SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
We have already encountered the concept of a group homomorphism,
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
We have already encountered the concept of a group homomorphism,
and seen that such homomorphisms respect inverses
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
We have already encountered the concept of a group homomorphism,
and seen that such homomorphisms respect inverses
(f (a−1) = f (a)−1 for all a),
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
We have already encountered the concept of a group homomorphism,
and seen that such homomorphisms respect inverses
(f (a−1) = f (a)−1 for all a),
and it turns out that there is a version of the Homomorphism Theorem
for groups also.
SECTION 14 : FACTOR GROUPS AND GROUP HOMOMORPHISMS
It is possible to make sense of the notion of a factor group, much as
one can speak of a factor ring.
We have already encountered the concept of a group homomorphism,
and seen that such homomorphisms respect inverses
(f (a−1) = f (a)−1 for all a),
and it turns out that there is a version of the Homomorphism Theorem
for groups also.
We consider all these things in this section.
14.1: Normal subgroups and factor groups
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
We know that by setting (a, b) ∈ σH if and only if ab−1 ∈ H,
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
We know that by setting (a, b) ∈ σH if and only if ab−1 ∈ H,
we obtain an equivalence relation on G (this is Proposition 12.1).
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
We know that by setting (a, b) ∈ σH if and only if ab−1 ∈ H,
we obtain an equivalence relation on G (this is Proposition 12.1).
This generalises congruence modulo an ideal I of a ring R,
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
We know that by setting (a, b) ∈ σH if and only if ab−1 ∈ H,
we obtain an equivalence relation on G (this is Proposition 12.1).
This generalises congruence modulo an ideal I of a ring R,
where one says a ≡I b if and only if a − b ∈ I,
14.1: Normal subgroups and factor groups
Let G be a group with H ≤ G.
We know that by setting (a, b) ∈ σH if and only if ab−1 ∈ H,
we obtain an equivalence relation on G (this is Proposition 12.1).
This generalises congruence modulo an ideal I of a ring R,
where one says a ≡I b if and only if a − b ∈ I,
and in that case we know that we can go on to define ring structure
on the set of equivalence classes under ≡I and obtain the factor ring
R/I.
So we would like to be able to do the same things with the equivalence
classes of σH in G
So we would like to be able to do the same things with the equivalence
classes of σH in G
and be able to define a group G/H.
So we would like to be able to do the same things with the equivalence
classes of σH in G
and be able to define a group G/H.
We might try to define (Ha)(Hb) = H(ab) and hope to get a group
structure.
So we would like to be able to do the same things with the equivalence
classes of σH in G
and be able to define a group G/H.
We might try to define (Ha)(Hb) = H(ab) and hope to get a group
structure.
Unfortunately, unless G is abelian, this is not guaranteed to work.
So we would like to be able to do the same things with the equivalence
classes of σH in G
and be able to define a group G/H.
We might try to define (Ha)(Hb) = H(ab) and hope to get a group
structure.
Unfortunately, unless G is abelian, this is not guaranteed to work.
An additional property of H is needed.
If H ≤ G, a group, we say H is a normal subgroup if
If H ≤ G, a group, we say H is a normal subgroup if
for all h ∈ H, g ∈ G : ghg −1 ∈ H.
If H ≤ G, a group, we say H is a normal subgroup if
for all h ∈ H, g ∈ G : ghg −1 ∈ H.
So in an abelian group, every subgroup is normal.
If H ≤ G, a group, we say H is a normal subgroup if
for all h ∈ H, g ∈ G : ghg −1 ∈ H.
So in an abelian group, every subgroup is normal.
Notation: H / G (similar to the notation for an ideal of a ring).
Theorem 14.1. If G is a group, and H ≤ G, then the above “definition” of multiplication of cosets in G is well-defined if and only if H is
a normal subgroup.
Theorem 14.1. If G is a group, and H ≤ G, then the above “definition” of multiplication of cosets in G is well-defined if and only if H is
a normal subgroup.
In this case, G/H = {Hg | g ∈ G} is a group under this multiplication,
Theorem 14.1. If G is a group, and H ≤ G, then the above “definition” of multiplication of cosets in G is well-defined if and only if H is
a normal subgroup.
In this case, G/H = {Hg | g ∈ G} is a group under this multiplication,
with identity element H1 = H and inverse of Ha equal to Ha−1.
Theorem 14.1. If G is a group, and H ≤ G, then the above “definition” of multiplication of cosets in G is well-defined if and only if H is
a normal subgroup.
