Group Theory
V. V. Acharya
Department of Mathematics,
Fergusson College, Pune-4
September 9, 2025
2
Contents
1 Normal Subgroups and Homomorphism
5
1.1 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4 Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5 Centralizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3
4
Dr. V. V. Acharya
Chapter 1
Normal Subgroups and
Homomorphism
Normal subgroups form a special class of subgroups. It is a tribute to the
genius of Galois who recognized these subgroups and discovered relevant
concepts. In this chapter, we define normal subgroups, discuss elementary
properties of normal subgroups and study some examples. Then, we discuss
the notion of homomorphism.
1.1
Normal Subgroups
Definition 1.1 Let G be a group and N be a subgroup of G. N is said to
be a normal subgroup of G if for every g ∈ G and n ∈ N, gng −1 ∈ N.
Notation. If N is a normal subgroup of G, we denote it by N ◁ G.
Example 1.1 If G is an abelian group, then every subgroup of G is a normal
subgroup.
Solution. Let N be a subgroup of G. We have to show that for every g ∈ G
and n ∈ N, gng −1 ∈ N. Now, gng −1 = ngg −1 as G is an abelian group.
Hence, gng −1 = n ∈ N . Hence, every subgroup of an abelian group is a
normal subgroup.
Let G be a group and N be a subgroup of G. Suppose g ∈ G. By gN g −1 ,
we mean the set of all the elements of the type gng −1 where n ∈ N. Thus,
gN g −1 = {gng −1 |n ∈ N }.
Note that gN g −1 is also a subgroup of G. Further, from the definition of
normal subgroup it follows that N is a normal subgroup of G if and only if
gN g −1 ⊂ N for all g ∈ G.
5
6
Chapter 1. Normal Subgroups and Homomorphism
Lemma 1.1 Let G be a group and N be a subgroup of G. N is a normal
subgroup of G if and only if gN g −1 = N for every g ∈ G.
Proof. If gN g −1 = N for every g ∈ G, then gN g −1 ⊂ N for every g ∈ G.
Hence, N is a normal subgroup of G.
Conversely, assume that N is a normal subgroup of G. Hence, for every
g ∈ G, gN g −1 ⊂ N. It remains to show that N ⊂ gN g −1 for every g ∈ G.
Let n ∈ N . Since, N is a normal subgroup of G, g −1 ng ∈ N for every g ∈ G.
Hence, g(g −1 ng)g −1 ∈ gN g −1 . But, g(g −1 ng)g −1 = n. Hence, n ∈ gN g −1 and
N ⊂ gN g −1 . This establishes the lemma.
Lemma 1.2 The subgroup N of G is a normal subgroup of G if and only if
every left coset of N in G is a right coset of N in G.
Proof. If N is a normal subgroup of G, then for every g ∈ G, gN g −1 = N,
hence (gN g −1 )g = N g. But (gN g −1 )g = gN, hence the left coset gN equals
the right coset N g.
Suppose, conversely, that every left coset of N in G is a right coset of
N in G. Thus, for g ∈ G, gN, being a left coset, must be a right coset.
Since, g = ge ∈ gN, whatever the right coset gN turns out to be, it must
contain the element g; however g is in the right coset N g, and two distinct
right cosets have no element in common. So this right coset is unique. Thus,
gN = N g. Thus, for g ∈ G and n ∈ N, gng −1 = (gn)g −1 . Now gn ∈ gN.
But, gN = N g. Hence, gn ∈ N g. Hence, there exists n1 such that gn = n1 g.
Hence, gng −1 = (n1 g)g −1 = n1 ∈ N for every g ∈ G and n ∈ N. Hence, N is
a normal subgroup of G.
Lemma 1.3 A subgroup N of G is a normal subgroup of G if and only if
the product of two right cosets of N in G is again a right coset of N in G.
Proof. Suppose N is a normal subgroup of G, and that a, b ∈ G. Consider
(N a)(N b); since N is normal in G, aN = N a, and so
N aN b = N (aN )b = N (N a)b = N ab.
Lemma 1.4 If H is a subgroup of index 2 in G then H is a normal subgroup
of G.
