Section 11: Normal Subgroups

Abstract Algebra, Saracino
Second Edition
Section 11
Normal
Subgroups
"In mathematics the art of proposing
a question must be held of higher
value than solving it." – Georg Cantor -

Lehman College, Department of Mathematics
Galois and Abel
Niels Henrik Abel (1802-1829)
- Norwegian Mathematician
Évariste Galois (1811-1832)
- French Mathematician

Definition of Normal Subgroups (1 of 1)
Definition: Let 𝐺 be a group. A subgroup 𝐻 of 𝐺 is said
to be a normal subgroup of G if 𝑎𝐻 = 𝐻𝑎 for all 𝑎 ∈ 𝐺.
Note: For any group 𝐺, 𝑒 and 𝐺 are normal subgroups
since 𝑎 𝑒 = 𝑒 𝑎 = 𝑎 for any a ∈ 𝐺, and 𝑎𝐺 = 𝐺𝑎 = 𝐺.
It follows directly from the definition that every subgroup
of an abelian group is normal. However, the converse
of the last statement is not true. We can show that every
subgroup of the group of unit quaternions 𝒬8 is normal,
but 𝒬8 itself is nonabelian. Note also that if 𝐻 is a
normal subgroup of a group 𝐺, then it does not always
mean that 𝑎ℎ = ℎ𝑎 for every ℎ ∈ 𝐻 and every 𝑎 ∈ 𝐺.
This will be demonstrated in the following example.

Examples of Normal Subgroups (1 of 3)
Example 1. Consider the group 𝐺 = 𝑆3:
Let 𝐻 = 𝑒, 1 2 3 , (1 3 2) . Then 𝐻 is a subgroup of 𝐺.
Since 𝑒, (1 2 3), and (1 3 2) are elements of 𝐻, then
𝑒𝐻 = 𝐻𝑒, 1 2 3 𝐻 = 𝐻(1 2 3), and 1 3 2 𝐻 = 𝐻(1 3 2).
Now,
Hence, 1 2 𝐻 = 𝐻(1 2). Again,
Therefore 2 3 𝐻 = 𝐻 2 3 .
𝐺 = 𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)
1 2 𝐻 = 1 2 𝑒, 1 2 1 2 3 , 1 2 1 3 2 } = 1 2 , 2 3 , (1 3)
𝐻 1 2 = 𝑒 1 2 , 1 2 3 1 2 , 1 3 2 1 2 } = 1 2 , 1 3 , (2 3)
2 3 𝐻 = 2 3 𝑒, 2 3 1 2 3 , 2 3 1 3 2 } = 2 3 , 1 3 , (1 2)
𝐻 2 3 = 𝑒 2 3 , 1 2 3 2 3 , 1 3 2 2 3 } = 2 3 , 1 2 , (1 3)

Example 1 (cont’d). Similarly, we can show that
1 3 𝐻 = 𝐻 1 3 . It follows that 𝐻 is a normal subgroup
of 𝑆3. But for 1 2 3 ∈ 𝐺 and 2 3 ∈ 𝐺, we have:
Hence, 2 3 1 2 3 ≠ 1 2 3 2 3 . Now, consider the
subgroup 𝐾 = 𝑒, (1 2) of 𝐺 = 𝑆3. For this subgroup:
Hence, 1 3 𝐻 ≠ 𝐻(1 3), and so 𝐾 = 𝑒, (1 2) is not a
normal subgroup of 𝐺 = 𝑆3
2 3 1 2 3 = (1 3)
1 2 3 2 3 = (1 2)
1 3 𝐻 = 1 3 𝑒, (1 3)(1 2) = 1 3 , (1 2 3)
𝐻 1 3 = 𝑒 1 3 , (1 2) 1 3 = 1 3 , (1 3 2)

Example 2. Recall the group 𝐺𝐿 𝑛, ℝ of 𝑛 × 𝑛 real
matrices with nonzero determinant. The set S𝐿 𝑛, ℝ of
matrices in 𝐺𝐿 𝑛, ℝ with determinant 1 is a subgroup of
𝐺𝐿 𝑛, ℝ . First, S𝐿 𝑛, ℝ is a nonempty subset of
𝐺𝐿 𝑛, ℝ , because 𝐼 ∈ S𝐿 𝑛, ℝ . Next, if A, B ∈ 𝑆𝐿 𝑛, ℝ
then det 𝐴𝐵 = det 𝐴 det 𝐵 = 1 and
det 𝐴−1
= det 𝐴 −1
= 1. So 𝑆𝐿 𝑛, ℝ is closed under
both multiplication and inverses. In addition
since if A ∈ 𝑆𝐿 𝑛, ℝ and 𝐵 ∈ 𝐺𝐿 𝑛, ℝ then
Above we used the fact that det 𝐴𝐵 = det 𝐴 det 𝐵
and det 𝐴−1
= det 𝐴 −1
from linear algebra.
S𝐿 𝑛, ℝ ⊴ 𝐺𝐿 𝑛, ℝ
det 𝐵𝐴𝐵−1
= det 𝐵 det 𝐴 det 𝐵−1
= 1

