This is a review of some elementary criteria for a group to be abelian.
Lemma. Let be a group. Then the following are equivalent:
is abelian,
- The map
given by
is a group homomorphism;
- The map
given by
is a group homomorphism;
- The diagonal
is normal.
Proof. We prove that each criterion is equivalent to (1).
For (2), note that , which equals
if and only if
.
For (3), note that , which equals
if and only if
.
For (4), clearly is normal if
is abelian. Conversely, note that
, which is in the diagonal if and only if
.
The map
given by
is a group homomorphism for
.
Fascinating; I have never heard of this. Are you saying that this is true for any group
? Do you know where I can read more about this?
Hello, Remy 🙂
such that the three maps
,
,
are three group endomorphisms must be an abelian group.
Yes, I meant that any group
You can find conditions of this type among the exercises in Herstein’s algebra textbook (which is now a bit too old-fashioned, but contains a fantastic wealth of good exercises). For example you can prove that if
is an endomorphism for three consecutive values of
, then
must be abelian; or that if
and
are both endomorphisms, then
must be abelian.
I liked these exercises a lot as a student, so I was very pleased to see the topic addressed fully in this nice elementary paper:
Abelian Forcing Sets, Joseph A. Gallian and Michael Reid, The American Mathematical Monthly, Vol. 100, No. 6 (Jun. – Jul., 1993), pp. 580-582