \leq N_G(Q).
\(Q\) is fixed by every \(x \in P\) iff \(xQx^{-1}=Q\) for all \(x \in P\), i.e. \(P \leq N_G(Q)\), by definition of normaliser. B
20
P \leq N_G(Q) \ \Rightarrow\ P,\,Q \text{ are both Sylow } p\text{-subgroups of } N_G(Q), \text{ and } Q \trianglelefteq N_G(Q) \ \Rightarrow\ Q \text{ is the unique one} \ \Rightarrow\ P=Q.
\(Q\) is normal in its own normaliser by definition, so it is the only Sylow \(p\)-subgroup of \(N_G(Q)\) — any conjugate of \(Q\) inside \(N_G(Q)\) equals \(Q\) itself by normality, and conjugacy of Sylow subgroups (Step 15) applied inside \(N_G(Q)\) forces the Sylow \(p\)-subgroup \(P\) of \(N_G(Q)\) to be conjugate to, hence equal to, \(Q\). C
21
\therefore |\mathrm{Fix}_P(\mathrm{Syl}_p(G))| = 1 \ \Rightarrow\ n_p \equiv 1 \pmod p.
Combining Steps 18–20: \(P\) itself is always a fixed point, and Step 20 shows it is the only one; substituting into Step 18 gives the congruence. This completes the proof. B
Result
n_p \equiv 1 \pmod p, \qquad n_p \mid m, \qquad n_p = [G:N_G(P)], \qquad \mathrm{Syl}_p(G) \text{ is a single conjugacy class}
Reading. Every finite group has subgroups realising the largest possible power of \(p\) dividing its order; all such subgroups look the same up to relabelling (conjugation), every smaller \(p\)-subgroup sits inside one of them, and the number of them is tightly constrained: it divides the "co-\(p\)-part" \(m\) of \(|G|\) and leaves remainder \(1\) on division by \(p\).
Scope. Applies to any finite group \(G\) and any prime \(p\) dividing \(|G|\) (if \(p \nmid |G|\), take \(n=0\) and \(P=\{1\}\) trivially). Does not directly extend to infinite groups, nor to primes not dividing \(|G|\) in any nontrivial way, nor to non-maximal prime-power subgroups as a *conjugacy* statement (only as a containment statement, via Step 14).
Corollaries & converses
- Cauchy's theorem is the case \(n=1\): if \(p \mid |G|\) then \(G\) has an element of order \(p\) (any non-identity element of a Sylow \(p\)-subgroup, of order a power of \(p\), has some power of order exactly \(p\)).
- Normality criterion. \(P \trianglelefteq G\) for \(P \in \mathrm{Syl}_p(G)\) if and only if \(n_p = 1\); this is the standard route to proving a group of a given order is not simple.
- Every \(p\)-subgroup is subconjugate to every Sylow \(p\)-subgroup (Step 14), and a Sylow \(p\)-subgroup is a maximal \(p\)-subgroup of \(G\) (if \(P \lneq K\) with \(K\) a \(p\)-group then \(|K| \gt p^n\), impossible).
- Converse fails: the numerical constraints \(n_p \equiv 1 \pmod p\) and \(n_p \mid m\) are necessary but not sufficient — not every divisor \(d\) of \(m\) with \(d \equiv 1 \pmod p\) is realised as \(n_p\) for some group of that order; the theorems narrow the candidates for \(n_p\) but do not by themselves guarantee every candidate occurs.
- If \(G\) is abelian, \(n_p = 1\) always (conjugation is trivial, so the single conjugacy class of Step 15 has one element), recovering existence and uniqueness of the \(p\)-primary component from the structure theorem for finite abelian groups.
Fails without
- Drop finiteness: in \((\mathbb{Q},+)\), every element is infinitely \(p\)-divisible and there is no subgroup that plays the role of a "maximal \(p^n\)-subgroup" for any finite \(n\); more instructively, in the Prüfer group \(\mathbb{Z}(p^\infty)\) (a countable \(p\)-group), the group itself has no maximal subgroup at all, so "Sylow subgroups form one conjugacy class of maximal \(p\)-subgroups" has no analogue — the whole finite-index/finite-orbit machinery (Steps 1–4, 10–12) breaks down without a finite group to act on finite sets.