Normal Subgroups (1 of 8)
Theorem 1. Let 𝐺 be a group and let 𝐻 be a subgroup
of 𝐺. The following conditions are equivalent:
a) 𝐻 is a normal subgroup of 𝐺
b) 𝑔𝐻𝑔−1
⊆ 𝐻 for all 𝑔 ∈ 𝐺
c) 𝑔𝐻𝑔−1 = 𝐻 for all 𝑔 ∈ 𝐺
Proof. 𝑎) ⇒ 𝑏) Let 𝑔 ∈ 𝐺. Since 𝐻 is a normal subgroup
of 𝐺, then 𝑔𝐻 = 𝐻𝑔. Hence for any ℎ ∈ 𝐻, 𝑔ℎ ∈ 𝐻𝑔, so
𝑔ℎ = ℎ′
𝑔 for some ℎ′
∈ 𝐻. Consequently, we have
𝑔ℎ𝑔−1
= ℎ′
𝑔𝑔−1
= ℎ′
∈ 𝐻, and 𝑔𝐻𝑔−1
⊆ 𝐻.
𝑏) ⇒ 𝑐) Let ℎ ∈ 𝐻, then 𝑔−1
ℎ 𝑔−1 −1
= 𝑔−1
ℎ𝑔 ∈ 𝐻.

Theorem 1 (cont’d). 𝑏) ⇒ 𝑐) Let ℎ ∈ 𝐻, then
𝑔−1
ℎ 𝑔−1 −1
= 𝑔−1
ℎ𝑔 ∈ 𝐻. This means that we have
𝑔 𝑔−1
ℎ𝑔 𝑔−1
∈ 𝑔𝐻𝑔−1
or ℎ ∈ 𝑔𝐻𝑔−1
. Hence, it follows
that 𝐻 ⊆ 𝑔𝐻𝑔−1
, and 𝑔𝐻𝑔−1
= 𝐻.
𝑐) ⇒ 𝑎) Since 𝑔𝐻𝑔−1
= 𝐻 for all 𝑔 ∈ 𝐺, then 𝑔𝐻 = 𝐻𝑔
for all 𝑔 ∈ 𝐺. Hence, 𝐻 is a normal subgroup of 𝐺. ∎
Theorem 11.2. Let 𝐺 be a group, then any subgroup of
the center 𝑍(𝐺) is a normal subgroup of 𝐺.
Proof. Let 𝐻 be any subgroup of 𝑍(𝐺). Let ℎ ∈ 𝐻 and let
𝑔 ∈ 𝐺. Since we have 𝐻 ⊆ 𝑍(𝐺) then 𝑔ℎ = ℎ𝑔 and
𝑔ℎ𝑔−1
= ℎ𝑔𝑔−1
= ℎ ∈ 𝐻. It follows that 𝑔𝐻𝑔−1
⊆ 𝐻 and
𝐻 is normal in 𝐺, by Theorem 1. ∎

Example 2. Let 𝐺 be a group and let 𝐻 be a subgroup
of 𝐺. If for all 𝑎, 𝑏 ∈ 𝐺, 𝑎𝑏 ∈ 𝐻 implies 𝑏𝑎 ∈ 𝐻, show that
𝐻 is normal in 𝐺.
Solution. Let ℎ ∈ 𝐻 be arbitrary. Then for any 𝑔 ∈ 𝐺
Since 𝑎𝑏 ∈ 𝐻 implies 𝑏𝑎 ∈ 𝐻 for any 𝑎𝑏 ∈ 𝐻, It follows
that 𝑔ℎ 𝑔−1
= 𝑔ℎ𝑔−1
∈ 𝐻 for all ℎ ∈ 𝐻 and for all 𝑔 ∈ 𝐺.
This means that 𝑔𝐻𝑔−1
⊆ 𝐻 and 𝐻 is normal in 𝐺, by
Theorem 1. ∎
ℎ = 𝑒ℎ = 𝑔−1
𝑔 ℎ = 𝑔−1
𝑔ℎ ∈ 𝐻

Corollary 1. If 𝐺 is abelian then every subgroup of 𝐺 is
normal.
Proof. If 𝐺 is abelian then 𝑍 𝐺 = 𝐺 and every subgroup
of 𝐺 is normal by Theorem 11.3. ∎
Theorem 11.3. Let 𝐺 be a group and let 𝐻 be a
subgroup of 𝐺 such that 𝐺: 𝐻 = 2, then 𝐻 is normal in
𝐺. (index 2 subgroups are normal).
Proof. Since there are exactly two distinct right cosets
of 𝐻 in 𝐺, one of them is 𝐻 and the other must be 𝐺 − 𝐻
(set difference) since right cosets of 𝐻 partition 𝐺.
Likewise, the distinct left cosets must be 𝐻 and 𝐺 − 𝐻.

