**Definition 1**. Let be an integer. A group is called –**abelian** if for all

In other words, is -abelian if the map defined by is a group homomorphism.

**Definition 2**. If is both -abelian and -abelian, for some integers then is also -abelian because then for we will have

So the set

i.e., the set of those integers for which is -abelian, is a multiplicative subset of Clearly The set is called the **exponent semigroup** of

**Remark 1**. Since we have if and only if So a group is -abelian if and only if or, equivalently, for all So a group is -abelian if and only if it is -abelian. In other words, if and only if

**Example 1**. Every abelian group is obviously -abelian for all It is also clear that -abelian groups are abelian because gives As the next two examples show, there exists a non-abelian -abelian group for any

**Example 2**. Let be an odd integer and consider the Heisenberg group which is a non-abelian group (why?). We show that i.e. is both -abelian and -abelian. To see that, let

An easy induction shows that

for all integers Thus the identity matrix, and so for all

Hence and for all proving that is both -abelian and -abelian.

**Remark 2**. Notice that, in Example 2, if is even, then (as elements of and so is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that can also be viewed as the group of triples

with multiplication defined by

In the next example, we modify the above multiplication such that we get for all

**Example 3**. Let be any integer and consider the set Define multiplication in by

It’s easy to see that is a non-abelian group. We show that Let A quick induction shows that

for all integers So for all and thus for all proving that is -abelian.

Also, since for all we have and so is -abelian.

**Problem 1**. Let be a group. Show that

i) if for some integer then for all

ii) if for some integer then is abelian.

**Solution**. i) Let We have

and so

ii) Let By i), we have and Thus

and so

**Problem 2**. Let be an -abelian group and let Show that

i)

ii)

**Solution**. i) By Remark 1, for all Thus

and the result follows.

ii) Again, using the Remark 1, we have

**Remark 3**. Let be an -abelian group for some integer An immediate result of Problem 2, ii), is that if either is torsion-free or is finite and then is abelian.