Proof. Suppose H is a subgroup of index 2 in G and a ∈ G. If a ∈ H
then aH = H = Ha. Suppose a ̸∈ H. Thus, H and Ha are distinct right
cosets of H in G. Also, H and aH are distinct left cosets of H in G. Since
G = H ∪ Ha = H ∪ aH, we get Ha = aH. This shows that every left coset
of H in G is a right coset of H in G. Hence H is a normal subgroup of G.
1.1. Normal Subgroups
7
Lemma 1.5 If H is the only subgroup of G of a fixed finite order then H is
a normal subgroup of G.
Proof. We note that if H is a subgroup of G then gHg −1 is also a subgroup
of G. Also, both H and gHg −1 have equal number of elements. Since, H
is the only subgroup of G of a fixed finite order we get H = gHg −1 for all
g ∈ G. Hence, H is a normal subgroup of G.
Remark 1.1 Let
\ {H}i∈I be a family of normal subgroups of G. Then the
subgroup H =
Hi is a normal subgroup. Indeed, if y ∈ H, and x ∈ G,
i∈I
then xyx−1 lies in each Hi , whence in H.
Examples
1. If G is the group of quaternions, then H = {±1, ±i} is a normal subgroup of G as index of H in G is 2. Similarly, {±1, ±j} and {±1, ±k}
are also normal subgroups of G. Observe that N = {1, −1} is also a
normal subgroup of G for if g ∈ G and n ∈ N then gng −1 = n for all
n ∈ N and g ∈ G. Hence, N is a normal subgroup of G.
2. If G = S3 then H = {ρ0 , ρ1 , ρ2 } is a normal subgroup as index of H in
G equals 2. We observe that H = {ρ0 , µ1 } is not a normal subgroup
of G for ρ1 H = {ρ1 , µ2 } and Hρ1 = {ρ1 , µ3 }. Thus, ρ1 H ̸= Hρ1 .
Similarly, ρ2 H = {ρ2 , µ3 } and Hρ2 = {ρ2 , µ2 }. Similarly, we can show
that {ρ0 , µ2 } and {ρ0 , µ3 } are also not normal subgroups of S3 .
Example 1.2 Show that the intersection of two normal subgroups of G is a
normal subgroup of G.
Solution. Let N1 and N2 be normal subgroups of G. Let N = N1 ∩ N2 . We
note that N is a subgroup of G. Suppose n ∈ N and g ∈ G. Then n ∈ N1
and n ∈ N2 . Since, N1 and N2 are normal subgroups of G, gng −1 ∈ N1 and
gng −1 ∈ N2 for all g ∈ G. Thus, for every n ∈ N, gng −1 ∈ N. Hence, N is a
normal subgroup of G.
Example 1.3 Prove that there exist three groups K ⊂ H ⊂ G, where K is
normal in H, H is normal in G, but K is not normal in G.
Solution. Consider the group D4 given by
D4 = {ρ0 , ρ1 , ρ2 , ρ3 , µ1 , µ2 , δ1 , δ2 }.
8
Chapter 1. Normal Subgroups and Homomorphism
The group D4 has three subgroups of order 4 given by
H1 = {ρ0 , ρ1 , ρ2 , ρ3 }, H2 = {ρ0 , ρ2 , µ1 , µ2 }, H3 = {ρ0 , ρ2 , δ1 , δ2 }.
Index of H1 , H2 and also of H3 in D4 is 2. Hence, these are normal
subgroups of D4 . Further, D4 has five subgroups of order 2 given by
K1 = {ρ0 , ρ2 }, K2 = {ρ0 , µ1 }, K3 = {ρ0 , µ2 },
K4 = {ρ0 , δ1 }, K5 = {ρ0 , δ2 }.
Observe that K1 is a normal subgroup of H1 as well as that of H2 and H3 as
it is of index 2 in each of these subgroups. Also, K1 is a normal subgroup of
D4 as both ρ0 , ρ2 commute with every element of D4 .
Also, K2 and K3 are normal subgroups of H2 as they are of index 2 in
H2 . But ρ1 K2 ̸= K2 ρ1 and ρ1 K3 ̸= K3 ρ1 . Hence, both K2 and K3 are not
normal subgroups of D4 .
