716
Modeling Arithmetic, Computation, and Languages
so it is not an isomorphism. The function g is a bijection because g(x) = g( y)
implies x + 1 = y + 1, or x = y, so g is one-to-one; g is also onto because for
any z [ Z, z − 1 [ Z and g(z − 1) = z. But g is not a homomorphism because
g(x + y) = (x + y) + 1 ∙ (x + 1) + ( y + 1) = g(x) + g( y). Hence g is not an
isomorphism.
What can be said about functions from a group to a group that are homomorphisms but not isomorphisms? More specifically, let 3G, # 4 be a group with identity
i G , 3H, + 4 be a group with identity i H , and f be a homomorphism, f: G S H . The
range of f, f (G), is a subset of H (remember that f might not be an onto function, so
it isn’t necessarily true that f (G) = H ). In fact, f (G) is actually a subgroup of H.
It is easy enough to prove the three properties needed for f (G) to be a subgroup,
and the proof will show that f (i G ) = i H and −f (x) = f (x
−1
) (Exercise 1). We can
define a binary relation r on G by
x r y 4 f (x) = f ( y)
and r is an equivalence relation.
PrACtiCe 35 Prove that r is an equivalence relation on G.
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If f is a one-to-one function, x r y would mean that f (x) = f ( y) and therefore
x = y. The equivalence classes formed by r would then be trivial, each containing
only a single element of G. But in general, because f is not a bijection, it might not
be a one-to-one function. Let 3i G 4 be the equivalence class determined by r that
contains the identity of G. Because f (i G ) = i H , 3i G 4 = 5x [ G 0 f (x) = i H 6. This
set is called the kernel of the homomorphism f.
The kernel K is a subset of G, and in addition it is a subgroup of G. We need
to prove the three properties necessary for a subgroup:
1. K is closed: Let x and y be elements in K. Then f (x) = f ( y) = i H . And
f (x # y) = f (x) + f ( y) = i H + i H = i H , so x # y is in K.
2. i G belongs to K: Yes, because f (i G ) = i H .
3. Let x be an element in K: Then f (x) = i H . Also f (x
−1
) = −f (x) =
−i H = i H so x
−1
is in K.
ReminDeR
Given a homomorphism f
from 3G, # 4 to 3H, +4, the
range f (G) is a subgroup
of 3H, +4 and the kernel K
is a subgroup of 3G, # 4.
example 18
Let R* = R − 506 and let R
+
be the set of all positive real numbers. Then
the function f defined by f (x) = 0 x 0 is a homomorphism from the group 3R*, # 4
to the group 3R
+
, # 4. The kernel K of f is 5x [ R* 0 f (x) = 0 x 0 = 16. Therefore
K = 51, −16.
PrACtiCe 36 The function f defined by f (x) = x # 3 1 is a homomorphism from the group 3Z, + 4 to the
group 3Z 3 , + 3 4. Find the kernel K.
For our discussion of cosets, we begin with a definition.
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Modeling Arithmetic, Computation, and Languages
so it is not an isomorphism. The function g is a bijection because g(x) = g( y)
implies x + 1 = y + 1, or x = y, so g is one-to-one; g is also onto because for
any z [ Z, z − 1 [ Z and g(z − 1) = z. But g is not a homomorphism because
g(x + y) = (x + y) + 1 ∙ (x + 1) + ( y + 1) = g(x) + g( y). Hence g is not an
isomorphism.
What can be said about functions from a group to a group that are homomorphisms but not isomorphisms? More specifically, let 3G, # 4 be a group with identity
i G , 3H, + 4 be a group with identity i H , and f be a homomorphism, f: G S H . The
range of f, f (G), is a subset of H (remember that f might not be an onto function, so
it isn’t necessarily true that f (G) = H ). In fact, f (G) is actually a subgroup of H.
It is easy enough to prove the three properties needed for f (G) to be a subgroup,
and the proof will show that f (i G ) = i H and −f (x) = f (x
−1
) (Exercise 1). We can
define a binary relation r on G by
x r y 4 f (x) = f ( y)
and r is an equivalence relation.
PrACtiCe 35 Prove that r is an equivalence relation on G.
■
If f is a one-to-one function, x r y would mean that f (x) = f ( y) and therefore
x = y. The equivalence classes formed by r would then be trivial, each containing
only a single element of G. But in general, because f is not a bijection, it might not
be a one-to-one function. Let 3i G 4 be the equivalence class determined by r that
contains the identity of G. Because f (i G ) = i H , 3i G 4 = 5x [ G 0 f (x) = i H 6. This
set is called the kernel of the homomorphism f.
The kernel K is a subset of G, and in addition it is a subgroup of G. We need
to prove the three properties necessary for a subgroup:
1. K is closed: Let x and y be elements in K. Then f (x) = f ( y) = i H . And
f (x # y) = f (x) + f ( y) = i H + i H = i H , so x # y is in K.
2. i G belongs to K: Yes, because f (i G ) = i H .
3. Let x be an element in K: Then f (x) = i H . Also f (x
−1
) = −f (x) =
−i H = i H so x
−1
is in K.
ReminDeR
Given a homomorphism f
from 3G, # 4 to 3H, +4, the
range f (G) is a subgroup
of 3H, +4 and the kernel K
is a subgroup of 3G, # 4.
example 18
Let R* = R − 506 and let R
+
be the set of all positive real numbers. Then
the function f defined by f (x) = 0 x 0 is a homomorphism from the group 3R*, # 4
to the group 3R
+
, # 4. The kernel K of f is 5x [ R* 0 f (x) = 0 x 0 = 16. Therefore
K = 51, −16.
PrACtiCe 36 The function f defined by f (x) = x # 3 1 is a homomorphism from the group 3Z, + 4 to the
group 3Z 3 , + 3 4. Find the kernel K.
For our discussion of cosets, we begin with a definition.
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