The Jordan-Hölder Theorem
1 Definition. Let G be a group. For subgroups Gi , i = 0, 1, 2, . . . , n of G such that
(i) G0 = G and Gn = { e };
(ii) Gi+1 C Gi for i = 0, 1, . . . , n − 1 (that is, Gi is a proper normal subgroup of Gi+1 );
the chain
G0 B G 1 B G 2 B · · · B G n
is called a subnormal series for G. Moreover, if for each i = 0, 1, . . . , n − 1, the quotient group
Gi /Gi+1 is a simple group (that is, for each i, Gi+1 is a maximal normal subgroup of Gi ), then
the subnormal series is called a composition series for G, and the factor groups G i /Gi+1 , i =
0, 1, . . . , n − 1 are called the factors of the composition series. The length of the composition series
G = G0 B G1 B G2 B · · · B Gn = { e } is n.
2 Lemma. Let G be a group with composition series G = G0 B G1 B G2 B · · · B Gn = { e }. Then
for any normal subgroup K of G, if we remove duplicates from the series K = K ∩ G0 D K ∩ G1 D
K ∩ G2 D · · · D K ∩ Gn = { e }, the result is a composition series for K of length at most n.
Proof. We need to show that for each i = 0, 1, . . . , n − 1, K ∩ Gi+1 C K ∩ Gi and that the
quotient group (K ∩ Gi )/(K ∩ Gi+1 ) is simple. Let x ∈ K ∩ Gi and g ∈ K ∩ Gi+1 . Then xgx−1 ∈ K
since K C G, and xgx−1 ∈ Gi+1 since Gi+1 C Gi . Thus xgx−1 ∈ K ∩ Gi+1 , which proves that
K ∩ Gi+1 C K ∩ Gi .
We now examine the quotient group (K ∩ Gi )/(K ∩ Gi+1 ). For any i, the simplicity of Gi /Gi+1
implies that the only normal subgroups of Gi that contain Gi+1 are Gi and Gi+1 . Since both Gi+1
and K ∩ Gi are normal in Gi , it follows that Gi+1 C Gi+1 (K ∩ Gi ) C Gi and so
Gi+1 (K ∩ Gi )/Gi+1 C Gi /Gi+1 .
Thus either Gi+1 (K ∩ Gi ) = Gi+1 or else Gi+1 (K ∩ Gi ) = Gi . Now, by the second isomorphism
theorem, we have
Gi+1 (K ∩ Gi )/Gi+1 ' (K ∩ Gi )/(K ∩ Gi ∩ Gi+1 ) = (K ∩ Gi )/(K ∩ Gi+1 ).
If Gi+1 (K ∩ Gi ) = Gi+1 , then K ∩ Gi = K ∩ Gi+1 and so we would have a duplicate to remove,
while if Gi+1 (K ∩ Gi ) = Gi , then Gi /Gi+1 ' (K ∩ Gi )/(K ∩ Gi+1 ) and thus (K ∩ Gi )/(K ∩ Gi+1 )
is simple.
3 Theorem (The Jordan-Hölder Theorem). Let G be a group that has a composition series. Then
any two composition series for G have the same length. Moreover, if
G = G 0 B G1 B G2 B · · · B G n = { e }
and
G = H 0 B H1 B H2 B · · · B H n = { e }
are two composition series for G, there exists a permutation τ of { 0, 1, . . . , n − 1 } such that for each
i = 0, 1, 2, . . . , n − 1, Gi /Gi+1 ' Hτ (i) /Hτ (i)+1 .
Proof. The proof will be by induction on the length of a composition series. Suppose that G is a
group with a composition series of length 1. Then the subnormal series G B { e } can’t be refined,
so it must be a composition series. In particular, G ' G/{ e }, so G is simple. Thus G B { e } is the
only composition series for G, and so the assertions is valid for all groups with a composition series
of length 1.
Suppose now that n > 1 is an integer with the property that the assertion is valid for any group
that has a composition series of length less than n, and let G be a group with a composition series of
length n, say G = G0 B G1 B G2 B · · · B Gn = { e } (so that Gi 6= Gi+1 for each i = 0, 1, . . . , n − 1).