In this case, G/H = {Hg | g ∈ G} is a group under this multiplication,
with identity element H1 = H and inverse of Ha equal to Ha−1.
Moreover, for all g ∈ G, Hg = gH = {gh | h ∈ H} (left cosets are
right cosets and vice versa).
Proof: Suppose H is normal.
Proof: Suppose H is normal.
Suppose Ha1 = Hb1, Ha2 = Hb2.
Proof: Suppose H is normal.
Suppose Ha1 = Hb1, Ha2 = Hb2.
−1
So a1 σH b1 and a2 σH b2, and so a1b−1
∈
H
and
a
b
2
1
2 ∈ H.
Proof: Suppose H is normal.
Suppose Ha1 = Hb1, Ha2 = Hb2.
−1
So a1 σH b1 and a2 σH b2, and so a1b−1
∈
H
and
a
b
2
1
2 ∈ H.
Hence
−1
(a1a2)(b1b2)−1 = a1(a2b−1
)b
2
1
Proof: Suppose H is normal.
Suppose Ha1 = Hb1, Ha2 = Hb2.
−1
So a1 σH b1 and a2 σH b2, and so a1b−1
∈
H
and
a
b
2
1
2 ∈ H.
Hence
−1
(a1a2)(b1b2)−1 = a1(a2b−1
)b
2
1
−1
−1
= a1(a2b−1
)a
a
b
1
2
1
1 .
−1 −1
But a2b−1
∈
H,
so
a
(a
b
1
2
2
2 )a1 ∈ H by normality of H,
−1 −1
But a2b−1
∈
H,
so
a
(a
b
1
2
2
2 )a1 ∈ H by normality of H,
−1
−1
and so a1(a2b−1
)a
a
b
1
2
1
1 ∈H
−1 −1
But a2b−1
∈
H,
so
a
(a
b
1
2
2
2 )a1 ∈ H by normality of H,
−1
−1
and so a1(a2b−1
)a
a
b
1
2
1
1 ∈H
since H is a subgroup and so is closed under multiplication.
−1 −1
But a2b−1
∈
H,
so
a
(a
b
1
2
2
2 )a1 ∈ H by normality of H,
−1
−1
and so a1(a2b−1
)a
a
b
1
2
1
1 ∈H
since H is a subgroup and so is closed under multiplication.
This shows that a1a2 σH b1b2, and so H(a1a2) = H(b1b2)
−1 −1
But a2b−1
∈
H,
so
a
(a
b
1
2
2
2 )a1 ∈ H by normality of H,
−1
−1
and so a1(a2b−1
)a
a
b
1
2
1
1 ∈H
since H is a subgroup and so is closed under multiplication.
This shows that a1a2 σH b1b2, and so H(a1a2) = H(b1b2)
and so right coset multiplication is well-defined.
Conversely, suppose right coset multiplication is well-defined.
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
= (H(g1))(Hg −1)
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
= (H(g1))(Hg −1)
= ((Hg)(H1))(Hg −1)
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
= (H(g1))(Hg −1)
= ((Hg)(H1))(Hg −1)
= ((Hg)(Hh))(Hg −1)
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
= (H(g1))(Hg −1)
= ((Hg)(H1))(Hg −1)
= ((Hg)(Hh))(Hg −1)
= ···
Conversely, suppose right coset multiplication is well-defined.
Note that H = H1 = Hh for any h ∈ H (exercise!).
Then for any g ∈ G and h ∈ H,
H = H1
= H(gg −1)
= H(g1g −1)
= (H(g1))(Hg −1)
= ((Hg)(H1))(Hg −1)
= ((Hg)(Hh))(Hg −1)
= ···
= H(ghg −1).
So (ghg −1, 1) ∈ σH , so ghg −1 = (ghg −1)1−1 ∈ H.
So (ghg −1, 1) ∈ σH , so ghg −1 = (ghg −1)1−1 ∈ H.
Hence H is normal.
So (ghg −1, 1) ∈ σH , so ghg −1 = (ghg −1)1−1 ∈ H.
Hence H is normal.
In this case, (Hg)(H1) = H(g1) = Hg,
So (ghg −1, 1) ∈ σH , so ghg −1 = (ghg −1)1−1 ∈ H.
Hence H is normal.
In this case, (Hg)(H1) = H(g1) = Hg,
and similarly (H1)(Hg) = Hg,
So (ghg −1, 1) ∈ σH , so ghg −1 = (ghg −1)1−1 ∈ H.
Hence H is normal.