Also, K4 and K5 are normal subgroups of H3 as they are of index 2 in
H3 . But ρ1 K4 ̸= K4 ρ1 and ρ1 K5 ̸= K3 ρ5 . Hence, both K4 and K5 are not
normal subgroups of D4 .
Thus, we get that normality is not transitive.
1.2
Quotient Groups
Suppose that N is a normal subgroup of G. Let G/N denote the collection
of right cosets of N in G. Since, N is a normal subgroup of G, product of
two right cosets of N in G is again a right coset of N in G. We can make
use of this product to make the collection of right cosets into a group. We
prove this in the following theorem:
Theorem 1.1 If G is a group, N a normal subgroup of G, then G/N is also
a group.
Proof. Note that G/N denotes the collection of right cosets of N in G.
1. Suppose X, Y ∈ G/N. Then X = N a and Y = N b for some a, b ∈ G.
Hence, XY = (N a)(N b) = N ab ∈ G/N. Thus, the product is a binary
operation.
2. Suppose X, Y, Z ∈ G/N. Then X = N a, Y = N b and Z = N c for some
a, b, c ∈ G. Hence,
(XY )Z = (N aN b)N c = N (ab)N c = N (ab)c = N a(bc)
1.2. Quotient Groups
9
(since product in G is associative). Also,
X(Y Z) = N a(N bN c) = N aN (bc) = N a(bc).
Thus, (XY )Z = X(Y Z). Hence, the product in G/N is associative.
3. Consider the element N = N e ∈ G/N. If X ∈ G/N, X = N a, a ∈
G, so XN = N aN e = N ae = N a = X, and similarly N X = X.
Consequently, N is an identity element for G/N.
4. Suppose X = N a ∈ G/N (where a ∈ G); thus N a−1 ∈ G/N, and
N aN a−1 = N aa−1 = N e = N. Similarly, N a−1 N a = N a−1 a = N e =
N. Hence, N a−1 is the inverse of N a in G/N.
This shows that G/N is a group.
Definition 1.2 If G is a group, N a normal subgroup of G, then the group
G/N is called the quotient group or factor group of G by N .
If G is a finite group and N a normal subgroup of G, then G/N has
as its elements the right cosets of N in G, and since there are precisely
iG (N ) = o(G)/o(N ) such cosets. Hence, we have
Lemma 1.6 If G is a finite group and N is a normal subgroup of G then
o(G/N ) = o(G)/o(N ).
Examples of Quotient groups.
1. Let G be the group of integers under addition and let N be the set of
all multiples of 5. We note that N is a subgroup of G. Further, as G
is abelian N is a normal subgroup of G. Since the operation in G is
addition, we shall write the cosets of N in G as N + a instead of N a.
Consider the five cosets N, N + 1, N + 2, N + 3, N + 4. These are all
the cosets of N in G, for given a ∈ G, a = 5b + c and 0 ≤ c ≤ 4 (
c is the remainder of a on division by 5). We note that the formula
N anb = N ab translates into: (N + a) + (N + b) = N + (a + b). We
write the addition table for G/N.
+
N
N +1
N +2
N +3
N +4
N
N
N +1
N +2
N +3
N +4
N +1
N +1
N +2
N +3
N +4
N
N +2
N +2
N +3
N +4
N
N +1
N +3
N +3
N +4
N
N +1
N +2
N +4
N +4
N
N +1
N +2
N +3
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Chapter 1. Normal Subgroups and Homomorphism
2. Let G = S3 and N = {ρ0 , ρ1 , ρ2 }. Note that N is a subgroup of index
2 and hence N is a normal subgroup of G. The cosets of N in G are
given by N and µ1 N. Observe that N = ρ0 N = ρ1 N = ρ2 N and
µ1 N = µ2 N = µ3 N. Note that the group multiplication table is given
by
+
N
µ1 N
N
N
µ1 N
µ1 N
µ1 N
N
Thus, G/N is a group of order 2 and hence cyclic.
1.3
Homomorphism
The notion of homomorphism is one central idea which is common to all
aspects of modern algebra. By homomorphism one means a mapping from
one algebraic system to a like algebraic system which preserves structure. In
this section, we study the notion of homomorphism.