Proof (cont’d). Thus, the left cosets are the right
cosets, and for any 𝑔 ∈ 𝐺, we have 𝑔𝐻 = 𝐻𝑔. ∎
Example 3. We know that every subgroup of an abelian
group is normal. Is the converse true? That is, if every
subgroup of a group is normal, is the group abelian?
Solution. Consider the group of unit quaternions,
where 𝒬8 = 1, −1, 𝑖, −𝑖, 𝑗, −𝑗, 𝑘, −𝑘 under multiplication,
with 𝑖2
= 𝑗2
= 𝑘2
= 𝑖𝑗𝑘 = −1. The subgroups of 𝒬8 are:
1 , −1 , 𝑖 , 𝑗 , 𝑘 , and 𝒬8. It follows that 𝒬8 is a
normal subgroup of itself. The subgroups 𝑖 , 𝑗 , and
𝑘 all have order 4, so they are index 2 subgroups and
are therefore normal by Theorem 11.3.

Solution (cont’d). The subgroups 1 and −1 are both
contained in the center 𝑍 𝒬 and are therefore normal
by Theorem 11.2. Thus, every subgroup of 𝒬8 is normal,
but 𝒬8 itself is nonabelian.
Example 4. We have 𝑆 𝑛 = 𝑛! and 𝐴 𝑛 = 𝑛!/2 so the
index 𝑆 𝑛: 𝐴 𝑛 = 2 and 𝐴 𝑛 is a normal subgroup of 𝑆 𝑛for
all 𝑛 by Theorem 11.3.
Theorem 11.4. Let 𝐺 be a group and let 𝐻 be a
subgroup of 𝐺. Suppose 𝑔 ∈ 𝐺, then 𝑔𝐻𝑔−1
is a
subgroup of 𝐺 with the same number of elements as 𝐻.
Proof. Let 𝑔 ∈ 𝐺, then 𝑔𝐻𝑔−1
⊆ 𝐺. Now, 𝑔𝐻𝑔−1
≠ ∅

Proof (cont’d). Now, 𝑔𝐻𝑔−1
≠ ∅ since we have
𝑔𝑒𝑔−1
= 𝑔𝑔−1
= 𝑒 ∈ 𝑔𝐻𝑔−1
. We will now show that
𝑔𝐻𝑔−1
is closed under multiplication. Let 𝑔ℎ1 𝑔−1
and
𝑔ℎ2 𝑔−1
be elements of 𝑔𝐻𝑔−1
for some ℎ1, ℎ2 ∈ 𝐻. Then
since ℎ1ℎ2 ∈ 𝐻 because 𝐻 is a subgroup of 𝐺. For
closure under inverses, let 𝑔ℎ𝑔−1
for some
ℎ ∈ 𝐻. Then
since ℎ−1
∈ 𝐺 because 𝐻 is a subgroup of 𝐺. It follows
that 𝑔𝐻𝑔−1 is a subgroup of 𝐺.
𝑔ℎ1 𝑔−1 𝑔ℎ1 𝑔−1 = 𝑔ℎ1 𝑔−1 𝑔 ℎ1 𝑔−1
= 𝑔ℎ1ℎ2 𝑔−1
𝑔ℎ𝑔−1 −1
= 𝑔−1 −1
ℎ−1
𝑔−1
= 𝑔ℎ−1
𝑔−1

Proof (cont’d). To prove that |𝑔𝐻𝑔−1
| = |𝐻| we will
construct a one-to-one and onto function from 𝐻 to
𝑔𝐻𝑔−1
. Let 𝑓: 𝐻 → 𝑔𝐻𝑔−1
be given by 𝑓 ℎ = 𝑔ℎ𝑔−1
for
all ℎ ∈ 𝐻. To show that 𝑓 is one-to-one, let ℎ1, ℎ2 ∈ 𝐻
and let 𝑓 ℎ1 = 𝑓(ℎ2) then 𝑔ℎ1 𝑔−1
= 𝑔ℎ2 𝑔−1
. By left
and right cancellation, ℎ1 = ℎ2 and 𝑓 is one-to-one. To
show that 𝑓 is onto 𝑔𝐻𝑔−1
, let 𝑔ℎ𝑔−1
be an arbitrary
element of 𝑔𝐻𝑔−1
, then 𝑔ℎ𝑔−1
= 𝑓(ℎ) for some ℎ ∈ 𝐻. It
follows that 𝑓 is onto 𝑔𝐻𝑔−1
, and |𝑔𝐻𝑔−1
| = |𝐻| ∎
Corollary 11.5. If 𝐺 is a group and 𝐻 a subgroup of 𝐺
and no other subgroup has the same number of
elements as 𝐻, then 𝐻 is normal in 𝐺.

