Lagrange's theorem states that for any finite group G and subgroup H of G, the order of H divides the order of G. The document provides the proof of Lagrange's theorem and several examples. It also discusses corollaries, including that every group of prime order is cyclic, every group of order less than 6 is abelian, and the order of an element must divide the group order. However, the converse of Lagrange's theorem is false - there can exist groups where not every divisor of the group order is a possible subgroup order.

1. Abstract Algebra, Saracino
Second Edition
Section 10
Lagrange’s
Theorem
"I have had my results for a long
time: but I do not yet know how I am
to arrive at them." - Carl Friedrich Gauss -

2. Lehman College, Department of Mathematics
Lagrange and Klein
Christian Felix Klein (1849-1925)
- German Mathematician
Joseph-Louis Lagrange (1736-1813)
- Franco-Italian Mathematician

3. Lehman College, Department of Mathematics
Lagrange’s Theorem (1 of 4)
Theorem 10.1 (Lagrange). Let 𝐺 be a finite group and
let 𝐻 be a subgroup of 𝐺. Then 𝐻 divides 𝐺 .
Lemma 10.2. Let 𝐺 be any group (not necessarily finite)
and let 𝐻 be a subgroup of 𝐺. Let 𝐻𝑎 and 𝐻𝑏 be two
right cosets of 𝐻 in 𝐺. Then there is a one-to-one
correspondence between the elements of 𝐻𝑎 and 𝐻𝑏.
That is 𝐻𝑎 = |𝐻𝑏|. In other words, the cardinality of
every right coset is the same.
Proof. To show that 𝐻𝑎 = 𝐻𝑏 , we will show the
existence of a one-to-one and onto function from 𝐻𝑎 to
𝐻𝑏. Define 𝑓: 𝐻𝑎 → 𝐻𝑏 by 𝑓 ℎ𝑎 = ℎ𝑏 for all ℎ ∈ 𝐻.

4. Lehman College, Department of Mathematics
Lagrange’s Theorem (2 of 4)
Proof (cont’d). To show that 𝐻𝑎 = 𝐻𝑏 , we will show
that there is a one-to-one and onto function from 𝐻𝑎 to
𝐻𝑏. Define 𝑓: 𝐻𝑎 → 𝐻𝑏 by 𝑓 ℎ𝑎 = ℎ𝑏 for all ℎ ∈ 𝐻. To
show that 𝑓 is one-to-one, let ℎ1, ℎ2 ∈ 𝐻. Then, we have
𝑓 ℎ1 𝑎 = ℎ1 𝑏 and we also have 𝑓 ℎ2 𝑎 = ℎ2 𝑏. If
𝑓 ℎ1 𝑎 = 𝑓 ℎ2 𝑎 , then ℎ1 𝑏 = ℎ2 𝑏. By right-cancellation
of the element 𝑏, we have ℎ1 = ℎ2, so ℎ1 𝑎 = ℎ2 𝑎, and 𝑓
is one-to-one. To show 𝑓 is onto the set 𝐻𝑏, pick an
arbitrary element ℎ𝑏 ∈ 𝐻𝑏 (every element of 𝐻𝑏 is of the
form ℎ𝑏, for some ℎ ∈ 𝐻). But ℎ𝑏 = 𝑓(ℎ𝑎). It follows that
𝑓 is onto the set 𝐻𝑏. ∎
The lemma means that every right coset of 𝐻 in 𝐺 has
the same number of elements.

5. Lehman College, Department of Mathematics
Lagrange’s Theorem (3 of 4)
For example, if one right coset of 𝐻 in 𝐺 has 16,
elements then all right cosets have 16 elements. We
say that any two right cosets of 𝐻 in 𝐺 have the same
cardinality.
Proof of Lagrange’s Theorem. Let 𝐺 and 𝐻 be as in
the statement of the theorem. Define an equivalence
relation ≡ 𝐻 on 𝐺 by the following:
By theorem 9.1, the equivalence classes under ≡ 𝐻
partition 𝐺 into mutually disjoint equivalence classes. By
Theorem 9.3, these equivalence classes are the right
cosets of 𝐻 in 𝐺. That is, for any 𝑎 ∈ 𝐺, 𝑎 = 𝐻𝑎.
𝑎 ≡ 𝐻 𝑏 if and only if 𝑎𝑏−1
∈ 𝐻

