Section 9.1 Algebraic Structures
703
to the identity of the other, that inverses map to inverses, and that if one group
is commutative, so is the other. Indeed, we can prove that these expectations are
correct. (The proofs do not make use of the one-to-one property of the isomorphism, so an onto homomorphism also maps the identity to the identity, inverses
to inverses, and preserves commutativity.)
Suppose, then, that f is an isomorphism from the group 3S, # 4 to the group
3T, + 4 and that i S and i T are the identities in the respective groups. Under the
function f, i S maps to an element f (i S ) in T. Let t be any element in T. Then, because
f is an onto function, t = f (s) for some s [ S. It follows that
f (i S ) + t = f (i S ) + f (s)
= f (i S # s)
(because f is a homomorphism)
= f (s)
(because i S is the identity in S)
= t
Therefore
f (i S ) + t = t
Similarly,
t + f (i S ) = t
The element f (i S ) acts like an identity element in 3T, + 4, and because the identity
is unique, f (i S ) = i T .
PrACtiCe 28 Prove that if f is an isomorphism from the group 3S, # 4 to the group T, + 4, then for any
s [ S, f (s
−1
) = −f (s) (inverses map to inverses). (Hint: Show that f (s
−1
) acts like the
inverse of f (s).)
■
PrACtiCe 29 Prove that if f is an isomorphism from the commutative group 3S, # 4 to the group 3T, + 4,
then 3T, + 4 is a commutative group.
■
example 14
3R
+
, # 4 and 3R, + 4 are both groups. Let b be a positive real number, b ∙ 1, and let
f be the function from R
+
to R defined by
f (x) = log b x
Then f is an isomorphism. To prove it, we must show that f is a bijection
( one-to-one and onto) and that f is a homomorphism (preserves the operation).
We can show that f is onto: For r [ R, b
r [ R
+
and f (b
r
) = log b b
r
= r. Also, f
is one-to-one: If f (x 1 ) = f (x 2 ), then log b x 1 = log b x 2 . Let p = log b x 1 = log b x 2 .
Then b
p
= x 1 and b
p
= x 2 , so x 1 = x 2 . Finally f is a homomorphism: For
x 1 , x 2 [ R
+
, f (x 1 # x 2 ) = log b (x 1 # x 2 ) = log b x 1 + log b x 2 = f (x 1 ) + f (x 2 ). Note
that log b 1 = 0, so f maps 1, the identity of 3R
+
, # 4 to 0, the identity of 3R, + 4.
703
to the identity of the other, that inverses map to inverses, and that if one group
is commutative, so is the other. Indeed, we can prove that these expectations are
correct. (The proofs do not make use of the one-to-one property of the isomorphism, so an onto homomorphism also maps the identity to the identity, inverses
to inverses, and preserves commutativity.)
Suppose, then, that f is an isomorphism from the group 3S, # 4 to the group
3T, + 4 and that i S and i T are the identities in the respective groups. Under the
function f, i S maps to an element f (i S ) in T. Let t be any element in T. Then, because
f is an onto function, t = f (s) for some s [ S. It follows that
f (i S ) + t = f (i S ) + f (s)
= f (i S # s)
(because f is a homomorphism)
= f (s)
(because i S is the identity in S)
= t
Therefore
f (i S ) + t = t
Similarly,
t + f (i S ) = t
The element f (i S ) acts like an identity element in 3T, + 4, and because the identity
is unique, f (i S ) = i T .
PrACtiCe 28 Prove that if f is an isomorphism from the group 3S, # 4 to the group T, + 4, then for any
s [ S, f (s
−1
) = −f (s) (inverses map to inverses). (Hint: Show that f (s
−1
) acts like the
inverse of f (s).)
■
PrACtiCe 29 Prove that if f is an isomorphism from the commutative group 3S, # 4 to the group 3T, + 4,
then 3T, + 4 is a commutative group.
■
example 14
3R
+
, # 4 and 3R, + 4 are both groups. Let b be a positive real number, b ∙ 1, and let
f be the function from R
+
to R defined by
f (x) = log b x
Then f is an isomorphism. To prove it, we must show that f is a bijection
( one-to-one and onto) and that f is a homomorphism (preserves the operation).
We can show that f is onto: For r [ R, b
r [ R
+
and f (b
r
) = log b b
r
= r. Also, f
is one-to-one: If f (x 1 ) = f (x 2 ), then log b x 1 = log b x 2 . Let p = log b x 1 = log b x 2 .
Then b
p
= x 1 and b
p
= x 2 , so x 1 = x 2 . Finally f is a homomorphism: For
x 1 , x 2 [ R
+
, f (x 1 # x 2 ) = log b (x 1 # x 2 ) = log b x 1 + log b x 2 = f (x 1 ) + f (x 2 ). Note
that log b 1 = 0, so f maps 1, the identity of 3R
+
, # 4 to 0, the identity of 3R, + 4.