Quotient (Factor) Groups (1 of 5)
Proof . For any 𝑔 ∈ 𝐺, 𝑔𝐻𝑔−1
is a subgroup of 𝐺 with
the same number of elements as 𝐻 . By the hypothesis,
𝑔𝐻𝑔−1
= 𝐻, so by Theorem 1, 𝐻 is normal in 𝐺. ∎
Notation: If 𝐻 is a normal subgroup of a group 𝐺, we
write:
Notation: If 𝐻 ⊴ 𝐺, then 𝐺/𝐻 is defined as the set of all
left (= right) cosets of 𝐻 in 𝐺. That is 𝐺/𝐻 = 𝑎𝐻 | 𝑎 ∈ 𝐺
Theorem 11.6. Let 𝐺 be a group and let 𝐻 be a normal
subgroup of 𝐺, then 𝐺/𝐻 is a group under the operation
∗ defined below:
For all 𝑎𝐻, 𝑏𝐻 ∈ 𝐺/𝐻, 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻
𝐻 ⊴ 𝐺

Proof. We want to show that the operation ∗ is a well-
defined binary operation on the set 𝐺/𝐻. That is, if we
have 𝑎𝐻 = 𝑎1 𝐻 and 𝑏𝐻 = 𝑏1 𝐻, then
Now 𝑎𝐻 = 𝑎1 𝐻 means 𝑎1 𝑎−1
∈ 𝐻 and 𝑏𝐻 = 𝑏1 𝐻 means
𝑏1 𝑏−1
∈ 𝐻. It follows that we want:
That is, we want 𝑎1 𝑏1 𝑎𝑏 −1
∈ 𝐻. But
𝑎𝐻 ∗ 𝑏𝐻 = 𝑎1 𝐻 ∗ 𝑏1 𝐻
𝑎𝐻 ∗ 𝑏𝐻 = 𝑎𝑏 𝐻 = 𝑎1 𝑏1 𝐻 = 𝑎1 𝐻 ∗ 𝑏1 𝐻
𝑎1 𝑏1 𝑎𝑏 −1
= 𝑎1 𝑏1 𝑏−1
𝑎−1
= 𝑎1 𝑏1 𝑏−1
𝑎1
−1
𝑎1 𝑎−1
= 𝑎1 𝑏1 𝑏−1
𝑎1
−1
𝑎1 𝑎−1

Proof (cont’d). We have
Since 𝑎1 𝑎−1
∈ 𝐻 and 𝑏1 𝑏−1
∈ 𝐻, then 𝑎1 𝑏1 𝑎𝑏 −1
is an
element of 𝐻, if and only if 𝑎1 𝐻𝑎1
−1
is a subset of 𝐻.
That is, if 𝐻 is normal in 𝐺.
We will now check if ∗ is associative. Let 𝑎𝐻, 𝑏𝐻, 𝑐𝐻 be
elements of 𝐻/𝐺, then
Hence, ∗ is an associative binary operation on 𝐺/𝐻.
𝑎1 𝑏1 𝑎𝑏 −1
= 𝑎1 𝑏1 𝑏−1
𝑎−1
= 𝑎1 𝑏1 𝑏−1
𝑎1
−1
𝑎1 𝑎−1
= 𝑎1 𝑏1 𝑏−1 𝑎1
−1
𝑎1 𝑎−1
𝑎𝐻 ∗ 𝑏𝐻 ∗ 𝑐𝐻 = 𝑎𝑏 𝐻 ∗ 𝑐𝐻 = 𝑎𝑏 𝑐𝐻 = 𝑎𝑏𝑐 𝐻
𝑎𝐻 ∗ 𝑏𝐻 ∗ 𝑐𝐻 = 𝑎𝐻 ∗ 𝑏𝑐 𝐻 = 𝑎 𝑏𝑐 𝐻 = 𝑎𝑏𝑐 𝐻