6. Lehman College, Department of Mathematics
Lagrange’s Theorem (4 of 4)
Proof of Lagrange’s Theorem (cont’d). Since 𝐺 is finite,
only finitely many distinct right cosets of 𝐻 can fit into 𝐺:
For some integer 𝑘 and elements 𝑎1, 𝑎2, 𝑎3, … , 𝑎 𝑘 ∈ 𝐺.
Now, by Lemma 10.2 all right cosets of 𝐻 have the
same number of elements, namely 𝐻 . Recall that 𝐻 is
always one of the right cosets of 𝐻 in 𝐺. Now, counting
the elements in 𝐺, and noting that the union above is a
disjoint one, we have:
Thus, 𝐺 = 𝑘 ⋅ |𝐻| and 𝐻 divides 𝐺 . ∎
𝐺 = 𝐻𝑎1 ⊔ 𝐻𝑎2 ⊔ ⋯ ⊔ 𝐻𝑎 𝑘
𝐺 = 𝐻𝑎1 + 𝐻𝑎2 + ⋯ + |𝐻𝑎 𝑘|
= 𝐻 + 𝐻 + ⋯ + 𝐻 = 𝑘 ⋅ |𝐻|

7. Lehman College, Department of Mathematics
Cosets (1 of 1)
Theorem 1. Let 𝐺 be a group and let 𝐻 be a subgroup
of a group 𝐺. Then ℒ 𝐻 = ℛ 𝐻 , where ℒH (respectively,
ℛ 𝐻) denotes the set of all left (respectively, right) cosets
of 𝐻 in 𝐺.
Proof. We will show the existence of a one-to-one and
onto function from ℒ 𝐻 to ℛ 𝐻. We will define 𝑓: ℒ 𝐻 → ℛ 𝐻
by 𝑓 𝑎𝐻 = 𝐻𝑎−1
for all 𝑎𝐻 ∈ ℒ 𝐻. Firstly, we will show
that this function is well-defined. Let 𝑎𝐻, 𝑏𝐻 ∈ ℒ 𝐻 be two
representatives of the same coset, then 𝑎𝐻 = 𝑏𝐻 and
𝑎−1 𝑏 ∈ 𝐻. It follows that 𝑓 𝑎𝐻 = 𝐻𝑎−1 and that
𝑓 𝑏𝐻 = 𝐻𝑏−1
. Now, we have 𝑎−1
𝑏−1 −1
= 𝑎−1
𝑏 ∈ 𝐻.
Hence, 𝐻𝑎−1
= 𝐻𝑏−1
and 𝑓 is a well-defined function.

8. Lehman College, Department of Mathematics
Index of a Subgroup (1 of 1)
Proof. To show that 𝑓 is one-to-one, let 𝑎𝐻, 𝑏𝐻 ∈ ℒ 𝐻,
such that 𝑓 𝑎𝐻 = 𝑓(𝑏𝐻). Then, 𝐻𝑎−1
= 𝐻𝑏−1
and
𝑎−1
𝑏−1 −1
∈ 𝐻. It follows that 𝑎−1
𝑏−1 −1
= 𝑎−1
𝑏 ∈ 𝐻.
Since 𝑎−1
𝑏 ∈ 𝐻, then 𝑎𝐻 = 𝑏𝐻 and 𝑓 is one-to-one. To
show that 𝑓 is onto ℛH, let 𝐻𝑎 ∈ ℛ 𝐻 be an arbitrary
element. Since 𝐻𝑎 = 𝐻 𝑎−1 −1
= 𝑓 𝑎−1
𝐻 , it follows
that 𝑓 is onto ℛ 𝐻. ∎
Definition. If 𝐺 is a group (not necessarily finite) and 𝐻
is a subgroup of 𝐺, then the number of distinct right (or
left) cosets of 𝐻 in 𝐺 is called the index of 𝐻 in 𝐺 and is
denoted 𝐺: 𝐻 . Therefore, Lagrange’s Theorem for a
finite group 𝐺 can be written as:
𝐺 = 𝐺: 𝐻 ⋅ |𝐻|

