The Sylow Theorems

Lagrange's theorem says the order of a subgroup divides the order of the group. Its converse fails: $A_4$ has order $12$ but no subgroup of order $6$. The Sylow theorems recover as much of the converse as is true — for every prime power dividing $|G|$ to its maximal extent, a subgroup of that order exists — and then say precisely how many there are.¹ The count is constrained enough that, for many orders, it alone forces a normal subgroup, which is the standard route to proving a group is not simple.

Sylow subgroups

Write ${\mathrm{Syl}}_p(G)$ for the set of Sylow $p$-subgroups and $n_p=|{\mathrm{Syl}}_p(G)|$ for their number. The definition isolates the $p$-part of the order: factor $|G|=p^{a}m$ once, and a Sylow $p$-subgroup accounts for the entire $p^a$.

The theorems

Two independent constraints bound $n_p$: the congruence $n_p\equiv 1\mathrm{mod}p$ and the divisibility $n_p\mid m$. Their intersection is often a very short list.

Existence and the congruence both rest on a lemma controlling how a $p$-subgroup meets a Sylow normalizer.

If $H=N_G(P)\cap Q$, then $PH$ is a subgroup (because $H$ normalizes $P$) of order $|P||H|/|P\cap H|$, a power of $p$; since $P\le PH$ and $P$ already has the maximal $p$-power order, $PH=P$, forcing $H\le P$.²

The congruence

Let $G$ act by conjugation on the set $S=\{P_1,\dots ,P_r\}$ of conjugates of one Sylow $p$-subgroup $P$. Restrict the action to a $p$-subgroup $Q$ and split $S$ into $Q$-orbits. By orbit-stabilizer and the lemma, the orbit of $P_i$ has size $[Q:Q\cap P_i]$. Taking $Q=P_1$: the orbit of $P_1$ is a single point (since $P_1\cap P_1=P_1$), while every other orbit has size $[P_1:P_1\cap P_i]>1$, a positive power of $p$. Summing,

The conjugacy part then shows $S$ is all of ${\mathrm{Syl}}_p(G)$, so $n_p=r\equiv 1(\mathrm{mod}p)$, and orbit-stabilizer for the full group gives $n_p=[G:N_G(P)]$.³ Existence is a separate induction on $|G|$ using the class equation: if $p\mid |Z(G)|$, quotient by a central subgroup of order $p$; otherwise some non-central centralizer has index prime to $p$ and supplies the Sylow subgroup by induction.

Normality from uniqueness

If $n_p=1$ then $gP{g}^{-1}\in {\mathrm{Syl}}_p(G)=\{P\}$ for all $g$, so $P$ is normal. Conversely a normal $P$ is the only Sylow $p$-subgroup, since all are conjugate to it. A unique subgroup of its order is characteristic, hence normal.⁴ Proving $n_p=1$ therefore proves normality, the target of every non-simplicity argument below.

Reading the count in examples

Factor the order, list the candidate $n_p$, and let the two constraints prune.

For $A_4$: $n_3\mid 4$ and $n_3\equiv 1(\mathrm{mod}3)$ leave $n_3\in \{1,4\}$, and the four subgroups $⟨(123)⟩,⟨(124)⟩,⟨(134)⟩,⟨(234)⟩$ realize $n_3=4$; meanwhile $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$ force $n_2=1$, the normal Klein four-group. The largest prime tends to give the tightest constraint, since $n_p\equiv 1\mathrm{mod}p$ rules out more values as $p$ grows.

Proving non-simplicity

If some $n_p=1$, the corresponding Sylow subgroup is a proper nontrivial normal subgroup, so $G$ is not simple. Three techniques cover most small orders.

Direct pruning. Suppose $|G|=pq$ with primes $p<q$. Then $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$; since $p<q$ the only option is $n_q=1$, so the Sylow $q$-subgroup is normal and $G$ is not simple. Any group of order $pq$ has a normal Sylow subgroup for the larger prime.⁵

Element counting. When the constraints allow $n_p>1$, count elements of each prime order. Take $|G|=30=2\cdot 3\cdot 5$ and suppose neither the Sylow $5$- nor the Sylow $3$-subgroup is normal, so $n_5=6$ and $n_3=10$. Distinct Sylow $5$-subgroups meet only in the identity, so they contribute $6\cdot 4=24$ elements of order $5$; the ten Sylow $3$-subgroups contribute $10\cdot 2=20$ elements of order $3$. That is $44$ non-identity elements in a group of $30$ — a contradiction. One of the Sylow subgroups must be normal.

Changing primes and using cosets. For borderline orders the coset action supplies the last step. If $H\le G$ has index $k$ and $G$ is simple, then $G$ embeds in $S_k$ (the kernel of the coset action is a proper normal subgroup, hence trivial), so $|G|\mid k!$. This bounds how small an index a simple group can have, and combined with element counting it classifies the simple group of order $60$ as $A_5$.

Worked example: groups of order 12

The three techniques together settle the structure of every group of order $12=2^2\cdot 3$. The claim is that such a $G$ either has a normal Sylow $3$-subgroup or is isomorphic to $A_4$.⁵

Suppose the Sylow $3$-subgroup is not normal, so $n_3\ne 1$. The constraints $n_3\mid 4$ and $n_3\equiv 1(\mathrm{mod}3)$ leave $n_3=4$. The four Sylow $3$-subgroups have prime order $3$, so they intersect pairwise in the identity and account for $4\cdot 2=8$ elements of order $3$. Since $[G:N_G(P)]=n_3=4$, the normalizer $N_G(P)$ equals $P$ itself. Now let $G$ act by conjugation on its four Sylow $3$-subgroups; the resulting homomorphism $\phi :G\to S_4$ has kernel $K={\bigcap }_g{N}_G(P_g)\le N_G(P)=P$, and since $P$ is not normal, $K=1$. So $\phi$ is injective and $G$ embeds in $S_4$. The $8$ elements of order $3$ map to the $8$ three-cycles of $S_4$, all of which lie in $A_4$, so the image meets $A_4$ in a subgroup of order at least $8$; both groups have order $12$, forcing $\phi (G)=A_4$ and $G\cong A_4$. In that case the Sylow $2$-subgroup — the Klein four-group of double transpositions — is the unique one, hence normal, so $G$ still has some normal Sylow subgroup. The order-$12$ groups are thereby sorted into those with a normal Sylow $3$-subgroup and the single exception $A_4$.

The same $|G|\mid k!$ bound settles order $60$. A simple group of order $60$ can have no subgroup of index less than $5$, since an index $k\in \{2,3,4\}$ would embed $G$ in $S_k$ with $60\nmid k!$. Running the possible values $n_2\in \{3,5,15\}$ through this bound and an element count forces the Sylow $2$-normalizer to have index $5$; the coset action then embeds $G$ in $S_5$ as a subgroup that must coincide with $A_5$. Every simple group of order $60$ is isomorphic to $A_5$.