Proof (cont’d). We will now check if ∗ has an identity
element. Let 𝑎𝐻 be an element of 𝐻/𝐺, then
So 𝑒𝐻 = 𝐻 is the identity element in 𝐺/𝐻 under ∗. For
inverses, let 𝑎𝐻 ∈ 𝐺/𝐻, then
It follows that 𝐺/𝐻 is closed under inverses and 𝐺/𝐻 is
a group under the binary operation ∗. ∎
Definition: Let 𝐺 be a group and let 𝐻 be a normal
subgroup of 𝐺. Then the group 𝐺/𝐻 of all left cosets of
𝐻 in 𝐺 under the binary operation 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 is
called the quotient group or factor group of 𝐺 by 𝐻.
𝑎𝐻 ∗ 𝑒𝐻 = 𝑎𝑒 𝐻 = 𝑎𝐻 = 𝑒𝑎 𝐻 = 𝑒𝐻 ∗ 𝑎𝐻
𝑎𝐻 ∗ 𝑎−1
𝐻 = 𝑎𝑎−1
𝐻 = 𝑒𝐻 = 𝑎−1
𝑎 𝐻 = 𝑎−1
𝐻 ∗ 𝑎𝐻

Definition: Let 𝐺 be a group and let 𝐻 be a normal
subgroup of 𝐺. Then the group 𝐺/𝐻 of all left cosets of
𝐻 in 𝐺 under the binary operation 𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻 is
called the quotient group or factor group of 𝐺 by 𝐻.
We defined [𝐺: 𝐻] (the index of 𝐻 and 𝐺) as the number
of distinct left (or right) cosets. Since 𝐺/𝐻 is the set of
all left (or right) cosets of a normal subgroup 𝐺, then:
If the group 𝐺 is finite, then by Lagrange’s Theorem, we
have 𝐺: 𝐻 =
|𝐺|
|𝐻|
. It follows that for finite groups 𝐺:
|𝐺/𝐻| = [𝐺: 𝐻]
|𝐺/𝐻| =
|𝐺|
|𝐻|

Examples of Quotient Groups (1 of 3)
Example 5. Let 𝐺 = ℤ6 = 0, 1, 2, 3, 4, 5 under addition
modulo 6. Then 𝐺 is cyclic, hence abelian. From
Corollary 1, we know that every subgroup of 𝐺 is
normal. Let 𝐻 = 0, 3 be a subgroup of 𝐺. Then 𝐻 ⊴ 𝐺,
and the distinct left (right) cosets of 𝐻 in 𝐺 are:
𝐻 = 0 + 𝐻 = 0, 3 = 3 + 𝐻, 1 + 𝐻 = 1, 4 = 4 + 𝐻, and
2 + 𝐻 = 2, 5 = 5 + 𝐻. We see that |𝐺/𝐻| = 3, and
|𝐺| = 6 with |𝐻| = 2, so 𝐺 /|𝐻| = 6/2 = 3.
1 + 𝐻 and 2 + 𝐻 are inverses of each other. 1 + 𝐻 and
2 + 𝐻 both have order 3. This is actually the group ℤ3.
𝐺/𝐻 = 𝐻, 1 + 𝐻, 2 + 𝐻
1 + 𝐻 + 2 + 𝐻 = 3 + 𝐻 = 𝐻
1 + 𝐻 + 1 + 𝐻 + (1 + 𝐻 = 3 + 𝐻 = 𝐻

Example 6. In Example 1, we looked at 𝐺 = 𝑆3
We showed that the subgroup 𝐻 = 𝑒, 1 2 3 , (1 3 2) is
normal in 𝐺. Let us look at the set of distinct cosets of 𝐻
in 𝐺:
Here:
We see that |𝐺/𝐻| = 2, and |𝐺| = 6 with |𝐻| = 3, so
𝐺 /|𝐻| = 6/3 = 2. Now,
So 1 2 𝐻 is its own inverse. This is the group ℤ2.
𝐺/𝐻 = 𝐻, (1 2)𝐻
𝐺 = 𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)
1 2 𝐻 = 1 2 , 2 3 , (1 3) = 2 3 𝐻 = 1 3 𝐻
𝐻 = 𝑒𝐻 = 𝑒, 1 2 3 , (1 3 2) = 1 2 3 𝐻 = 1 3 2 𝐻
1 2 𝐻 ⋅ 1 2 𝐻 = 1 2 1 2 𝐻 = 𝑒𝐻 = 𝐻