In this case, (Hg)(H1) = H(g1) = Hg,
and similarly (H1)(Hg) = Hg,
so H = H1 is an (hence the) identity element for G/H.
We leave it as an exercise to verify that:
We leave it as an exercise to verify that:
(i) associativity of multiplication holds in G/H, and
We leave it as an exercise to verify that:
(i) associativity of multiplication holds in G/H, and
(ii) G/H is thus a monoid which is in fact a group in which
We leave it as an exercise to verify that:
(i) associativity of multiplication holds in G/H, and
(ii) G/H is thus a monoid which is in fact a group in which
(Hg)−1 = H(g −1) for all g ∈ G.
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Hence gH ⊆ Hg.
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Hence gH ⊆ Hg.
Conversely, if hg ∈ Hg (where ∈ H), then h2 = g −1hg ∈ H,
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Hence gH ⊆ Hg.
Conversely, if hg ∈ Hg (where ∈ H), then h2 = g −1hg ∈ H,
and gh2 = gg −1hg = hg, so hg ∈ gH.
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Hence gH ⊆ Hg.
Conversely, if hg ∈ Hg (where ∈ H), then h2 = g −1hg ∈ H,
and gh2 = gg −1hg = hg, so hg ∈ gH.
Hence Hg ⊆ gH.
Finally, if gh ∈ gH (where h ∈ H and g ∈ G), then h1 = ghg −1 ∈ H,
so h1g = ghg −1g = gh, so gh ∈ Hg.
Hence gH ⊆ Hg.
Conversely, if hg ∈ Hg (where ∈ H), then h2 = g −1hg ∈ H,
and gh2 = gg −1hg = hg, so hg ∈ gH.
Hence Hg ⊆ gH.
So Hg = gH as claimed.
For example, for any positive integer n, nZ ≤ Z,
For example, for any positive integer n, nZ ≤ Z,
∼ Z , something true when they are viewed as rings also.
and Z/nZ =
n
For example, for any positive integer n, nZ ≤ Z,
∼ Z , something true when they are viewed as rings also.
and Z/nZ =
n
Both Z and Zn are cyclic.
For example, for any positive integer n, nZ ≤ Z,
∼ Z , something true when they are viewed as rings also.
and Z/nZ =
n
Both Z and Zn are cyclic.
More generally we have the following.
For example, for any positive integer n, nZ ≤ Z,
∼ Z , something true when they are viewed as rings also.
and Z/nZ =
n
Both Z and Zn are cyclic.
More generally we have the following.
Proposition 14.2. Suppose G is a cyclic group, with N / G.
For example, for any positive integer n, nZ ≤ Z,
∼ Z , something true when they are viewed as rings also.
and Z/nZ =
n
Both Z and Zn are cyclic.
More generally we have the following.
Proposition 14.2. Suppose G is a cyclic group, with N / G.
Then G/N is cyclic.
Proof: Suppose G = hai for some a ∈ G, and N / G.
Proof: Suppose G = hai for some a ∈ G, and N / G.
Then every coset in G/N has the form N b for some b ∈ G,
Proof: Suppose G = hai for some a ∈ G, and N / G.
Then every coset in G/N has the form N b for some b ∈ G,
that is, N ak for some k > 0 (since G = hai).
Proof: Suppose G = hai for some a ∈ G, and N / G.
Then every coset in G/N has the form N b for some b ∈ G,
that is, N ak for some k > 0 (since G = hai).
But N ak = (N a)k , so all cosets in G/N are powers of N a.
Proof: Suppose G = hai for some a ∈ G, and N / G.
Then every coset in G/N has the form N b for some b ∈ G,
that is, N ak for some k > 0 (since G = hai).
But N ak = (N a)k , so all cosets in G/N are powers of N a.
So G/N = hN ai.
For a group G, {1} and G are normal subgroups,
For a group G, {1} and G are normal subgroups,
∼ G/{1}, and G/G =
∼ {1},
and like rings, we have that G =
For a group G, {1} and G are normal subgroups,
∼ G/{1}, and G/G =
∼ {1},
and like rings, we have that G =
the trivial group with a single element.
For a group G, {1} and G are normal subgroups,
∼ G/{1}, and G/G =
∼ {1},
and like rings, we have that G =
the trivial group with a single element.
A group is simple if its only normal subgroups are {1} and itself.
For a group G, {1} and G are normal subgroups,
∼ G/{1}, and G/G =
∼ {1},
and like rings, we have that G =
the trivial group with a single element.
A group is simple if its only normal subgroups are {1} and itself.