Definition 1.3 A mapping ϕ from a group G into a group G′ is said to be
a homomorphism if for all a, b ∈ G, ϕ(ab) = ϕ(a)ϕ(b).
Note that on the left side of the relation, namely, in the term ϕ(ab), the
product ab is computed in G using the product of elements of G, whereas on
the right side of this relation, namely, in the term ϕ(a)ϕ(b), the product is
that of elements in G′ . We now study some examples.
Examples.
1. Let G, G′ be groups. Define ϕ(x) = e′ for all x ∈ G, where e′ is the
identity element of G′ . This is trivially a homomorphism.
2. Let G be a group. Define ϕ from G to itself by ϕ(x) = x for every
x ∈ G. Note that ϕ is a homomorphism which is one-one and onto.
3. Let G be the group of all real numbers under addition and let G′ be
the group of non-zero real numbers with the product being ordinary
multiplication of real numbers. Define ϕ : G → G′ by ϕ(a) = 2a .
To verify that this mapping is a homomorphism we must see whether
ϕ(ab) = ϕ(a)ϕ(b). Note that by the product ab on the left hand side
1.3. Homomorphism
11
we mean the operation in G, that is, addition. Hence, we must check
if 2a+b = 2a 2b , which is indeed true. Since, 2a is always positive, the
image of ϕ is not all of G′ , so ϕ is not onto. Also, it is easy to see that
ϕ is one-one.
4. Let G be the group of integers under addition. Define ϕ : G → G by
ϕ(x) = 3x. Now, ϕ(x + y) = 3(x + y) = 3x + 3y = ϕ(x) + ϕ(y). Hence,
ϕ is a homomorphism. Note that ϕ is one-one but not onto.
5. Let G be the group of integers under addition and G′ = Z/nZ under
addition. Define ϕ : G → G by ϕ(x) = [x], residue class of x modulo
n. Now, ϕ(x + y) = [x + y] = [x] + [y] = ϕ(x) + ϕ(y). Hence, ϕ is a
homomorphism. Note that ϕ is an onto homomorphism.
6. Let G be the group of positive numbers with the product being ordinary multiplication of real numbers and let G′ be the group of all
real numbers under addition . Define ϕ : G → G′ by ϕ(a) = log a.
To verify that this mapping is a homomorphism we must see whether
ϕ(ab) = ϕ(a)ϕ(b). Note that by the product ϕ(a)ϕ(b) on the right hand
side we mean the operation in G′ , that is, addition. Hence, we must
check if log ab = log a + log b, which is indeed true. Also, it is easy to
see that ϕ is one-one and onto.
7. Let G be the group of non-zero complex numbers with the product
being multiplication and let G′ be the group of non-zero real numbers
under multiplication. Define ϕ : G → G′ by ϕ(z) = |z|. Note that
ϕ(z1 z2 ) = |z1 z2 | = |z1 ||z2 | = ϕ(z1 )ϕ(z2 ). Thus, ϕ is a homomorphism.
8. Let G denote the set of all 2 × 2 matrices having real coefficients and
non-zero determinant, i.e.,
a b
G= A=
; a, b, c, d ∈ R, ad − bc ̸= 0
c d
and G′ be the group of nonzero
real numbers under multiplication.
h a b Define ϕ : G → G′ by ϕ(
) = ad − bc.
c d
It is easy to see that ϕ is an onto homomorphism but ϕ is not one-one.
12
Chapter 1. Normal Subgroups and Homomorphism
9. Let G = Sn , the group of all permutations on n symbols. Let G′ be the
group consisting of 1 and −1 under multiplication. Define ϕ : G → G′
by ϕ(σ) = sign of σ. Note that ϕ is an onto homomorphism.
Lemma 1.7 If ϕ is a homomorphism of G into G′ , then
1. ϕ(e) = e′ , the identity element of G′ .
2. ϕ(x−1 ) = ϕ(x)−1 for all x ∈ G.
Proof. To prove 1, note that
ϕ(x)e′ = ϕ(x) = ϕ(xe) = ϕ(x)ϕ(e).
Hence, by cancellation law in G′ , we get ϕ(e) = e′ .
To prove 2, note that
e′ = ϕ(e) = ϕ(xx−1 ) = ϕ(x)ϕ(x−1 ),
so ϕ(x−1 ) = ϕ(x)−1 .