Example 7. Consider the group ℤ under addition and
consider the subgroup 3ℤ = … , −6, −3, 0, 3, 6, … . Since
ℤ is abelian, we know from Corollary 1 that every
subgroup of ℤ is normal. The left cosets of 3ℤ in ℤ are:
The distinct left cosets are: 3ℤ, 1 + 3ℤ and 2 + 3ℤ. Then
0 + 3ℤ = … , −9, −6, −3, 0, 3, 6, 9, … = 3ℤ
1 + 3ℤ = … , −8, −5, −2, 1, 4, 7, 10, …
2 + 3ℤ = … , −7, −4, −1, 2, 5, 8, 11, …
3 + 3ℤ = … , −6, −3, 0, 3, 6, 9, 12, … = 3ℤ
4 + 3ℤ = … , −5, −2, 1, 4, 7, 10, 13, … = 1 + 3ℤ
−1 + 3ℤ = … , −10, −7, −4, −1, 2, 5, 8, … = 2 + 3ℤ
−2 + 3ℤ = … , −11, −8, −5, −2, 1, 4, 7, … = 1 + 3ℤ
ℤ/3ℤ = 3ℤ, 1 + 3ℤ, 2 + 3ℤ

Corollaries (1 of 10)
Example 8. Let 𝐺 be a group and let 𝐻 be a normal
subgroup of 𝐺. Show that if 𝐺 is abelian then the
quotient group 𝐺/𝐻 is also abelian.
Solution. Let 𝑎𝐻 and 𝑏𝐻 be arbitrary elements of 𝐺/𝐻
for some 𝑎, 𝑏 ∈ 𝐺. Then
But 𝑎𝑏 = 𝑏𝑎 for all 𝑎, 𝑏 ∈ 𝐺 since 𝐺 is abelian. Hence,
Since 𝑎𝐻 ∗ 𝑏𝐻 = 𝑏𝐻 ∗ 𝑎𝐻 for all 𝑎𝐻, 𝑏𝐻 in 𝐺/𝐻 it follows
that 𝐺/𝐻 is abelian. ∎
𝑎𝐻 ∗ 𝑏𝐻 = (𝑎𝑏)𝐻
𝑎𝐻 ∗ 𝑏𝐻 = 𝑎𝑏 𝐻 = 𝑏𝑎 𝐻 = 𝑏𝐻 ∗ 𝑎𝐻

Example 9. Let 𝐺 be a group and let 𝐻 and 𝐾 be
subgroups of 𝐺 with 𝐻 normal in 𝐺. Show that 𝐻 ∩ 𝐾 is a
normal subgroup of 𝐾.
Solution. We know that 𝐻 ∩ 𝐾 is a subgroup of 𝐺. Since
𝐻 ∩ 𝐾 is contained in both 𝐻 and 𝐾 then 𝐻 ∩ 𝐾 is a
subgroup of 𝐻 and 𝐾, respectively. In order show that
𝐻 ∩ 𝐾 ⊴ 𝐾, it is sufficient to show that for all 𝑘 ∈ 𝐾:
Let 𝑎 ∈ 𝐻 ∩ 𝐾, then 𝑎 ∈ 𝐻 and 𝑎 ∈ 𝐾.Hence, 𝑘𝑎𝑘−1
∈ 𝐾
for all 𝑘 ∈ 𝐾. We want to show that 𝑘𝑎𝑘−1
∈ 𝐻 ∩ 𝐾.
Since 𝐻 ⊴ 𝐺, then 𝑔𝑎𝑔−1
∈ 𝐻 for all 𝑎 ∈ 𝐻 and all 𝑔 ∈ 𝐺.
But 𝑘 ∈ 𝐾 ⊆ 𝐺 so 𝑘 ∈ 𝐺. It follows that 𝑘𝑎𝑘−1 ∈ 𝐻.
𝑘 𝐻 ∩ 𝐾 𝑘−1
⊆ 𝐻 ∩ 𝐾

Solution (cont’d). Now, 𝑘𝑎𝑘−1
∈ 𝐾 and 𝑘𝑎𝑘−1
∈ 𝐻 for
all 𝑘 ∈ 𝐾. It follows that 𝑘𝑎𝑘−1
∈ 𝐻 ∩ 𝐾, and we have
𝑘 𝐻 ∩ 𝐾 𝑘−1
⊆ 𝐻 ∩ 𝐾 for all 𝑘 ∈ 𝐾, but this implies that
𝐻 ∩ 𝐾 is normal in 𝐾 by Theorem 1.
Example 10. Let 𝐺 be an abelian group of odd order 𝑛
and let 𝐺 = 𝑔1, 𝑔2, 𝑔3, … , 𝑔 𝑛 = 𝑒 . Show that the product
Solution. Since 𝐺 is of odd order, by Lagrange’s
Theorem 𝐺 can have no element of order 2 (the order of
an element must divide the order of the group). Thus,
each nonidentity element is distinct from its respective
inverse. Since 𝐺 is abelian, then the above product is a
product of (𝑛 − 1)/2 copies of the identity. ∎
𝑔1 𝑔2 𝑔3 ⋯ 𝑔 𝑛 = 𝑒

