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Sylow theorem
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Sylow theorem

The Sylow theorems of group theory, named after Ludwig Sylow, form a partial converse to the theorem of Lagrange, which states that if H is a subgroup of a finite group G, then the order of H divides the order of G. The Sylow theorems guarantee, for certain divisors of the order of G, the existence of corresponding subgroups, and give information about the number of those subgroups.

Table of contents
1 Definition
2 Sylow theorems
3 Example applications
4 Proofs


Let p be a prime number; then we define a Sylow p-subgroup of G to be a maximal p-subgroup of G (i.e., a subgroup which is a p-group, and which is not a proper subgroup of any other p-subgroup of G). The set of all Sylow p-subgroups for a given prime p is sometimes written Sylp(G).

Collections of subgroups which are each maximal in one sense or another are not uncommon in group theory. The surprising result here is that in the case of Sylp(G), all members are actually isomorphic to each other; and this property can be exploited to determine other properties of G.

Sylow theorems

The following facts were first proposed and proven by Norwegian mathematician Ludwig Sylow in 1872, and published in Mathematische Annalen. Given a finite group G and a prime p which divides the order of G, we can write the order of G as (pn · s), where n > 0 and p does not divide s. Then:

There exists a Sylow p-subgroup of G, of order pn.
All Sylow p-subgroups of G are conjugate to each other (and therefore isomorphic), i.e. if H and K are Sylow p-subgroups of G, then there exists an element g in G with g-1H.g = K.
Let np be the number of Sylow p-subgroups of G.
  • np divides s.
np = 1 mod p.

In particular, the above implies that every Sylow p-subgroup is of the same order, pn; and conversely, if a subgroup has order pn, then it is a Sylow p-subgroup, and so is isomorphic to every other Sylow p-subgroup. Because of the maximality condition, if H is any p-subgroup of G, then H is a subgroup of a p-subgroup of order pn.

Example applications

Let G be a group of order 15 = 3 · 5. We have that n3 must divide 5, and n3 = 1 mod 3. The only value satisfying these constraints is 1; therefore, there is only one subgroup of order 3, and it must be normal (since it has no distinct conjugates). Similarly, n5 divides 3, and n5 = 1 mod 5; thus it also has a single normal subgroup of order 5. Since 3 and 5 are coprime, the intersection of these two subgroups is trivial, and so G must be a cyclic group. Thus, there is only 1 group of order 15 (up to isomorphism); namely Z15.

For a more complex example, we can show that there are no simple groups of order 350. If |G| = 350 = 2 · 52 · 7, then n5 must divide 14 ( = 2 · 7), and n5 = 1 mod 5. Therefore n5 = 1 (since neither 6 nor 11 divides 14), and thus G must have a normal subgroup of order 52, and so cannot be simple.


The proofs of the Sylow theorem exploit the notion of group action in various creative ways. The group G acts on itself or on the set of its p subgroups in various ways, and each such action can be exploited to prove one of the Sylow theorem.

The proofs rely on several facts about the conjugacy classes of elements and subsets of G:

First, given a group of order pn·s, we can prove that G has a subgroup K of order pn using induction. K is necessarily maximal by Lagrange, so it is a Sylow p-subgroup; and therefore Sylp(G) is not empty.

Next, we use the existence of K to prove that Cl(K), the conjugacy class of K, is in fact Sylp(G); so every Sylow p-subgroup is conjugate to K. As a bonus, we simultaneously find that np = 1 mod p.

The other numerical fact about np follows almost immediately.

It is worth pointing out that the above argument as regards conjugacy is valid as long as |Cl(K)| = [G:N(K)] is finite; so we can state an infinite analog of the Sylow theorems: