Showing that a factor group is abelian

[valexodamium]

This problem is from Seth Warner's Modern Algebra; problem number 12.21 (so Google can find it.) It's actually in the free preview, find it http://books.google.com/books?id=jd... Algebra "12.21"&pg=PA90#v=onepage&q&f=false"

1. The problem is stated as:

If h is an endomorphism of a group G such that \kappa_a \circ h = h \circ \kappa_a for every a \in G, then the set H = {x\inG : h(x) = h(h(x))} is a normal subgroup of G, and G/H is an abelian group.​

2. \kappa_a is the inner automorphism defined by a, and is defined as

\kappa_a(x) = a\Deltax\Deltaa*​

where \Delta is the group's binary operator and a* is the inverse of a.

We are also given a theorem that states that G/H is abelian iff x\Delta y\Delta x*\Delta y* \in H \forall x, y \in G (which I also had to prove, but I'll spare you.)


3. Now, I managed to prove that H is a subgroup and that H is normal. I am at a loss, however, as to how to prove that G/H is abelian. I understand that I'm supposed to show that x\Deltay\Deltax*\Deltay* \in H \forall x,y , but I can only see a way to do that if either x or y is in H, for example:

if x is in H, then

x\DeltaH = H​

and since y\DeltaH\Deltay* = H \forall y because H is normal,

x\Delta(y\DeltaH\Deltay*) = H​

which contains x\Deltay\Deltax*\Deltay* because H contains x*.

but the challenge is to show that this is true \forall x,y\inG

 

[ystael]

First of all, Seth Warner should be beaten with a rubber hose for inflicting this notation on you. The operation in a group should be written xy, not x\triangle y, and the inversion is x^{-1}, not x^*. (Unless the group is abelian and written in additive notation, in which case it's x + y and -x, respectively.) Incidentally, the triangle symbol is \triangle, not \Delta which is a Greek letter.

Now that I've got that off my chest ... You need to show that xyx^{-1}y^{-1} \in H for every x, y \in G. The condition for z \in H is that h(h(z)) = h(z). Why not just play around with h(h(xyx^{-1}y^{-1})) and see what happens? The condition \kappa_a \circ h = h \circ \kappa_a for every a \in G gives you some clever tricks to manipulate this with; if you don't see how, write out the equation (\kappa_a \circ h)(z) = (h \circ \kappa_a)(z) for a given z\in G.

 

[valexodamium]

Thanks! That and h(xy) = h(x)h(y) did the trick.

(And yeah, the book's notation is weird. It seems that it starts using more conventional notation more often eventually, though.)