Theorem 3. Let 𝐺 be a group. Show that if 𝐺/𝑍(𝐺) is
cyclic, then 𝐺 is abelian.
Proof: Let 𝑍 𝐺 = 𝑍. We know that 𝑍 ⊴ 𝐺, so 𝐺/𝑍 is a
group under the operation 𝑎𝑍 ∗ 𝑏𝑍 = 𝑎𝑏 𝑍 for all
elements 𝑎, 𝑏 ∈ 𝐺. Since 𝐺/𝑍 is cyclic, then 𝐺/𝑍 = ⟨𝑔𝑍⟩
for some 𝑔𝑍 ∈ 𝐺/𝑍. Let 𝑎, 𝑏 ∈ 𝐺, then 𝑎𝑍, 𝑏𝑍 ∈ 𝐺/𝑍.
Since 𝐺/𝑍 is cyclic then 𝑎𝑍 = 𝑔𝑍 𝑛
= 𝑔 𝑛
𝑍 and, also
𝑏𝑍 = 𝑔𝑍 𝑚
= 𝑔 𝑚
𝑍 for some 𝑛, 𝑚 ∈ ℤ. It follows that we
have 𝑎 ∈ 𝑔 𝑛
𝑍 and 𝑏 ∈ 𝑔 𝑚
𝑍. Therefore, 𝑎 = 𝑔 𝑛
𝑧1 and
𝑏 = 𝑔 𝑚 𝑧2 for some 𝑧1, 𝑧2 ∈ 𝑍. Now,
𝑎𝑏 = (𝑔 𝑛
𝑧1) 𝑔 𝑚
𝑧2 = 𝑔 𝑛
𝑔 𝑚
𝑧1 𝑧2 = 𝑔 𝑛+𝑚
𝑧1 𝑧2
= 𝑔 𝑚+𝑛 𝑧2 𝑧1 = 𝑔 𝑚 𝑔 𝑛 𝑧2 𝑧1 = (𝑔 𝑚 𝑧2)(𝑔 𝑛 𝑧1) = 𝑏𝑎

Example 11. Let 𝐺 be a group and let 𝐻 ⊴ 𝐺. Suppose
𝐻 = 2. Show that 𝐻 ⊆ 𝑍 𝐺 .
Solution: Since 𝐻 = 2, then 𝐻 = 𝑒, 𝑎 for some 𝑎 ∈ 𝐺,
(𝑎 ≠ 𝑒). Since 𝐻 ⊴ 𝐺, then 𝑔𝐻𝑔−1
= 𝐻 for all 𝑔 ∈ 𝐺, but
So 𝑔𝑎𝑔−1
= 𝑎 for all 𝑔 ∈ 𝐺. Therefore 𝑔𝑎 = 𝑎𝑔 for all
𝑔 ∈ 𝐺. It follows that 𝑎 ∈ 𝑍 𝐺 . Since 𝑒 ∈ 𝑍(𝐺), then
Theorem 4. Let 𝐺 be a finite group and let 𝐻 and 𝐾 be
subgroups of 𝐺.
𝑔𝐻𝑔−1
= 𝑔𝑒𝑔−1
, 𝑔𝑎𝑔−1
= 𝑒, 𝑔𝑎𝑔−1
= 𝑒, 𝑎 = 𝐻
𝐻 = 𝑒, 𝑎 ⊆ 𝑍(𝐺)

Theorem 4. Let 𝐺 be a finite group and let 𝐻 and 𝐾 be
subgroups of 𝐺.
a) If |𝐻| and |𝐾| are relatively prime (gcd 𝐻 , |𝐾| = 1),
then 𝐻 ∩ 𝐾 = 𝑒 .
b) If 𝐻 and 𝐾 are two distinct subgroups, both of prime
order 𝑝, then 𝐻 ∩ 𝐾 = 𝑒 .
Proof (a). Now, 𝐻 ∩ 𝐾 is a subgroup of 𝐺 (cf. Homework
5, #1). Since 𝐻 ∩ 𝐾 is contained in both 𝐻 and 𝐾, then
𝐻 ∩ 𝐾 is subgroup of both 𝐻 and 𝐾. By Lagrange’s
Theorem, |𝐻 ∩ 𝐾| is a divisor of both 𝐻| and |𝐾|, but
gcd 𝐻 , |𝐾| = 1, so 𝐻 ∩ 𝐾 = 1 and 𝐻 ∩ 𝐾 = 𝑒 .

