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About LinkBacksLetbe 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
 = \{ g \in G \mid \varphi(g) = e_H\} )
so if we prove that = e_H )
we are finishedis homo which means
and
 = ( \varphi(a))^{-1} )
not the converse direction for allwe have
 )
wantabelian
let \Rightarrow \varphi^{-1} (a) , \varphi^{-1}(b) \in G )
\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1} \in Ker(\varphi) )
![\varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1}] = e_H](http://latex.codecogs.com/png.latex?\varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1}] = e_H )
but it is homo
![\varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a^{-1})) (\varphi^{-1}(b^{-1}))] = e_H](http://latex.codecogs.com/png.latex?\varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a^{-1})) (\varphi^{-1}(b^{-1}))] = e_H )
the proof ends
this is just the same thing as Amer said:
suppose thatis abelian. then for any x,y in G:
(since φ is a homomorphism)
(again, since φ is a homomorphism)
(since φ(G) is abelian).
showing that
on the other hand suppose thatfor 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 quantityis 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 writefor
, try proving that:
. this shows [G,G] is normal (why?).
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