Throughout, \(G\) is a finite group of order \(n\), \(p\) is a prime, \(n = p^k m\) with \(p\notdivides m\), and \(n_p\) is the number of Sylow \(p\)-subgroups of \(G\).
Constraints from the order
By Sylow’s third theorem, \[ n_p \equiv 1 \pmod p, \qquad n_p \divides m. \] If \(n_p = 1\), the unique Sylow \(p\)-subgroup \(P\) is normal: for \(g\in G\), \(gP\inverseof{g}\) is again a Sylow \(p\)-subgroup, so \(gP\inverseof{g} = P\).
Groups of prime-power order:
If \(n\) is prime, \(G\) is cyclic and simple.
If \(n = p^k\) with \(k\geq 2\), then \(Z(G) \neq 1\) by the class equation. If \(Z(G)\neq G\), then \(Z(G)\) is a proper nontrivial normal subgroup; if \(Z(G)=G\), then \(G\) is abelian and every subgroup of order \(p\) is proper, nontrivial, and normal.
A Sylow count with \(n_p = 1\)
If the only divisor of \(m\) congruent to \(1\) modulo \(p\) is \(1\), then \(n_p=1\) and the Sylow \(p\)-subgroup is normal.
For \(n = 20 = 2^2\cdot 5\): \(n_5 \equiv 1 \pmod 5\) and \(n_5 \divides 4\), so \(n_5 = 1\).
Counting elements
Distinct subgroups of prime order \(p\) intersect trivially, so if the Sylow \(p\)-subgroups have order \(p\), they contain \(n_p(p-1)\) elements of order \(p\). Elements of distinct prime orders are distinct and nonidentity, so summing over the primes \(p\) whose Sylow \(p\)-subgroups have order \(p\) gives \(\sum_p n_p(p-1)\leq n-1\); values of the \(n_p\) that violate this inequality do not occur. Sylow subgroups of order \(p^2\) can intersect in a subgroup of order \(p\), and then this count does not apply; see Two Sylow subgroups meeting nontrivially.
For \(n = 30\): if \(n_5 = 6\) and \(n_3 = 10\), then \(G\) has \(6\cdot 4 + 10 \cdot 2 = 44 > 30\) elements of order \(5\) or \(3\). Hence \(n_5 = 1\) or \(n_3 = 1\).
A subgroup of small index
If \(G\) is simple and \(H < G\) has index \(k > 1\), then the action of \(G\) on the \(k\) left cosets of \(H\) has kernel a proper normal subgroup, hence trivial, so \(G\) embeds in \(S_k\) and \[ \size G \divides k!. \] The subgroup \(H\) can be a Sylow subgroup or the normalizer \(N_G(P)\) of a Sylow subgroup \(P\).
For \(n = 24\): if \(n_2 = 3\), then \([G : N_G(P)] = 3\) for a Sylow \(2\)-subgroup \(P\), and \(24 \notdivides 3! = 6\), so \(G\) is not simple. If \(n_2 = 1\), the Sylow \(2\)-subgroup is normal.
A simple group with a proper subgroup of index $n$ embeds in $S_n$ (Tag PR-5FGA7)
Let \(G\) be a simple group and \(H < G\) a proper subgroup of finite index \([G:H] = n\). Then there is an injective group homomorphism \(\phi\colon G \injects S_n\).
The action of \(G\) on the set \(G/H\) of left cosets by \(g \cdot xH \coloneqq gxH\) gives a homomorphism \(\phi\colon G \to \operatorname{Sym}(G/H) \cong S_n\). If \(g \in \ker\phi\), then \(gH = H\), so \(\ker\phi \subseteq H \neq G\). Since \(\ker\phi\) is a normal subgroup of the simple group \(G\) and \(\ker\phi \neq G\), \(\ker\phi\) is trivial.
Index the smallest prime
If \(H \leq G\) has index \(q\), the smallest prime dividing \(\size G\), then \(H\) is normal in \(G\). In particular, a subgroup of index \(2\) is normal.
