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$$\in$$G : 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$$\Delta$$x$$\Delta$$a*​

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$$\Delta$$y$$\Delta$$x*$$\Delta$$y* $$\in$$ 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$$\Delta$$H = H​

and since y$$\Delta$$H$$\Delta$$y* = H $$\forall$$ y because H is normal,

x$$\Delta$$(y$$\Delta$$H$$\Delta$$y*) = H​

which contains x$$\Delta$$y$$\Delta$$x*$$\Delta$$y* because H contains x*.

but the challenge is to show that this is true $$\forall$$ x,y$$\in$$G

 

[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.)