Group maps are determined by
the image of
: The image is an
element whose order divides
, and all such
elements are the image of such a group map.
Theorem.
(a) If is a group map,
then
.
(b) If satisfies
, then there is a group map
such that
.
Proof. (a) Suppose is a group map. Now
in
, so
This shows that .
Since , I have
.
Hence, .
(b) Let , and suppose
. Define
by
Since , I have
for some
.
Now
Since g sends to 0, the Universal
Property of the Quotient produces a (unique) group map
defined by
Then , and
is the desired group map.
Corollary. The number of group maps is
.
Proof. The number of elements of order d in
a cyclic group is (where
denotes the Euler
-function). The divisor sum of the Euler
-function is the identity:
So the number of elements whose orders divide is
, and the theorem shows that each such
element gives rise to a group map
.
Example. (a) Enumerate the group maps .
(b) Show by direct computation that given by
is not a group map.
(a) Since , there are 6 such maps by the
Corollary. They are determined by sending
to an element whose order divides
6.
Thus, the possible group maps have
For example, the group map
It is easy to determine the kernel and the image. The image is the
unique subgroup of of order 3,
so
By the First Isomorphism Theorem, the kernel must have order . The unique subgroup of
of order 6 is
(b) Consider the function given by
.Then
Therefore, , so f is not
a group map.
Copyright 2016 by Bruce Ikenaga