Definition 1.4 If ϕ is a homomorphism of G into G′ , the kernel of ϕ, Kϕ ,
is defined by
Kϕ = {x ∈ G|ϕ(x) = e′ , e′ = identity element of G′ }.
The above lemma guarantees us that Kϕ is non-empty. But, we can actually
prove much more than this. In fact, in the following lemma, we prove that
Kϕ is a normal subgroup.
Lemma 1.8 If ϕ is a homomorphism of G into G′ with kernel K, then K is
a normal subgroup of G.
Proof. We first prove that K is a subgroup of G. Note that e ∈ K, hence,
K is non-empty. Suppose x, y ∈ K. Hence, ϕ(x) = ϕ(y) = e′ , where e′ is the
identity element of G′ . and so ϕ(xy) = ϕ(x)ϕ(y) = e′ e′ = e′ , whence xy ∈ K.
Also, if x ∈ K, ϕ(x) = e′ , so ϕ(x−1 ) = ϕ(x)−1 = e′−1 = e′ . Hence, K is a
subgroup of G.
To prove that K is a normal subgroup of G, we must prove that for every
g ∈ G and k ∈ K, gkg −1 ∈ K. In other words, we have to show that for every
g ∈ G and k ∈ K, ϕ(gkg −1 ) = e′ . Consider,
ϕ(gkg −1 ) = ϕ(g)ϕ(k)ϕ(g −1 ) = ϕ(g)e′ ϕ(g −1 )
= ϕ(g)ϕ(g −1 ) = ϕ(gg −1 ) = ϕ(e) = e′ ,
hence, K is a normal subgroup of G.
1.3. Homomorphism
13
Example 1.4
1. Note that the determinant is a homomorphism from the
multiplicative group of square matrices into the multiplicative group of
a field. The kernel is called the special linear group (and is normal).
2. Let G = {Ta,b : R → R | Ta,b (x) = ax + b, a ̸= 0, b ∈ R}. Then G is a
group under composition of mappings. Let A = {Ta,o | a ∈ R⋆ }. (Note
that A is a subgroup of G isomorphic to R⋆ ) and N = {T1,b | b ∈ R} (N
is a subgroup of G isomorphic to R) the group of translations. Verify
that Ta,b 7→ a is a homomorphism of G onto the multiplicative group,
whose kernel is N, Thus N is normal further, we have G = AN = N A,
and N ∩ A = {id}.
We say that G is the semidirect product of A and N.
Proposition 1.1 Let G, G′ be groups and ϕ : G → G′ be a homomorphism.
Then
1. ϕ(G) = {x ∈ G′ |x = ϕ(g) for g ∈ G} is a subgroup of G′ .
2. If G is an abelian group then ϕ(G) is an abelian group.
3. If G is a cyclic group then ϕ(G) is a cyclic group.
Proof.
1. We note that ϕ(G) is nonempty. Let x, y ∈ ϕ(G). Hence, x = ϕ(g1 ) and
y = ϕ(g2 ). Hence, xy = ϕ(g1 )ϕ(g2 ) = ϕ(g1 g2 ), as ϕ is a homomorphism.
Hence, xy ∈ ϕ(G). Also, if x ∈ ϕ(G) then x = ϕ(g) and x−1 = ϕ(g −1 ).
This implies that x−1 ∈ ϕ(G). Hence, ϕ(G) is a subgroup of G′ .
2. Suppose G is an abelian group and x, y ∈ ϕ(G). Hence, x = ϕ(g1 ) and
y = ϕ(g2 ) for some g1 , g2 ∈ G. Note that as G is an abelian group,
xy = ϕ(g1 )ϕ(g2 ) = ϕ(g1 g2 ) = ϕ(g2 g1 ) = ϕ(g2 )ϕ(g1 ) = yx.
Hence, ϕ(G) is an abelian group.
3. Suppose G is a cyclic group and a be a generator of G. Let x ∈ ϕ(G).
Hence, x = ϕ(g) for some g ∈ G. Since, G is cyclic, g = ak for some
integer k. Hence, x = ϕ(g) = ϕ(ak ) = ϕ(a)k . Thus, every element in
ϕ(G) is a power of ϕ(a). This shows that ϕ(G) is cyclic with ϕ(a) as a
generator.