9. Lehman College, Department of Mathematics
Examples (1 of 2)
Example 1. Let 𝐺 be the Klein 4-group. That is, we
have 𝐺 = 𝑒, 𝑎, 𝑏, 𝑐 with 𝑎2
= 𝑏2
= 𝑐2
= 𝑒, with the
following Cayley table:
If 𝐻 = 𝑎 = 𝑒, 𝑎 . We have 𝐻 = 2. Now, since 𝐺 = 4,
we have:
and 𝐺: 𝐻 = 2
⋅ 𝒆 𝒂 𝒃 𝒄
𝒆 𝒆 𝒂 𝒃 𝒄
𝒂 𝒂 𝒆 𝒄 𝒃
𝒃 𝒃 𝒄 𝒆 𝒂
𝒄 𝒄 𝒃 𝒂 𝒆
𝐺 = 𝐺: 𝐻 ⋅ 𝐻 4 = 𝐺: 𝐻 ⋅ 2so

10. Lehman College, Department of Mathematics
Examples (2 of 2)
Example 2. Let 𝐺 = 𝑆3. And let 𝐻 = ⟨ 1 2 ⟩. That is:
and 𝐻 = 𝑒, (1 2) . Then 𝐺 = 6 and 𝐻 = 2. It follows
that 𝐺 = 𝐺: 𝐻 ⋅ |𝐻|, so 6 = 𝐺: 𝐻 ⋅ 2, and 𝐺: 𝐻 = 3.
Theorem 10.4. Let 𝐺 be a finite group and let 𝑥 ∈ 𝐺.
Then, we obtain that o(𝑥) divides |𝐺|. Consequently,
𝑥|𝐺|
= 𝑒 for all 𝑥 ∈ 𝐺. That is, the order of an element
divides the order of a finite group.
Proof. Consider the subgroup 𝐻 = ⟨𝑥⟩. By Lagrange’s
theorem |𝐻| divides |𝐺|. But 𝐻 = o 𝑥 , so o(𝑥) divides
|𝐺|. By definition, 𝑥o(𝑥)
= 𝑒, since o(𝑥) divides |𝐺|, then
𝐺 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)

11. Lehman College, Department of Mathematics
Corollaries (1 of 3)
Theorem 10.4 (cont’d). Since o(𝑥) divides |𝐺|, then
𝐺 = 𝑘 ⋅ o(𝑥) for some positive integer 𝑘. It follows that
Theorem 10.5. Every group 𝐺 of prime order is cyclic.
Moreover, every non-identity element 𝑎 of 𝐺 is a
generator of the group.
Proof. Let 𝐺 be a group of prime order 𝑝. Since 𝑝 > 1,
then 𝐺 is not trivial, and 𝐺 has a non-identity element 𝑎.
Let 𝐻 be the cyclic subgroup of 𝐺 generated by 𝑎. That
is 𝐻 = ⟨𝑎⟩. By Lagrange’s Theorem | 𝑎 | divides 𝐺 = 𝑝.
Hence | 𝑎 | is either 1 or 𝑝. But 𝑎 ≠ 𝑒, so 𝑎 > 1. it
follows that 𝑎 = 𝑝 and 𝐺 = ⟨𝑎⟩ is cyclic. ∎
𝑥|𝐺|
= 𝑥 𝑘⋅o(𝑥)
= 𝑥o(𝑥) 𝑘
= 𝑒 𝑘
= 𝑒

12. Lehman College, Department of Mathematics
Corollaries (2 of 3)
Theorem 3. Every group 𝐺 of order less than six is
abelian. That is, if 𝐺 ≤ 5, then 𝐺 is abelian.
Proof. Let 𝐺 be a group, such that 𝐺 ≤ 5. If 𝐺 = 1,
then 𝐺 = 𝑒 = ⟨𝑒⟩ is cyclic, hence abelian. 𝐺 = 2 or 3
or 5, then 𝐺 is of prime order and is cyclic by Theorem
10.5, hence 𝐺 is abelian. Finally, if 𝐺 = 4, we will look
at two cases: If 𝐺 has an element of order 4, then 𝐺 is
cyclic, hence abelian. If 𝐺 has no element of order 4,
then by Lagrange’s Theorem, the order of an element
must divide the order of the group. So, every non-
identity element of 𝐺 must have order 2. It follows that
𝐺 = 𝑒, 𝑎 𝑏, 𝑐 , with 𝑎2
= 𝑏2
= 𝑐2
= 𝑒.