- Drop primality of \(p\): take \(G = A_4\), \(|G|=12\). For the composite "prime power target" \(6 = 2 \cdot 3\), \(A_4\) has no subgroup of order \(6\), even though \(6 \mid 12\) — Lagrange's converse is exactly what Sylow rescues, but only at prime-power indices; the theorem is silent (correctly) about the divisor \(6\).
- Confusing "maximal \(p^n\)-subgroup" with "any \(p^k\)-subgroup", \(k \lt n\): such subgroups exist (Cauchy plus induction, or subgroups of a Sylow subgroup) but are not asserted to form a single conjugacy class by this theorem as stated for non-maximal \(k\) in isolation from a Sylow subgroup — e.g. in \(G=\mathbb{Z}/p^2\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}\) taken non-abelian analogues, care is needed since abelian examples trivialise conjugacy; the genuinely instructive breakdown is that the *counting* formula \(n_p \equiv 1 \pmod p\) is proved (Steps 18–21) using the fact \(P\) is self-normalising-relative-uniqueness in \(N_G(P)\), an argument that used maximality of \(P\) as a \(p\)-subgroup essentially.
Common errors
- Writing "\(n_p \equiv 1 \pmod{p}\) and \(n_p \mid |G|\)" — the divisibility is by \(m = |G|/p^n\), the index, not by \(|G|\) itself (since \(p \mid |G|\) would often contradict \(n_p \equiv 1\)).
- Concluding \(P \trianglelefteq G\) directly from "\(P\) is a Sylow subgroup" — normality only follows when the extra fact \(n_p=1\) has been separately established; a Sylow subgroup is not normal in general (e.g. \(S_3\) has three Sylow \(2\)-subgroups).
- Believing all subgroups of order \(p^n\) style reasoning applies at non-maximal prime-power orders \(p^k\) (\(k \lt n\)) — those need not all be conjugate, only Sylow (maximal) ones are guaranteed to be.
- Computing \(n_p\) by solving \(n_p \equiv 1 \pmod p\), \(n_p \mid m\) and assuming the *unique* such divisor (when one exists) is automatically the right count — the constraints only narrow the possibilities; further group-specific argument (e.g. counting elements, using simplicity) is often required to pin down \(n_p\) exactly when several candidates survive.
- Forgetting \(N_G(P) \geq P\), and so computing \([G:N_G(P)]\) as if \(N_G(P))\) could be smaller than \(P\); by definition every element of \(P\) normalises \(P\), so \(P \leq N_G(P)\) always.
Discussion
The three parts of the theorem are traditionally attributed jointly to Ludwig Sylow (1872), though the proof given here — via group actions on sets, rather than Sylow's original more computational approach — is due to Helmut Wielandt (1959) and is now the standard textbook route because it proves all three parts uniformly from a single idea: count orbits modulo \(p\) and locate a fixed point or an orbit of size coprime to \(p\).
Conceptually, the Sylow theorems are a vast generalisation of Cauchy's theorem, and historically they were discovered as the natural strengthening needed to control the structure of finite groups once "does an element of order \(p\) exist" (Cauchy, 1845) was understood but "does a full \(p^n\)-subgroup exist, and how many are there" was not. The counting congruence \(n_p \equiv 1 \pmod p\) is what elevates the theorem from a mere existence statement to a genuinely computational tool: for many small orders \(|G|\), the constraints \(n_p \mid m\), \(n_p \equiv 1 \pmod p\) leave only \(n_p=1\) as a possibility, immediately producing a normal subgroup and hence non-simplicity — this is the engine behind the classification of groups of order up to (roughly) 100 taught in a first course.