Proof (b). Now, 𝐻 ∩ 𝐾 is a subgroup of 𝐻, which has
prime order. By Lagrange’s Theorem, 𝐻 ∩ 𝐾 = 1 or
𝐻 ∩ 𝐾 = 𝐻 . Since 𝐻 ∩ 𝐾 ⊆ 𝐻, then 𝐻 ∩ 𝐾 = 𝑒 or
𝐻 ∩ 𝐾 = 𝐻. Since 𝐻 ∩ 𝐾 ⊆ 𝐾, if 𝐻 ∩ 𝐾 = 𝐻, then 𝐻 ⊆ 𝐾.
But 𝐻 = 𝐾 and 𝐻 ≠ 𝐾, so there is some 𝑎 ∈ 𝐻 with
𝑎 ∉ 𝐾. Therefore 𝐻 cannot be a subset of 𝐾. It follows
then that 𝐻 ∩ 𝐾 = 𝑒 . ∎
Theorem 5. Let 𝐺 be a group of order 𝑝𝑞 for distinct
primes 𝑝 and 𝑞 (𝑝 < 𝑞). Then all proper subgroups of 𝐺
are cyclic and 𝐺 has an element of order 𝑝.
Proof. By Lagrange’s Theorem, the order of all
subgroups of 𝐺 is a divisor of 𝐺 = 𝑝𝑞. The only proper

Proof (cont’d). The only proper divisors of 𝑝𝑞 are 1, 𝑝,
and 𝑞. So, if 𝐻 is a proper subgroup of 𝐺, then we have
𝐻 = 1, 𝑝, or 𝑞. If 𝐻 = 1, then 𝐻 = 𝑒 = ⟨𝑒⟩ is cyclic.
Otherwise, if 𝐻 = 𝑝, or 𝑞 (primes) then 𝐻 is cyclic.
Let 𝑚 be the maximum order of all elements of 𝐺. Then,
by Lagrange’s Theorem 𝑚 = 1, 𝑝, 𝑞, or 𝑝𝑞. If 𝑚 = 𝑝𝑞,
then 𝐺 has an element 𝑎 of order 𝑝𝑞. Thus, 𝐺 = ⟨𝑎⟩ is
cyclic. Since 𝑎 𝑝𝑞
= 𝑒, then 𝑎 𝑞 𝑝
= 𝑒 and 𝑎 𝑞
is an
element of order 𝑝 in 𝐺. If 𝑚 = 𝑝, then 𝐺 has an element
of order 𝑝. However, if 𝑚 = 𝑞 then there exists 𝑏 ∈ 𝐺 of
order 𝑞. Consider the cyclic subgroup 𝐻 of 𝐺 generated
by 𝑏. 𝐻 = 𝑒, 𝑏, 𝑏2
, 𝑏3
, … , 𝑏 𝑞−1
. Every nonidentity
element of 𝐻 has order 𝑞.

Proof (cont’d). From Theorem 4(b), if 𝐾 is another
subgroup of 𝐺 of order 𝑞, then 𝐻 ∩ 𝐾 = 𝑒 . It follows
that 𝐻 has 𝑞 − 1 elements of order 𝑞 that are not in 𝐾,
and vice versa. Since 𝐺 has one element 𝑒 of order 1,
Then 𝐺 can have at most 𝑝𝑞 − 1 elements of order 𝑞. It
follows that 𝐺 can have at most 𝑝 distinct subgroups of
order 𝑞.
Thus, the maximum number of elements of order 𝑞 in 𝐺
is 𝑝 𝑞 − 1 . Since if 𝐺 had 𝑝 + 1 distinct subgroups of
order 𝑞, it would have 𝑝 + 1 𝑞 − 1 elements of order
𝑞. Now, 𝑝 + 1 𝑞 − 1 = 𝑝𝑞 + 𝑞 − 𝑝 − 1 = 𝑝𝑞 − 1 + (𝑞 −
𝑝). But 𝑞 > 𝑝, so 𝑞 − 𝑝 > 0 and the number of elements
of order 𝑞 would exceed 𝑝𝑞 − 1.

Proof (cont’d). But the total number of nonidentity
elements in 𝐺 is 𝑝𝑞 − 1. We know that 𝐺 has at most
𝑝 𝑞 − 1 = 𝑝𝑞 − 𝑝 elements of order 𝑞. So, the
difference:
Since the smallest prime is 2. It follows that 𝐺 has at
least one element of order 𝑝.
𝑝𝑞 − 1 − 𝑝𝑞 − 𝑝 = 𝑝 − 1 ≥ 1

Section 11: Normal Subgroups

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Section 11: Normal Subgroups