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Now let G = H0 B H1 B · · · B Hm = { e } be a composition series for G (again, Hi 6= Hi+1 for each
i = 0, 1, . . . , m − 1). If G1 = H1 , then by the induction hypothesis applied to G1 , n − 1 = m − 1
and there is a permutation τ of { 1, 2, . . . , n − 1 } such that Gi /Gi+1 ' Hτ (i) /Hτ (i)+1 for every
i = 1, 2, . . . , n − 1. We may extend τ to a permutation of { 0, 1, . . . , n − 1 } by defining τ (0) = 0.
Since G0 /G1 = H0 /H1 , it follows that Gi /Gi+1 ' Hτ (i) /Hτ (i)+1 for every i = 0, 1, 2, . . . , n − 1.
Suppose then that H1 6= G1 . Since both G1 and H1 are maximal normal in G with H1 6= G1 , we
see that H1 − G1 6= ∅ and G1 − H1 6= ∅, so H1 C H1 G1 C G and H1 6= H1 G1 . Thus H1 G1 = G,
from which we conclude that G/H1 ' G1 /(H1 ∩ G1 ). Since G/H1 is simple, we conclude that
G1 /(H1 ∩ G1 ) is simple as well. Now by Lemma 2, upon removing duplicates from the series
H1 = H1 ∩ G0 D H1 ∩ G1 D · · · D H1 ∩ Gn = { e },
the result is a composition series for H1 of length at most n, and thus upon removing duplicates,
H1 ∩ G 1 D · · · D H 1 ∩ G n = { e }
is a composition series for H1 ∩ G1 of length at most n − 1. Since G1 /(H1 ∩ G1 ) is simple, it follows
that
G1 B H1 ∩ G 1 B · · · B H 1 ∩ G n = { e }
is a composition series for G1 . But then
G1 B G2 B · · · B Gn = { e } and
G1 D H1 ∩ G 1 D H1 ∩ G 2 D · · · D H 1 ∩ G n = { e }
are both composition series for G1 , with the first series of length n − 1. By our induction hypothesis,
both series have the same length. Since G1 6= H1 ∩ G1 , any duplication must occur later in the
series. Let
G1 = K1 B K2 = H1 ∩ G 1 B K3 B · · · B K n = { e }
denote the composition series for G1 of length n − 1 that results from the remove of duplicates. By
hypothesis, there exists a permutation α of { 1, 2, . . . , n − 1 } such that Gi /Gi+1 ' Kα(i) /Kα(i)+1
for each i = 1, 2, . . . , n − 1. Extend α to { 0, 1, . . . , n − 1 } by defining α(0) = 0. Then
G = G0 B G1 B G2 B · · · B Gn = { e } and
G = K 0 B K1 = G 1 B K2 = H1 ∩ G 1 B · · · B K n = { e }
are composition series of length n for G, and α is a permutation of { 0, 1, . . . , n − 1 } such that
Gi /Gi+1 ' Kα(i) /Kα(i)+1 for each i = 0, 1, . . . , n − 1. Moreover, we have found a composition series
for H1 ∩ G1 of length n − 2.
Let us now carry out similar computations for the composition series G = H0 B H1 B · · · B
Hm = { e } and the normal subgroup G1 of G. Again by Lemma 2, upon the removal of duplicates
from the series
G1 = H0 ∩ G1 D H1 ∩ G1 D H2 ∩ G1 D · · · D Hm ∩ G1 = { e },
we obtain a composition series for G1 , so that upon the removal of duplicates from the series
H1 ∩ G1 D H2 ∩ G1 D · · · D Hm ∩ G1 = { e },
we obtain a composition series for H1 ∩ G1 . Now since H1 ∩ G1 has a composition series of length
n − 2, we may apply the induction hypothesis to H1 ∩ G1 to conclude that all composition series of
H1 ∩ G1 have length n − 2, and so in particular, the preceding composition series for H1 ∩ G1 has
length n − 2.
Since H1 /(H1 ∩ G1 ) ' H1 G1 /G1 = G0 /G1 , which is a simple group, it follows that H1 /(H1 ∩ G1 )
is simple. Thus upon the removal of duplicates from
H1 B H1 ∩ G1 D H2 ∩ G1 D · · · D Hm ∩ G1 = { e },
the result is a composition series for H1 of length n − 1.