13. Lehman College, Department of Mathematics
Corollaries (3 of 3)
Proof (cont’d). Now, we have 𝑎𝑏 ≠ 𝑒, since 𝑏 ≠ 𝑎; and
𝑎𝑏 ≠ 𝑏 since 𝑎 ≠ 𝑒; also 𝑎𝑏 ≠ 𝑎 since 𝑏 ≠ 𝑒. It follows
that 𝑎𝑏 = 𝑐. Similar reasoning shows that 𝐺 is the Klein
4-group, which is abelian from its Cayley table below. ∎
From the above, we can conclude that the smallest
nonabelian group is 𝑆3 which has order 6.
⋅ 𝒆 𝒂 𝒃 𝒄
𝒆 𝒆 𝒂 𝒃 𝒄
𝒂 𝒂 𝒆 𝒄 𝒃
𝒃 𝒃 𝒄 𝒆 𝒂
𝒄 𝒄 𝒃 𝒂 𝒆

14. Lehman College, Department of Mathematics
Converse of Lagrange’s Theorem (1 of 2)
Theorem 3. The converse of Lagrange’s Theorem is
false.
Proof. We will show that the alternating group 𝐴4 of
even permutations of the symmetric group 𝑆4, which has
order 12 can have no subgroup of order 6. Homework 9,
question 2 asks for a list of the elements of 𝐴4 in cycle
notation:
𝐴4 has eight 3-cycles of order 3, as well as 3 double
transpositions of order 2. Suppose 𝐻 is a subgroup of
𝐴4 of order 6. Then 𝐻 cannot contain all eight 3-cycles.
𝐴4 =
𝑒, 1 2 3 , 1 3 2 , 1 2 4 , 1 4 2 , 1 3 4 , 1 4 3 ,
2 3 4 , 2 4 3 , 1 2 3 4 , 1 3 2 4 , 1 4 (2 3)

15. Lehman College, Department of Mathematics
Converse of Lagrange’s Theorem (2 of 2)
Theorem 3 (cont’d). Let 𝜎 ∉ 𝐻 be a 3-cycle that is not
in 𝐻, then we have 𝜎3
= 𝑒. Now, the cyclic subgroup
𝐾 = 𝜎 of 𝐴4 generated by 𝜎 is given by:
Consider the left cosets of 𝐻 in 𝐴4 generated by the
elements of 𝐾: 𝑒𝐻 = 𝐻, 𝜎𝐻, and 𝜎2
𝐻. They cannot all
be disjoint since 𝐻 = 𝜎𝐻 = |𝜎2
𝐻| and that would total
18 elements, but 𝐴4 = 12. So, either we have 𝜎H = 𝐻,
or 𝜎2
H = 𝐻 or 𝜎2
H = 𝜎𝐻. Now, if 𝜎H = 𝐻 then 𝜎 ∈ 𝐻, a
contradiction. If 𝜎2
H = 𝐻, then 𝜎2
∈ 𝐻, which implies
that (𝜎2
)2
= 𝜎4
= 𝜎3
𝜎 = 𝜎 ∈ 𝐻, another contradiction.
Finally, if 𝜎2
H = 𝜎𝐻, then 𝜎2
𝜎−1
= 𝜎 ∈ 𝐻, contradiction.
It follows that 𝐴4 cannot have a subgroup 𝐻 of order 6.
𝐾 = 𝑒, 𝜎, 𝜎2

16. Lehman College, Department of Mathematics
Corollaries (1 of 5)
Theorem 4. Let 𝐺 be a group, such that 𝐺 > 1. Show
that 𝐺 has only trivial subgroups if and only if |𝐺| is a
prime.
Proof (⟹). Let 𝐺 = 𝑝 for some prime 𝑝 and let 𝐻 be a
subgroup of 𝐺. By Lagrange’s Theorem, |𝐻| divides |𝐺|.
So 𝐻 = 1 or 𝑝. Thus, 𝐻 = 𝑒 or 𝐻 = 𝐺.
Conversely (⟸), suppose 𝐺 has only trivial subgroups.
Let 𝑎 ∈ 𝐺 (𝑎 ≠ 𝑒). Then 𝐻 = ⟨𝑎⟩ is a cyclic subgroup of
𝐺. But 𝐻 ≠ 𝑒 , since 𝑎 ≠ 𝑒, so 𝐻 = 𝐺 = ⟨𝑎⟩ and 𝐺 is a
finite cyclic group of order 𝑛. Suppose 𝑛 is not prime,
then 𝑛 = 𝑚𝑘 for some positive integers 0 < 𝑚, 𝑘 < 𝑛.
Since 𝑚 divides 𝑛 and 𝐺 is cyclic, then 𝐺 has a
subgroup of order 𝑚, generated by 𝑎 𝑛/𝑚
= 𝑎 𝑘
.