More structurally, the Sylow theorems are the finite-group shadow of a much broader phenomenon: the existence and conjugacy of maximal tori / Borel subgroups in algebraic groups, and more generally the theory of "Hall subgroups" for soluble groups (P. Hall's generalisation replaces the single prime \(p\) with an arbitrary set of primes \(\pi\), but the clean divisibility+conjugacy package only survives in full for soluble groups — for general finite groups, Hall \(\pi\)-subgroups need not exist). Seen this way, Sylow's theorem is the base case \(\pi = \{p\}\), which always works because every finite group is trivially "\(p\)-soluble" for the counting argument's purposes.
Common misconception: that "Sylow subgroup" means "normal subgroup of prime-power order" — it is neither required to be normal, nor is normality part of its definition; conflating the two is the single most common source of invalid proofs by students first meeting these theorems.
Worked examples
1
\text{Show every group of order } 15 \text{ is cyclic.}
Set-up: \(|G|=15=3 \cdot 5\). A
2
n_5 \mid 3,\ n_5 \equiv 1 \pmod 5 \ \Rightarrow\ n_5 = 1.
Only divisor of \(3\) congruent to \(1 \bmod 5\) is \(1\) (candidates \(1,3\); \(3 \not\equiv 1 \bmod 5\)). Counting theorem applied with \(p=5\), \(m=3\). B
3
n_3 \mid 5,\ n_3 \equiv 1 \pmod 3 \ \Rightarrow\ n_3 = 1.
Divisors of \(5\) are \(1,5\); \(5 \equiv 2 \bmod 3\), so only \(n_3=1\) survives. Counting theorem with \(p=3\), \(m=5\). B
4
P_5 \trianglelefteq G,\ P_3 \trianglelefteq G,\ P_5 \cap P_3 = \{1\},\ |P_5 P_3| = |P_5||P_3| = 15 = |G| \ \Rightarrow\ G = P_5 \times P_3.
\(n_p=1 \Rightarrow\) unique Sylow subgroup is normal (Corollaries); intersection is trivial by Lagrange (\(|P_5 \cap P_3|\) divides \(\gcd(5,3)=1\)); the internal direct product criterion (two normal subgroups, trivial intersection, product equal to \(G\)) applies. B
5
P_5 \cong \mathbb{Z}/5\mathbb{Z},\ P_3 \cong \mathbb{Z}/3\mathbb{Z} \ \Rightarrow\ G \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}.
Groups of prime order are cyclic (generated by any non-identity element, Lagrange); \(\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}\) since \(\gcd(5,3)=1\) (CRT for cyclic groups). A
\text{Every group of order } 15 \text{ is cyclic, i.e. } \cong \mathbb{Z}/15\mathbb{Z}
Reading. Sylow counting forces both Sylow subgroups to be unique, hence normal, hence the group splits as their direct product, which is forced to be cyclic.
1
\text{Show no group of order } 56 \text{ is simple.}
Set-up: \(|G|=56=2^3 \cdot 7\). A
2
n_7 \mid 8,\ n_7 \equiv 1 \pmod 7 \ \Rightarrow\ n_7 \in \{1, 8\}.
Divisors of \(8\) are \(1,2,4,8\); of these, \(1 \equiv 1\) and \(8 \equiv 1 \pmod 7\) both survive, while \(2,4\) do not. Counting theorem with \(p=7\), \(m=8\). B
3
\text{Suppose } G \text{ simple} \ \Rightarrow\ n_7 \neq 1 \ \Rightarrow\ n_7 = 8.
If \(n_7=1\) the unique Sylow \(7\)-subgroup would be normal (Corollaries), contradicting simplicity (assuming \(G\) nontrivial and not itself of prime order, clear here since \(56\) is composite and \(\neq 7\)). A
4
\text{Each Sylow 7-subgroup has order 7 (prime)} \ \Rightarrow\ \text{any two distinct ones intersect trivially.}
A subgroup of prime order has no nontrivial proper subgroups (Lagrange); the intersection of two distinct such subgroups is a proper subgroup of each, hence trivial. B
5
\text{Elements of order 7 counted: } 8 \times (7-1) = 48 \text{ distinct non-identity elements.}
Each of the \(8\) Sylow \(7\)-subgroups contributes \(6\) elements of order \(7\) (all non-identity elements, since the group has prime order), and by Step 4 these sets are pairwise disjoint across the \(8\) subgroups. B
6
56 - 48 = 8 \text{ elements remain for a single Sylow 2-subgroup of order } 8 \ \Rightarrow\ n_2 = 1.
