Isomorphism Theorem
Proposition (First Isomorphism Theorem) Let
and
be groups, and let
be a homomorphism from
onto
with
Then the mapping
defined by
for each
is an isomorphism of
onto
Proof.
To show that
is well-defined let
Then
for some
so
Thus,
implies
and so
is a mapping.
Since
for all
preserves the group operations.
Clearly,
is onto because
is onto
To show that
is one-to-one we will show that
If
then
and so
Therefore,
as desired.
The natural homomorphism
shows that each quotient group is the homomorphic image of
and the First Isomorphism Theorem states that the converse is also true; that is, each homomorphic image of
is a quotient group.
Indeed, if a group homomorphism is not onto then you can replace
by
and thus we have,
in the following commutative diagram.
Example (First Isomorphism Theorem) For
let
and
denote the congruence classes determined by
in
and
respectively.
Define
by
Then
is well-defined because if
then
and therefore,
and
Since
is a homomorphism.
Clearly,
is onto and since
the First Isomorphism Theorem yields
Example (First Isomorphism Theorem) Consider the multiplicative group
of nonzero complex numbers.
Let
be the set of all complex numbers of absolute value
Then
is a normal subgroup of
and the quotient group of
is isomorphic to the multiplicative group
of all positive real numbers.
To see this, use the First Homomorpism Theorem applied to the onto homomorphism
defined by
with
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Cite this as: Isomorphism Theorem Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/isomorphism-theorem.html
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