17. Lehman College, Department of Mathematics
Corollaries (2 of 5)
Theorem 4 (cont’d). This contradicts the assumption
that 𝐺 only has trivial subgroups. It follows that 𝑚 is
prime and 𝐺 is a group of prime order. ∎
Theorem 5. Let 𝐺 ≠ 𝑒 be a group of order 𝑝 𝑛
for some
prime 𝑝 and positive integer 𝑛. Then 𝐺 has a subgroup
(and an element) of order 𝑝.
Proof. Let 𝑎 ∈ 𝐺 (𝑎 ≠ 𝑒). Consider the cyclic subgroup 𝐻
of 𝐺 generated by 𝑎: 𝐻 = ⟨𝑎⟩ . Now, by Lagrange’s
Theorem |𝐻| divides 𝐺 = 𝑝 𝑛
. It follows that 𝐻 = 𝑝 𝑚
for some integer 0 < 𝑚 ≤ 𝑛. Since 𝐻 is cyclic, it has a
subgroup of order every divisor of 𝑝 𝑚
. Thus, 𝐺 has a
subgroup 𝐾 of order 𝑝, generated by 𝑎 𝑝 𝑚/𝑝
= 𝑎 𝑝 𝑚−1
. ∎

18. Lehman College, Department of Mathematics
Corollaries (3 of 5)
Theorem 6. Let 𝐺 be a group and let 𝐻 and 𝐾 be
subgroups of 𝐺, with 𝐻 = 𝑝 and 𝐾 = 𝑞 for distinct
primes 𝑝 and 𝑞. Show that 𝐻 ∩ 𝐾 = 𝑒 .
Proof. Let 𝑎 ∈ 𝐻 ∩ 𝐾. Now, 𝐻 ∩ 𝐾 is a subgroup of both
𝐻 and 𝐾. Consider the cyclic subgroup of ⟨𝑎⟩ generated
by 𝑎. By Lagrange’s Theorem, |⟨𝑎⟩| divides 𝐻 = 𝑝. It
follows that |⟨𝑎⟩| = 1 or 𝑝. Similarly, we have |⟨𝑎⟩|
divides 𝐾 = 𝑞. It follows that |⟨𝑎⟩| = 1 or 𝑞. Therefore,
|⟨𝑎⟩| = 1 and 𝑎 = 𝑒. Hence, 𝐻 ∩ 𝐾 = 𝑒 . ∎
Definition: The set 𝐻𝐾 ⊆ 𝐺 for subgroups 𝐻 and 𝐾 of a
group 𝐺 is defined as the set:
𝐻𝐾 = ℎ𝑘 | ℎ ∈ 𝐻, 𝑘 ∈ 𝐾 .

19. Lehman College, Department of Mathematics
Corollaries (4 of 5)
Theorem 6. Let 𝐺 be a group and let 𝐻 and 𝐾 be finite
subgroups of 𝐺. Then
Proof. Now, 𝐻𝐾 = ℎ𝑘 | ℎ ∈ 𝐻, 𝑘 ∈ 𝐾 . Since, by
definition, the left coset ℎ𝐾 = ℎ𝑘 | 𝑘 ∈ 𝐾 , it follows that
is a union of left cosets of 𝐾. Now, each left coset of 𝐾
has |𝐾| elements. To find 𝐻𝐾, it suffices to find the
number of distinct left cosets in the form ℎ𝐾 for some
ℎ ∈ 𝐻. For ℎ1, ℎ2 ∈ 𝐻, ℎ1 𝐾 = ℎ2 𝐾 iff ℎ1
−1
ℎ2 ∈ 𝐾 iff
ℎ1
−1
ℎ2 ∈ 𝐻 ∩ 𝐾 iff ℎ1 𝐻 ∩ 𝐾 = ℎ2 𝐻 ∩ 𝐾 .
|𝐻𝐾| =
𝐻 |𝐾|
|𝐻 ∩ 𝐾|
𝐻𝐾 =
ℎ∈𝐻
ℎ𝐾

