Let be a group homomorphism. Show that is abelian if and only if for all we have
let [tex]\varphi (G) [tex] be abelian want for [tex]x,y \in G [tex] we have
you should know that
so if we prove that
we are finished is homo which means and
not the converse direction for all we have
but it is homo
the proof ends
this is just the same thing as Amer said:
suppose that is abelian. then for any x,y in G:
(since φ is a homomorphism)
(again, since φ is a homomorphism)
(since φ(G) is abelian).
on the other hand suppose that for all x,y in G.
for any g,h in φ(G), we have φ(x) = g and φ(y) = h, for SOME x,y in G (because φ is onto φ(G)). but we know that ,
so . but:
, which quickly gives:
the quantity is called the commutator of x and y, and is written [x,y]. and here's a sneak peek of what lies ahead:
the subgroup of G generated by all the commutators, is called the commutator subgroup [G,G] (or sometimes G'). and what you (we?) have just proved is that
G/[G,G] is an abelian group (think about this for a second until it dawns on you what i am saying is true).
well, we've almost proved that. we haven't yet shown that [G,G] is normal in G, but it is, and if φ:G-->H is a homomorphism, and H is abelian, then ker(φ) contains [G,G]. [G,G] measures how "non-abelian" G is, in a natural way (the center of G,Z(G), does the same thing, but in a way that is "not natural").
if we write for , try proving that:
. this shows [G,G] is normal (why?).