These are just like simple rings: they have no interesting factor groups.
Proposition 14.3. Suppose G is a non-trivial abelian group.
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proof: Since every subgroup of G is normal,
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proof: Since every subgroup of G is normal,
G is simple if and only if it has no proper non-trivial subgroups.
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proof: Since every subgroup of G is normal,
G is simple if and only if it has no proper non-trivial subgroups.
Since every a 6= 1 in G generates a cyclic subgroup,
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proof: Since every subgroup of G is normal,
G is simple if and only if it has no proper non-trivial subgroups.
Since every a 6= 1 in G generates a cyclic subgroup,
G will be simple if and only if every cyclic subgroup (aside from {1})
equals G.
Proposition 14.3. Suppose G is a non-trivial abelian group.
Then G is simple if and only if it is isomorphic to (Zp, +) for a prime
p.
Proof: Since every subgroup of G is normal,
G is simple if and only if it has no proper non-trivial subgroups.
Since every a 6= 1 in G generates a cyclic subgroup,
G will be simple if and only if every cyclic subgroup (aside from {1})
equals G.
So G itself is cyclic.
∼ (Z, +) then of course G has many subgroups,
If G =
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
then for any g ∈ G, either g a = 1, in which case
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
then for any g ∈ G, either g a = 1, in which case
hgi has at most a elements, a contradiction,
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
then for any g ∈ G, either g a = 1, in which case
hgi has at most a elements, a contradiction,
or g a 6= 1 but (g a)b = g ab = 1,
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
then for any g ∈ G, either g a = 1, in which case
hgi has at most a elements, a contradiction,
or g a 6= 1 but (g a)b = g ab = 1,
so hg ai has at most b elements, a contradiction.
∼ (Z, +) then of course G has many subgroups,
If G =
so is not simple, a contradiction.
If G is finite and has non-prime order n = ab say, where a, b > 1,
then for any g ∈ G, either g a = 1, in which case
hgi has at most a elements, a contradiction,
or g a 6= 1 but (g a)b = g ab = 1,
so hg ai has at most b elements, a contradiction.
∼ (Z , +) by Theorem 10.7.
So the order of G is prime and so G =
p
Conversely, if G has a prime number of elements and a ∈ H ≤ G
with a 6= 1,
Conversely, if G has a prime number of elements and a ∈ H ≤ G
with a 6= 1,
then G must be simple, since any subgroup H must be such that
Conversely, if G has a prime number of elements and a ∈ H ≤ G
with a 6= 1,
then G must be simple, since any subgroup H must be such that
|H| divides p by Lagrange’s Theorem (12.3),
Conversely, if G has a prime number of elements and a ∈ H ≤ G
with a 6= 1,
then G must be simple, since any subgroup H must be such that
|H| divides p by Lagrange’s Theorem (12.3),
so either H = G or H = {1}, and so G is simple.
14.2: Group homomorphisms
14.2: Group homomorphisms
We have already met group homomorphisms, in Theorem 5.2 of Section 5.
14.2: Group homomorphisms
We have already met group homomorphisms, in Theorem 5.2 of Section 5.
They are simply semigroup homomorphisms f : G → H between
groups G and H,
14.2: Group homomorphisms
We have already met group homomorphisms, in Theorem 5.2 of Section 5.
They are simply semigroup homomorphisms f : G → H between
groups G and H,
and automatically satisfy the properties that f (1) = 1 and f (a−1) =
(f (a))−1 for all a ∈ G.
Given the similarity between normal subgroups and ideals,
Given the similarity between normal subgroups and ideals,
it is not surprising that we get a version of the Homomorphism Theorem for Rings (Theorem 5.4) that applies instead to groups.
Given the similarity between normal subgroups and ideals,
it is not surprising that we get a version of the Homomorphism Theorem for Rings (Theorem 5.4) that applies instead to groups.
For f : G → H, it makes sense to define
Given the similarity between normal subgroups and ideals,
it is not surprising that we get a version of the Homomorphism Theorem for Rings (Theorem 5.4) that applies instead to groups.
For f : G → H, it makes sense to define
Ker(f ) = {a ∈ G | f (a) = 1}.
Given the similarity between normal subgroups and ideals,
it is not surprising that we get a version of the Homomorphism Theorem for Rings (Theorem 5.4) that applies instead to groups.
For f : G → H, it makes sense to define
Ker(f ) = {a ∈ G | f (a) = 1}.
(This coincides with the ring definition when applied to the additive
abelian group of the ring.)