A subgroup whose index is the smallest prime dividing $\abs{G}$ is normal (Tag PR-PADL7)
Let \(G\) be a finite group, let \(p\) be the smallest prime dividing \(\abs{G}\), and let \(H\leq G\) be a subgroup of index \([G:H]=p\). Then \(H\) is normal in \(G\).
Let \(G\) act on the set \(G/H\) of left cosets by left translation, \(g\cdot xH=gxH\). This action is a homomorphism \(\varphi\colon G\to\operatorname{Sym}(G/H)\cong S_p\); let \(K\coloneqq\ker\varphi\). If \(g\in K\) then \(gH=H\), so \(K\subseteq H\), and \(K\normal G\). By the first isomorphism theorem \(G/K\) embeds in \(S_p\), so \([G:K]\) divides \(p!\). Also \([G:K]\) divides \(\abs{G}\), so every prime factor of \([G:K]\) is at least \(p\); the only such prime dividing \(p!\) is \(p\), and \(p^2\nmid p!\). Hence \([G:K]\in\theset{1,p}\). Since \(K\subseteq H\), \([G:K]\geq[G:H]=p\), so \([G:K]=p\) and \(K=H\). Therefore \(H=\ker\varphi\) is normal in \(G\).
The normalizer of a Sylow subgroup
For a Sylow \(p\)-subgroup \(P\), orbit-stabilizer for the conjugation action on the Sylow \(p\)-subgroups gives \(n_p = [G : N_G(P)]\). Hence a bound on \(n_p\) is a bound on the index of the subgroup \(N_G(P)\), to which A subgroup of small index and Index the smallest prime apply.
Two Sylow subgroups meeting nontrivially
Suppose the Sylow \(p\)-subgroups have order \(p^2\) and two of them, \(P \neq Q\), satisfy \(\size{P\intersect Q} = p\). Groups of order \(p^2\) are abelian, so \(D \da P\intersect Q\) is normal in both \(P\) and \(Q\), and \(N_G(D)\) contains the subgroup \(\langle P, Q\rangle\), whose order is a multiple of \(p^2\) greater than \(p^2\). Hence \([G:N_G(D)]\) is a proper divisor of \(m\), and either \(N_G(D) = G\), so that \(D\) is a proper nontrivial normal subgroup, or \(N_G(D)\) is a proper subgroup whose index \(k\) is a proper divisor of \(m\), and if \(G\) is simple then \(\size G\divides k!\) by A subgroup of small index. If instead any two distinct Sylow \(p\)-subgroups intersect trivially, Counting elements applies with \(n_p(p^2-1)\) nonidentity elements.
The action on the Sylow subgroups
\(G\) acts by conjugation on its \(n_p\) Sylow \(p\)-subgroups, giving a homomorphism \(\rho\colon G \to S_{n_p}\). By Sylow’s second theorem the action is transitive, so \(\ker\rho\) is a proper normal subgroup when \(n_p > 1\). If \(G\) is simple and \(n_p>1\), then \(\rho\) is injective and \(\size G \divides n_p!\). This is the action of \(G\) on the cosets of \(N_G(P)\).
Arguments and conclusions
| Argument | Input | Conclusion |
|---|---|---|
| Sylow count | the factorization of \(n\) | \(n_p = 1\), so a Sylow \(p\)-subgroup is normal |
| element count | Sylow subgroups of prime order | some \(n_p = 1\) |
| small index | a proper subgroup of index \(k\) | \(G\) simple implies \(\size G \divides k!\) |
| smallest prime index | a subgroup of index the least prime dividing \(n\) | that subgroup is normal |
| normalizer | \(n_p = [G:N_G(P)]\) | a subgroup of index \(n_p\) |
| intersecting Sylow subgroups | \(\size{P\intersect Q} = p\) for Sylow subgroups of order \(p^2\) | a subgroup \(N_G(P\intersect Q)\) of index a proper divisor of \(m\) |
| action on Sylow subgroups | \(n_p > 1\) | \(G\) simple implies \(\size G \divides n_p!\) |
Simple groups
The groups of prime order and the alternating groups \(A_n\) for \(n\geq 5\) are simple. The group \(A_5\) of order \(60\) is the smallest nonabelian simple group, and every simple group of order \(60\) is isomorphic to \(A_5\).