Any Sylow \(2\)-subgroup (order \(8\)) consists of elements whose order is a power of \(2\), hence cannot contain any of the \(48\) elements of order \(7\); only \(8\) non-order-7 elements (including identity) remain, exactly filling one Sylow \(2\)-subgroup, leaving no room for a second distinct one. C
7
n_2 = 1 \ \Rightarrow\ \text{the Sylow 2-subgroup is normal in } G, \text{ contradicting simplicity.}
Corollaries: unique Sylow subgroup is normal; a proper nontrivial normal subgroup contradicts \(G\) simple. A
\text{No group of order } 56 \text{ is simple.}
Reading. Forcing \(n_7=8\) uses up too many elements of order \(7\) to leave room for more than one Sylow \(2\)-subgroup, so \(n_2=1\) is forced instead, giving a normal subgroup either way.
Problems
- Find \(n_p\) for \(p=3\) in a group of order \(99 = 9 \cdot 11\), and deduce the group is abelian.
Solution
\(|G|=99=3^2 \cdot 11\). For \(p=11\): \(n_{11} \mid 9\), \(n_{11}\equiv 1 \pmod{11}\); divisors of \(9\) are \(1,3,9\), none of \(3,9\) is \(\equiv 1 \bmod 11\), so \(n_{11}=1\). For \(p=3\): \(n_3 \mid 11\), \(n_3 \equiv 1 \pmod 3\); divisors of \(11\) are \(1,11\), and \(11 \equiv 2 \bmod 3\), so \(n_3=1\). Both Sylow subgroups are unique hence normal, intersect trivially (coprime orders, Lagrange), and their product has order \(99=|G|\), so \(G \cong P_3 \times P_{11}\) (internal direct product). \(P_{11}\) has prime order \(11\), so is cyclic, hence abelian. \(P_3\) has order \(9=3^2\); every group of order \(p^2\) is abelian (a standard corollary of the class equation applied to \(p\)-groups). A direct product of two abelian groups is abelian, so \(G\) is abelian. - Prove that a group of order \(30\) has a normal subgroup of order \(15\).
Solution
\(|G|=30=2\cdot3\cdot5\). For \(p=5\): \(n_5\mid 6\), \(n_5\equiv1\pmod5\); divisors of \(6\) are \(1,2,3,6\), only \(1\) and \(6\) are \(\equiv1\bmod5\) (\(6\equiv1\)), so \(n_5\in\{1,6\}\). For \(p=3\): \(n_3\mid10\), \(n_3\equiv1\pmod3\); divisors of \(10\) are \(1,2,5,10\), and \(1,10\) are \(\equiv1\bmod3\) (\(10\equiv1\)), so \(n_3\in\{1,10\}\). Suppose both \(n_5=6\) and \(n_3=10\): the six Sylow \(5\)-subgroups (prime order, pairwise trivial intersection) contribute \(6\times4=24\) elements of order \(5\); the ten Sylow \(3\)-subgroups contribute \(10\times2=20\) elements of order \(3\); these element-sets are disjoint from each other (orders \(3\) vs \(5\)) and from the identity, giving at least \(24+20+1=45\gt30\) elements — contradiction. So \(n_5=1\) or \(n_3=1\); either way one of \(P_5,P_3\) is normal. Say \(P_5\trianglelefteq G\) (the other case is symmetric). Then \(P_5P_3\) is a subgroup (product of a normal subgroup with any subgroup is a subgroup) of order \(|P_5||P_3|/|P_5\cap P_3| = 5\cdot3/1=15\) (trivial intersection by Lagrange, coprime orders). A subgroup of index \(2\) is automatically normal (its two cosets are itself and the complement, and conjugation permutes cosets fixing the subgroup itself). Since \([G:P_5P_3]=30/15=2\), \(P_5P_3\trianglelefteq G\) has order \(15\). - Let \(|G|=p^2\) for a prime \(p\). Show directly from the Sylow theorems (not the class equation) that \(G\) has a normal subgroup of every order dividing \(|G|\).
