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