Given the similarity between normal subgroups and ideals,
it is not surprising that we get a version of the Homomorphism Theorem for Rings (Theorem 5.4) that applies instead to groups.
For f : G → H, it makes sense to define
Ker(f ) = {a ∈ G | f (a) = 1}.
(This coincides with the ring definition when applied to the additive
abelian group of the ring.)
Define Im(f ) in the usual way for functions.
Theorem 14.4: (Homomorphism Theorem for Groups)
Theorem 14.4: (Homomorphism Theorem for Groups)
For f : G → H a group homomorphism,
Theorem 14.4: (Homomorphism Theorem for Groups)
For f : G → H a group homomorphism,
Ker(f ) / G, Im(f ) ≤ H, and
Theorem 14.4: (Homomorphism Theorem for Groups)
For f : G → H a group homomorphism,
Ker(f ) / G, Im(f ) ≤ H, and
ψ : G/Ker(f ) → Im(f ) given by ψ(Ker(f )g) = f (g) is an isomorphism.
Theorem 14.4: (Homomorphism Theorem for Groups)
For f : G → H a group homomorphism,
Ker(f ) / G, Im(f ) ≤ H, and
ψ : G/Ker(f ) → Im(f ) given by ψ(Ker(f )g) = f (g) is an isomorphism.
∼ Im(f ).
Hence G/Ker(f ) =
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Then f (a−1b) = f (a−1)f (b) = f (a)−1f (b) = 1−1 · 1 = 1.
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Then f (a−1b) = f (a−1)f (b) = f (a)−1f (b) = 1−1 · 1 = 1.
So a−1b ∈ Ker(f ) and so it is a subgroup.
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Then f (a−1b) = f (a−1)f (b) = f (a)−1f (b) = 1−1 · 1 = 1.
So a−1b ∈ Ker(f ) and so it is a subgroup.
And for g ∈ G,
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Then f (a−1b) = f (a−1)f (b) = f (a)−1f (b) = 1−1 · 1 = 1.
So a−1b ∈ Ker(f ) and so it is a subgroup.
And for g ∈ G,
f (gag −1) = f (g)f (a)f (g)−1 = f (g)1f (g)−1 = 1,
Proof: Since f (1) = 1, Ker(f ) 6= ∅.
Suppose a, b ∈ Ker(f ).
Then f (a−1b) = f (a−1)f (b) = f (a)−1f (b) = 1−1 · 1 = 1.
So a−1b ∈ Ker(f ) and so it is a subgroup.
And for g ∈ G,
f (gag −1) = f (g)f (a)f (g)−1 = f (g)1f (g)−1 = 1,
so gag −1 ∈ Ker(f ). So Ker(f ) is a normal subgroup.
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
And for f (a) ∈ Im(f ), f (a)−1 = f (a−1) ∈ Im(f ).
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
And for f (a) ∈ Im(f ), f (a)−1 = f (a−1) ∈ Im(f ).
So Im(f ) is a subgroup of H.
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
And for f (a) ∈ Im(f ), f (a)−1 = f (a−1) ∈ Im(f ).
So Im(f ) is a subgroup of H.
If Ker(f )a = Ker(f )b, then a−1b ∈ Ker(f ),
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
And for f (a) ∈ Im(f ), f (a)−1 = f (a−1) ∈ Im(f ).
So Im(f ) is a subgroup of H.
If Ker(f )a = Ker(f )b, then a−1b ∈ Ker(f ),
so f (a)−1f (b) = f (a−1)f (b) = f (a−1b) = 1, so f (a) = f (b).
We know from Proposition 11.7 that Im(f ) is a subsemigroup of H.
But also, since f (1) = 1 in H, 1 ∈ Im(f ).
And for f (a) ∈ Im(f ), f (a)−1 = f (a−1) ∈ Im(f ).
So Im(f ) is a subgroup of H.
If Ker(f )a = Ker(f )b, then a−1b ∈ Ker(f ),
so f (a)−1f (b) = f (a−1)f (b) = f (a−1b) = 1, so f (a) = f (b).
So setting ψ(Ker(f )a) = f (a) is indeed well-defined.
We show ψ is a bijection.
We show ψ is a bijection.
If ψ(Ker(f )a) = ψ(Ker(f )b), then f (a) = f (b),
We show ψ is a bijection.
If ψ(Ker(f )a) = ψ(Ker(f )b), then f (a) = f (b),
so f (a−1b) = f (a)−1f (b) = 1,
We show ψ is a bijection.