Solution
Here \(G\) itself is its own (unique) Sylow \(p\)-subgroup, since \(n=2\), \(m=1\), and \(n_p \mid 1 \Rightarrow n_p=1\), consistent with \(G\) being a \(p\)-group. This shows \(G\trianglelefteq G\) (trivial) and existence of a subgroup of order \(p^2=|G|\), but the Sylow theorems for the single prime \(p\) do not by themselves produce subgroups of order \(p^1\) or classify normality of *proper* subgroups — that requires the class equation (a genuinely different tool: \(|G|=|Z(G)|+\sum[G:C_G(x_i)]\) forces \(Z(G)\) nontrivial, hence \(|Z(G)|\in\{p,p^2\}\), and if \(|Z(G)|=p\) then \(G/Z(G)\) is cyclic of order \(p\), forcing \(G\) abelian — a contradiction with \(|Z(G)|=p\) — so \(|Z(G)|=p^2\), \(G\) is abelian, and any subgroup of an abelian group is normal, giving normal subgroups of orders \(1,p,p^2\) via Cauchy for the order-\(p\) case). This problem is a caution: Sylow's theorem alone (for the single prime dividing \(|G|\)) gives existence of the top subgroup but the "every order, every subgroup normal" conclusion needs the class equation as a separate ingredient. - Compute the number of Sylow \(2\)-subgroups of \(S_4\) (order \(24=2^3\cdot3\)) and verify it satisfies the Sylow counting constraints.
Solution
\(|S_4|=24=2^3\cdot3\), so Sylow \(2\)-subgroups have order \(8\); these are exactly the dihedral subgroups \(D_4\) generated by symmetries of a labelling of the \(4\) points as a square (stabilisers of a partition into two pairs of "opposite" points under a chosen square structure). Constraint: \(n_2\mid3\), \(n_2\equiv1\pmod2\); candidates from divisors of \(3\) are \(1,3\), and both are odd hence both \(\equiv1\bmod2\) — the constraint alone does not decide between them. Direct count: there are exactly \(3\) ways to partition \(\{1,2,3,4\}\) into two pairs \(\{\{a,b\},\{c,d\}\}\), each giving rise to one Sylow \(2\)-subgroup (the stabiliser of that pairing structure, of order \(8\)); these three subgroups are distinct because each pairing gives a different subgroup, so \(n_2=3\). Check: \(3\mid3\) and \(3\equiv1\pmod2\), consistent. (This illustrates Problem-type situations where the congruence alone under-determines \(n_p\) and a structural count is needed.) - Show that a group \(G\) of order \(pq\), with \(p \lt q\) primes and \(p \nmid (q-1)\), is cyclic.
Solution
\(n_q\mid p\), \(n_q\equiv1\pmod q\); divisors of \(p\) are \(1,p\); since \(p\lt q\), \(p\not\equiv1\pmod q\) unless \(p=1\) (impossible, \(p\) prime), so \(n_q=1\): the Sylow \(q\)-subgroup \(Q\) is normal. \(n_p\mid q\), \(n_p\equiv1\pmod p\); divisors of \(q\) are \(1,q\). If \(n_p=q\), then \(q\equiv1\pmod p\), i.e. \(p\mid(q-1)\), contradicting the hypothesis; so \(n_p=1\), and \(P\) (Sylow \(p\)-subgroup) is also normal. Both \(P,Q\) normal, prime orders so cyclic, trivial intersection (coprime orders, Lagrange), product has order \(pq=|G|\): as in Problem 1, \(G\cong P\times Q\cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/q\mathbb{Z}\cong\mathbb{Z}/pq\mathbb{Z}\) by CRT, since \(\gcd(p,q)=1\). Hence \(G\) is cyclic.