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Showing that a factor group is abelian

  1. Apr 14, 2010 #1
    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 [tex]\kappa[/tex]a [tex]\circ[/tex] h = h [tex]\circ[/tex] [tex]\kappa[/tex]a for every a [tex]\in[/tex] G, then the set H = {x[tex]\in[/tex]G : h(x) = h(h(x))} is a normal subgroup of G, and G/H is an abelian group.​

    2. [tex]\kappa[/tex]a is the inner automorphism defined by a, and is defined as

    [tex]\kappa[/tex]a(x) = a[tex]\Delta[/tex]x[tex]\Delta[/tex]a*​

    where [tex]\Delta[/tex] 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 [tex]x\Delta y\Delta x*[/tex][tex]\Delta y* \in H \forall x, y \in G [/tex] (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[tex]\Delta[/tex]y[tex]\Delta[/tex]x*[tex]\Delta[/tex]y* [tex]\in[/tex] H [tex]\forall[/tex] 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[tex]\Delta[/tex]H = H​
    and since y[tex]\Delta[/tex]H[tex]\Delta[/tex]y* = H [tex]\forall[/tex] y because H is normal,

    x[tex]\Delta[/tex](y[tex]\Delta[/tex]H[tex]\Delta[/tex]y*) = H​
    which contains x[tex]\Delta[/tex]y[tex]\Delta[/tex]x*[tex]\Delta[/tex]y* because H contains x*.

    but the challenge is to show that this is true [tex]\forall[/tex] x,y[tex]\in[/tex]G
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 15, 2010 #2
    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 [tex]xy[/tex], not [tex]x\triangle y[/tex], and the inversion is [tex]x^{-1}[/tex], not [tex]x^*[/tex]. (Unless the group is abelian and written in additive notation, in which case it's [tex]x + y[/tex] and [tex]-x[/tex], 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 [tex]xyx^{-1}y^{-1} \in H[/tex] for every [tex]x, y \in G[/tex]. The condition for [tex]z \in H[/tex] is that [tex]h(h(z)) = h(z)[/tex]. Why not just play around with [tex]h(h(xyx^{-1}y^{-1}))[/tex] and see what happens? The condition [tex]\kappa_a \circ h = h \circ \kappa_a[/tex] for every [tex]a \in G[/tex] gives you some clever tricks to manipulate this with; if you don't see how, write out the equation [tex](\kappa_a \circ h)(z) = (h \circ \kappa_a)(z)[/tex] for a given [tex]z\in G[/tex].
    Last edited: Apr 15, 2010
  4. Apr 15, 2010 #3
    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.)
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