14
Chapter 1. Normal Subgroups and Homomorphism
Corollary 1.1 Let G, G′ be groups and ϕ : G → G′ be an onto homomorphism. Then
1. If G is an abelian group then G′ is an abelian group.
2. If G is a cyclic group then G′ is a cyclic group.
1.4
Isomorphism
Definition 1.5 A homomorphism ϕ from G onto G′ is said to be an isomorphism if ϕ is one-one. And, the two groups G and G′ are said to be
isomorphic. In this case we write G ≈ G′ .
Thus, an isomorphism is a homomorphism such that ϕ is one-one and onto.
Lemma 1.9 Let ϕ be a group homomorphism from G to G′ . Then, ϕ is
one-one if and only if the kernel of homomorphism, Kϕ , is {e}.
Proof. Suppose ϕ is one-one and x ∈ Kϕ . Hence, ϕ(x) = e′ . But ϕ(e) = e′
and ϕ is one-one. This implies that x = e. Hence, Kϕ = {e}.
Conversely, assume that Kϕ = {e}. Suppose ϕ(x) = ϕ(y). Multiplying
both sides by ϕ(x−1 ), we get ϕ(x−1 x) = ϕ(x−1 y) = e′ . Hence, x−1 y ∈ Kϕ .
But Kϕ = {e}. Hence, x−1 y = e. This implies that x = y. Hence, ϕ is one-one.
The following corollary follows easily from the above lemma.
Corollary 1.2 Let G, G′ be groups. A homomorphism ϕ : G → G′ from G
onto G′ is an isomorphism if and only if the kernel of homomorphism, Kϕ , is
{e}.
Theorem 1.2 Any infinite cyclic group is isomorphic to Z.
Proof. Let G be an infinite cyclic group and a be a generator of G. Define
ϕ : Z → G by ϕ(n) = an .
Note that ϕ(m + n) = am+n = (am )(an ) = ϕ(m)ϕ(n). Hence, ϕ is a
homomorphism. To prove that ϕ is onto, let x ∈ G, then x = an for some
integer n. Now, ϕ(n) = an = x. Hence, ϕ is onto.
To show that ϕ is one-one, let ϕ(m) = ϕ(n). Hence, ϕ(m − n) = 0. Hence,
am−n = e. Since, a is a generator of infinite cyclic group, we get m = n.
Hence, ϕ is an isomorphism.
1.4. Isomorphism
15
Theorem 1.3 Any finite cyclic group of order n is isomorphic to (Z/nZ).
Proof. Let G be a finite cyclic group of order n and a be a generator of G.
Define ϕ : Zn → G by ϕ(m̄) = am .
Note that
¯ n) = am+n = (am )(an ) = ϕ(m̄)ϕ(n̄).
ϕ(m̄ + n̄) = ϕ(m +
Hence, ϕ is a homomorphism. To prove that phi is onto, let x ∈ G, then
x = ak for some integer k, 0 ≤ k < n. Now, ϕ(k̄) = ak = x. Hence, ϕ is onto.
To show that ϕ is one-one, let ϕ(m) = ϕ(k). Hence, ϕ(m − k) = 0. Hence,
m−k
a
= e. Since, a is a generator of the cyclic group, we get n|m − k. Hence,
m̄ = k̄. Hence, ϕ is an isomorphism.
Remark 1.2 The following results follow easily. We leave the proofs of these
results as an exercise to the reader.
1. Let G and G′ be group which are isomorphic to each other. Then G is
abelian if and only if G′ is abelian. Thus, if G is an abelian group and
G′ is a nonabelian group then G and G′ can not be isomorphic to each
other.
2. Let G and G′ be group which are isomorphic to each other. Then G is
cyclic group if and only if G′ is a cyclic.
3. If two groups G and G′ are isomorphic to each other, then G and G′ have
equal number of elements. The converse is not true. For example, S3
and Z/6Z have equal number of elements but they are not isomorphic
to each other.
Example 1.5 Let G be a group. Define a mapping ϕ from G to G by
ϕ(x) = x−1 . Show that ϕ is an isomorphism if and only if G is an abelian
group.