20. Lehman College, Department of Mathematics
Corollaries (5 of 5)
Theorem 6 (cont’d). It follows that the number of
distinct left cosets of the form ℎ𝐾 is the same as the
number of distinct left cosets of the form ℎ(𝐻 ∩ 𝐾) for all
ℎ ∈ 𝐻. By Lagrange’s Theorem,
Since 𝐻 is finite, it follows that:
So, 𝐻𝐾 consists of
𝐻
|𝐻∩𝐾|
distinct cosets of 𝐾, each of
which contains |𝐾| elements. We conclude that
𝐻: 𝐻 ∩ 𝐾 =
𝐻
|𝐻 ∩ 𝐾|
|𝐻𝐾| =
𝐻 |𝐾|
|𝐻 ∩ 𝐾|
𝐻 = 𝐻: 𝐻 ∩ 𝐾 ⋅ |𝐻 ∩ 𝐾|

21. Lehman College, Department of Mathematics
Examples (1 of 2)
Example 4. Let 𝐺 be a group of order 𝑝𝑞𝑟, for distinct
primes 𝑝, 𝑞, and 𝑟. If 𝐻 and 𝐾 are subgroups of 𝐺 with
𝐻 = 𝑝𝑞 and 𝐾 = 𝑞𝑟, show that 𝐻 ∩ 𝐾 = 𝑞.
Solution. Now, 𝐻 ∩ 𝐾 is a subgroup of both 𝐻 and 𝐾, so
by Lagrange’s Theorem, 𝐻 ∩ 𝐾 divides both 𝐻 and
𝐾 . It follows that 𝐻 ∩ 𝐾 = 1 or 𝑝, or 𝑞, or 𝑝𝑞 and
𝐻 ∩ 𝐾 = 1 or 𝑞, or 𝑟, or 𝑞𝑟. Hence 𝐻 ∩ 𝐾 = 1 or 𝑞.
Now, 𝐻𝐾 ⊆ 𝐺, so 𝐻𝐾 ≤ 𝐺 = 𝑝𝑞𝑟. By Theorem 6,
If 𝐻 ∩ 𝐾 = 1, then
|𝐻𝐾| =
𝐻 |𝐾|
|𝐻 ∩ 𝐾|
𝐻𝐾 = 𝐻 𝐾 = 𝑝𝑞 𝑞𝑟 = 𝑝𝑞2
𝑟 > 𝑝𝑞𝑟

22. Lehman College, Department of Mathematics
Examples (2 of 2)
Example 4 (cont’d). If 𝐻 ∩ 𝐾 = 1, then
This is a contradiction, so 𝐻 ∩ 𝐾 = 𝑞.
Example 5. Let 𝐺 be a group of order 𝑝2
, where 𝑝 is
prime. Show that 𝐺 must have a subgroup of order 𝑝.
Solution. If 𝐺 is cyclic, then 𝐺 = 𝑎 = 𝑝2
for some
𝑎 ∈ 𝐺. Since 𝐺 has a subgroup of order every divisor of
𝑝2
, then 𝐺 has a subgroup 𝐻 or order 𝑝, generated by
𝑎 𝑝2/𝑝 = 𝑎 𝑝. If 𝐺 is not cyclic, let 𝑎 ∈ 𝐺 with 𝑎 ≠ 𝑒. By
Lagrange’s Theorem, o(𝑎) divides 𝐺 = 𝑝2
. So, o 𝑎 =
1, 𝑝 or 𝑝2
. But o 𝑎 ≠ 1 (𝑎 ≠ 𝑒) and, o 𝑎 ≠ 𝑝2
, since 𝐺
is not cyclic. So o 𝑎 = 𝑝, and 𝐾 = ⟨𝑎⟩ has order 𝑝.
𝐻𝐾 = 𝐻 𝐾 = 𝑝𝑞 𝑞𝑟 = 𝑝𝑞2
𝑟 > 𝑝𝑞𝑟