We have that
H1 B H2 B · · · B Hm = { e } and
H1 D H 1 ∩ G 1 D H 2 ∩ G 1 D · · · D H m ∩ G 1 = { e }
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are composition series for H1 . But then
K2 = H1 ∩ G1 D · · · B Kn = { e } and
H1 ∩ G 1 D H 2 ∩ G 1 D · · · D H m ∩ G 1 = { e }
are composition series for H1 ∩ G1 , and the first of these two composition series has length n − 2.
By our induction hypothesis, all composition series for H1 ∩ G1 have the same length, so the second
composition series also has length n − 2. Since H1 /(H1 ∩ G1 ) is simple, then
H1 B H2 B · · · B Hm = { e } and
H1 D H 1 ∩ G 1 D H 2 ∩ G 1 D · · · D H m ∩ G 1 = { e }
are two composition series for H1 with the second series having length n − 1, so by our induction
hypothesis, the first series must also have length n − 1. Since its length is m − 1, it follows that
m = n.
Let Li , i = 1, 2, . . . , n − 1 denote the distinct groups in the series
H1 D H1 ∩ G1 D H2 ∩ G1 D · · · D Hn ∩ G1 = { e },
so that L1 = H1 and L2 = H1 ∩ G1 . Then we have composition series
G = H0 B H1 B H2 B · · · B Hn = { e } and
G = L 0 B L1 B L2 B · · · B L n = { e }
of length n for G, and there exists a permutation β of { 0, 1, . . . , n − 1 } such that Hi /Hi+1 '
Lβ(i) /Lβ(i)+1 for each i = 0, 1, . . . , n − 1.
Finally, since K2 = L2 = H1 ∩ G1 , we have the two composition series for G
K3
G1
K4
Kn−1
{e}
H1 ∩ G 1
G
H1
L3
L4
Ln−1
We may apply the induction hypothesis to H1 ∩ G1 to obtain the existence of a permutation γ of
the set { 2, 3, . . . , n − 1 } such that for each i = 2, 3, . . . , n − 1, Ki /Ki+1 ' Li /Li+1 (recall that
K2 = L2 = H1 ∩ G1 and Kn = Ln = { e }). We have already observed that G/G1 ' H1 /(H1 ∩ G1 )
and G/H1 ' G1 /(H1 ∩ G1 ), so we may extend γ to a permutation of { 0, 1, 2, . . . , n − 1 } by setting
γ(0) = 1 and γ(1) = 0. Then, since K0 = G = L0 , K1 = G1 , L1 = H1 , and K2 = L2 = H1 ∩ G1 , we
have Ki /Ki+1 ' Li /Li+1 for all i = 0, 1, . . . , n − 1.
In summary, we have m = n, and for τ = β −1 γα, we have Gi /Gi+1 ' Hτ (i) /Hτ (i)+1 for all
i = 0, 1, . . . , n − 1. This completes the proof of the inductive step.
One very nice application of the Jordan-Hölder theorem is the fundamental theorem of arithmetic.
4 Theorem (The Fundamental Theorem of Arithmetic). Let n > 1 be a positive integer. Then
there exist unique primes p1 < p2 < · · · < pk and unique positive integers r1 , r2 , . . . , rk such that
n = pr11 pr22 · · · prkk .
Proof. Let G = hgi be a cyclic group of order n. Then every subgroup of G is normal, and there
is a unique subgroup of size d for each positive divisor d of n. Let d be the largest proper divisor
of n, and let G1 be the unique subgroup of G of size d. Then G/H is simple and cyclic, hence of
prime order. We may repeat this construction on the cyclic subgroup H, so by induction, we obtain
a composition series
G = G 0 B G1 B G2 B · · · B G m = { e }
for G with Gi /Gi+1 of prime order pi for each i. Thus
n = |G| = |G/G1 | |G1 | = |G/G1 | |G1 /G2 | · · · |Gm−1 /Gm ||Gm |
= p1 p2 · · · pm−1 (1)
The uniqueness of the prime decomposition of n follows now from the Jordan-Hölder theorem applied
to G.
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