Solution. Suppose G is an abelian group. Then
ϕ(xy) = (xy)−1 = y −1 x−1 = x−1 y −1
as G is abelian. Hence, ϕ(xy) = ϕ(x)ϕ(y). Hence, ϕ is a homomorphism.
Assume that ϕ(x) = ϕ(y). Hence, x−1 = y −1 which implies that x = y, hence
ϕ is one-one. ϕ is onto as ϕ(x−1 ) = x. Hence, ϕ is an isomorphism.
16
Chapter 1. Normal Subgroups and Homomorphism
Conversely, assume that ϕ is an isomorphism. Hence,
ϕ(x−1 y −1 ) = ϕ(x−1 )ϕ(y −1 ). Thus, (x−1 y −1 )−1 = xy. But,
(x−1 y −1 )−1 = yx. Thus, xy = yx for all x, y ∈ G. Hence, G is an abelian
group.
Example 1.6 Let G be a group and g ∈ G. Define a mapping ϕ from G to
G by ϕ(x) = gxg −1 . Show that ϕ is an isomorphism.
Solution. To show that ϕ is an isomorphism, we require to show that ϕ is
a homomorphism, ϕ is one-one and onto. We prove this one by one.
Note that ϕ(xy) = g(xy)g −1 = (gxg −1 )(gyg −1 ) = ϕ(x)ϕ(y). Hence, ϕ is a
homomorphism.
Suppose ϕ(x) = ϕ(y). Hence, gxg −1 = gyg −1 . By cancellation laws in
groups, we get x = y. Hence, ϕ is one-one.
Let x ∈ G. Note that ϕ(g −1 xg) = x. Hence, ϕ is onto. Thus, ϕ is an
isomorphism.
Example 1.7 Show that the Klein’s four group is isomorphic to (Z/8Z)⋆ .
Solution. Note that the Klein’s four group V and the group (Z/8Z)⋆ are
given by the following multiplication tables respectively,
⋆ e a b c
e e a b c
a a e c b
b b c e a
c c b a e
⋆
1̄
3̄
5̄
7̄
1̄
1̄
3̄
5̄
7̄
3̄
3̄
1̄
7̄
5̄
5̄
5̄
7̄
1̄
3̄
7̄
7̄
5̄
3̄
1̄
Define ϕ : V → Z/8Z⋆ by ϕ(e) = 1̄, ϕ(a) = 3̄, ϕ(b) = 5̄ and ϕ(c) = 7̄.
We can easily verify that ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ V. Hence, ϕ is a
homomorphism. Also, by definition, ϕ is one-one and onto. Hence, ϕ is an
isomorphism.
Example 1.8 Let G be a group and H be a subgroup of G. Let N be a
normal subgroup of G. Prove that
HN = {x ∈ G|x = hn, h ∈ H, n ∈ N }
is a subgroup of G. Also, prove that H ∩ N is a normal subgroup of G.
Further, prove that f : H → G/N given by f (h) = hN is a homomorphism.
What is the kernel of this homomorphism?
1.4. Isomorphism
17
Solution. We note that a non-empty subset K is a subgroup of G if and
only if for every x, y ∈ K xy −1 ∈ K.
Note that HN is nonempty. Let x, y ∈ HN. Hence, x = hn and y = h1 n1 .
Hence,
−1
xy −1 = hn(h1 n1 )−1 = hn(n−1
1 h1 )
−1
−1
= h(nn−1
1 )h1 = h(n2 )h1
−1
−1
−1
where n2 = nn−1
1 . Now, n2 h1 ∈ N h1 = h1 N, N being a normal subgroup
−1
of G. Hence, n2 h−1
= h−1
= hh−1
1
1 n3 , where n3 ∈ N. Hence, xy
1 n3 =
−1
(hh1 )n3 ∈ HN. This proves that HN is a subgroup of G.
Let h1 , h2 ∈ H. Now f (h1 h2 ) = h1 h2 N = (h1 N )(h2 N ), N being a normal
subgroup of G and h1 , h2 ∈ G. Hence, f (h1 h2 ) = f (h1 )f (h2 ). Thus, f is a
homomorphism. Note that h ∈ Ker(f ) if and only if h ∈ N. Hence, h ∈
Ker(f ) implies that h ∈ H ∩ N. Conversely if h ∈ H ∩ N then h ∈ Ker(f ).