If ψ(Ker(f )a) = ψ(Ker(f )b), then f (a) = f (b),
so f (a−1b) = f (a)−1f (b) = 1,
and so a−1b ∈ Ker(f ), so Ker(f )a = Ker(f )b.
We show ψ is a bijection.
If ψ(Ker(f )a) = ψ(Ker(f )b), then f (a) = f (b),
so f (a−1b) = f (a)−1f (b) = 1,
and so a−1b ∈ Ker(f ), so Ker(f )a = Ker(f )b.
So ψ is one-to-one.
We show ψ is a bijection.
If ψ(Ker(f )a) = ψ(Ker(f )b), then f (a) = f (b),
so f (a−1b) = f (a)−1f (b) = 1,
and so a−1b ∈ Ker(f ), so Ker(f )a = Ker(f )b.
So ψ is one-to-one.
For f (a) ∈ Im(f ), we have (ψ(Ker(f )a) = f (a) by definition, so
ψ is surjective.
Finally, we show ψ is a group homomorphism.
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
= f (ab)
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
= f (ab)
= f (a)f (b)
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
= f (ab)
= f (a)f (b)
= ψ(Ker(f )a)ψ(Ker(f )b),
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
= f (ab)
= f (a)f (b)
= ψ(Ker(f )a)ψ(Ker(f )b),
as required.
Finally, we show ψ is a group homomorphism.
For any Ker(f )a, Ker(f )b ∈ R/Ker(f ), we have
ψ((Ker(f )a)(Ker(f )b)) = ψ(Ker(f )(ab))
= f (ab)
= f (a)f (b)
= ψ(Ker(f )a)ψ(Ker(f )b),
as required.
∼ Im(f ).
So ψ is an isomorphism and so G/Ker(f ) =
For example, consider the (abelian) groups (R, +) and (C∗, ·),
For example, consider the (abelian) groups (R, +) and (C∗, ·),
where C∗ is the set of non-zero complex numbers,
For example, consider the (abelian) groups (R, +) and (C∗, ·),
where C∗ is the set of non-zero complex numbers,
a group under multiplication.
For example, consider the (abelian) groups (R, +) and (C∗, ·),
where C∗ is the set of non-zero complex numbers,
a group under multiplication.
Define f : (R, +) → (C∗, ·) by setting
For example, consider the (abelian) groups (R, +) and (C∗, ·),
where C∗ is the set of non-zero complex numbers,
a group under multiplication.
Define f : (R, +) → (C∗, ·) by setting
f (a) = eia for all a ∈ R.
Then f is a homomorphism:
Then f is a homomorphism:
f (a + b) = ei(a+b) = eiaeib = f (a)f (b).
Then f is a homomorphism:
f (a + b) = ei(a+b) = eiaeib = f (a)f (b).
Now Im(f ) consists of all complex numbers of the form eix for some
real x,
Then f is a homomorphism:
f (a + b) = ei(a+b) = eiaeib = f (a)f (b).
Now Im(f ) consists of all complex numbers of the form eix for some
real x,
which is just the set of complex numbers of length 1:
Then f is a homomorphism:
f (a + b) = ei(a+b) = eiaeib = f (a)f (b).
Now Im(f ) consists of all complex numbers of the form eix for some
real x,
which is just the set of complex numbers of length 1:
Im(f ) = {z ∈ C | |z| = 1}.
Also,
Ker(f ) = {a ∈ R | eia = 1} = {2kπ | k ∈ Z} = h2πi,
Also,
Ker(f ) = {a ∈ R | eia = 1} = {2kπ | k ∈ Z} = h2πi,
the cyclic subgroup generated by 2π in (R, +).
Also,
Ker(f ) = {a ∈ R | eia = 1} = {2kπ | k ∈ Z} = h2πi,
the cyclic subgroup generated by 2π in (R, +).
So by Theorem 14.4,
Also,
Ker(f ) = {a ∈ R | eia = 1} = {2kπ | k ∈ Z} = h2πi,
the cyclic subgroup generated by 2π in (R, +).
So by Theorem 14.4,
∼ R/Ker(f ) = R/h2πi.
Im(f ) =
Also,
Ker(f ) = {a ∈ R | eia = 1} = {2kπ | k ∈ Z} = h2πi,
the cyclic subgroup generated by 2π in (R, +).
So by Theorem 14.4,
∼ R/Ker(f ) = R/h2πi.
Im(f ) =
In other words, multiplying complex numbers of length 1 is equivalent
to adding their arguments “mod 2π”.