Let $\displaystyle \varphi :G\rightarrow H$ be a group homomorphism. Show that $\displaystyle \varphi[G]$ is abelian if and only if for all $\displaystyle x,y\in G,$ we have $\displaystyle xyx^{-1}y^{-1}\in Ker\left ( \varphi \right )$
Let $\displaystyle \varphi :G\rightarrow H$ be a group homomorphism. Show that $\displaystyle \varphi[G]$ is abelian if and only if for all $\displaystyle x,y\in G,$ we have $\displaystyle xyx^{-1}y^{-1}\in Ker\left ( \varphi \right )$
let [tex]\varphi (G) [tex] be abelian want for [tex]x,y \in G [tex] we have $\displaystyle xyx^{-1}y^{-1}\in Ker(\varphi) $
you should know that
$\displaystyle Ker(\varphi) = \{ g \in G \mid \varphi(g) = e_H\} $
so if we prove that $\displaystyle \varphi (xyx^{-1}y^{-1} ) = e_H $
we are finished $\displaystyle \varphi $ is homo which means $\displaystyle \varphi (ab)= \varphi (a) \varphi(b) $ and $\displaystyle \varphi(a^{-1}) = ( \varphi(a))^{-1} $
not the converse direction for all $\displaystyle x,y \in G $ we have $\displaystyle xyx^{-1}y^{-1} \in Ker(\varphi) $
want $\displaystyle \varphi(G) $ abelian
let $\displaystyle a,b \in \varphi (G) \Rightarrow \varphi^{-1} (a) , \varphi^{-1}(b) \in G $
$\displaystyle \varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1} \in Ker(\varphi) $
$\displaystyle \varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a))^{-1} (\varphi^{-1}(b))^{-1}] = e_H $
but it is homo
$\displaystyle \varphi [\varphi^{-1} (a)\varphi^{-1}(b) (\varphi^{-1}(a^{-1})) (\varphi^{-1}(b^{-1}))] = e_H $
$\displaystyle aba^{-1}b^{-1} = e_H \Rightarrow ab = ba $
the proof ends
this is just the same thing as Amer said:
suppose that $\displaystyle \varphi(G)$ is abelian. then for any x,y in G:
$\displaystyle \varphi(xyx^{-1}y^{-1}) = \varphi(x)\varphi(y)\varphi(x^{-1})\varphi(y^{-1})$ (since φ is a homomorphism)
$\displaystyle =\varphi(x)\varphi(y)(\varphi(x))^{-1}(\varphi(y))^{-1}$ (again, since φ is a homomorphism)
$\displaystyle = \varphi(x)(\varphi(x))^{-1}\varphi(y)(\varphi(y))^{-1}$ (since φ(G) is abelian).
$\displaystyle = e_He_H = e_H$ showing that $\displaystyle xyx^{-1}y^{-1} \in ker(\varphi)$
on the other hand suppose that $\displaystyle xyx^{-1}y^{-1} \in ker(\varphi)$ 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 $\displaystyle xyx^{-1}y^{-1} \in ker(\varphi)$,
so $\displaystyle \varphi(xyx^{-1}y^{-1}) = e_H$. but:
$\displaystyle \varphi(xyx^{-1}y^{-1}) = \varphi(x)\varphi(y)(\varphi(x))^{-1}(\varphi(y))^{-1}$
$\displaystyle = ghg^{-1}h^{-1} = e_H$, which quickly gives:
$\displaystyle gh = hg$
the quantity $\displaystyle xyx^{-1}y^{-1}$ 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 $\displaystyle a^g$ for $\displaystyle gag^{-1}$, try proving that:
$\displaystyle [a,b]^g = [a^g,b^g]$. this shows [G,G] is normal (why?).