Hence, Ker(f ) = H ∩ N.
This proves that H ∩ N is a normal subgroup of H as kernel of the
homomorphism is a normal subgroup of the group.
Exercise 1.1
1. If N is a normal subgroup of G and a ∈ G is of finite order m, prove
that N a is of finite order k and k is a divisor of m.
2. Let N and M be two normal subgroups of G such that N ∩ M = (e).
Show that for any n ∈ N and m ∈ M, nm = mn.
3. Let G be a group and H be a subgroup of G. Let
NH = {x ∈ G|xHx−1 = H}.
Show that NH is a group containing H, and H is normal in NH . Also,
prove that H is normal in G if and only if NH = G.
4. Show that every coset of (Z, +) in (R, +) has a unique coset representative x such that 0 ≤ x < 1.
5. Find all the quotient groups of the group Z12 under addition.
18
Chapter 1. Normal Subgroups and Homomorphism
6. Let R⋆ be the multiplicative group of non-zero real numbers. Define
ϕ : R⋆ → R⋆ by ϕ(x) = |x|. Show that ϕ is a homomorphism. What
is the image of this homomorphism? What is the kernel of this homomorphism?
7. Let C⋆ be the multiplicative group of nonzero complex numbers. Define ϕ : C⋆ → C⋆ by ϕ(z) = |z|. Show that ϕ is a homomorphism.
What is the image of this homomorphism? What is the kernel of this
homomorphism?
8. Let f : G → G′ be an isomorphism of groups. Let a ∈ G. Show that
the order of a is the same as the order of f (a).
9. Let G be an abelian group and n a positive integer. Show that the map
f : G → G given by f (x) = xn is a homomorphism of G into itself.
10. Let G = Sn , the group of all permutations on n symbols. Let G′ be the
group consisting of 1 and −1 under multiplication. Define ϕ : G → G′
by ϕ(σ) = sign of σ. Show that ϕ is an onto homomorphism. What is
the kernel of this homomorphism?
11. Let f : G → G′ be an onto homomorphism of groups. Let N be a
normal subgroup of G. Show that f (N ) is a normal subgroup of G′ .
12. Let G = Z12 , the group of residue classes modulo 12 and G′ = Z⋆13 , the
group of reduced residue classes modulo 13. Show that G and G′ are
isomorphic.
13. Prove that, if G is a cyclic group of order n and p divides n, then there
is a homomorphism of G onto a cyclic group of order p. What is the
kernel of this homomorphism?
14. Let G denote the set of all 2 × 2 matrices having integer coefficients,
that is,
a b
G=
|a, b, c, d ∈ Z .
c d
Prove that G is an abelian group with respect to the usual operation
of matrix addition. Let ϕ : G → Z be defined by
a b
ϕ
= a + d.
c d
1.5. Centralizer
19
Prove that ϕ is a homomorphism of G onto the additive group of integers and find its kernel.
15. Let G = {z ∈ C|z n = 1 for some n ≥ 1}. Show that G is isomorphic to
Q/Z.
1.5
Centralizer
Let S be a subset of G and let N = NS be the set of all elements x ∈ G such
that xSx−1 = S. Then one can easily prove that N is a subgroup of G. N
is called the normalizer of S. If S consists of one element a, then N is also
called the centralizer of a.
More generally, let
ZS = {x ∈ G : xyx−1 = y for all y ∈ S}.
Then ZS is a subgroup of G called the centralizer of S. The centralizer of G
itself is called the centre of G. Note that the center of G is the subgroup of
G consisting of all elements of G commuting will all other elements, and is a
normal subgroup of G. Further, every subgroup of the center of the group is
a normal subgroup.
Remark 1.3 Let H be a subgroup of G. Then H is a normal subgroup of
its normalizer NH . Further, if K is any subgroup of G containing H and such
that H is normal in K, then K ⊂ NH .
If K is a subgroup of NH , then KH is a group and H is normal in KH.
The normalizer of H is the largest subgroup of G in which H is normal.