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Factor Group Computations

  1. Nov 17, 2009 #1
    For those of you with "A First Course in Abstract Algebra 7th ed.," I'm on chapter 15 and in particular, the first 12 questions. Basically, I understand very well how to classify any factor groups of finite groups, but it's these factor groups of infinite groups that just seem so arbitrarily classified. Here's a few examples that exemplify my problem:

    8. (Z x Z x Z) / <(1,1,1)>. That is, Z x Z x Z "over" the subgroups generated by (1,1,1). Now the answer according to the book is that this factor group is isomorphic to Z x Z since every coset of the factor group contains a unique element of the form (0,m,n) where m and n are integers. This reasoning makes sense but then why can't I use the same reasoning and say that each coset also has a unique element of the form (0,0,m) or (m,n,q) thus making the factor group isomorphic to Z or Z x Z x Z respectively?

    11. (Z x Z) / <(2,2)>. According to the book, this factor group is isomorphic to Z2 x Z. If I'm given the answer I can come up with a reason why it's true: the factor group contains an element of order 2 and an element of infinite order... but not only is this logic flimsy at best, I also would have no idea how to come up with the answer in the first place.

    12. (Z x Z x Z) / <(3,3,3)> The answer here is Z3 x Z x Z with much of the same logic/problems as number 11. above.

    Any help is appreciated, these problems have been bugging me.
  2. jcsd
  3. Nov 18, 2009 #2
  4. Nov 18, 2009 #3
    8.- Let [tex](a,b,c)+H[/tex] be a coset induced by [tex]H:=<(1,1,1)> \subset \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[/tex] we want [tex](0,d,e) \in (a,b,c)+H[/tex] but this happens iff [tex](0,d,e)-(a,b,c) \in H[/tex] iff [tex](a,b-d,c-e) \in H[/tex] so we have [tex]b-d=a[/tex] and [tex]c-e=a[/tex] so taking [tex]d=b-a[/tex] and [tex]e=c-a[/tex] we get our element and the fact that it's unique is rather easy: Assume [tex](0,x,y) +H=(0,z,w)+H[/tex] then [tex]x-z=0=y-w[/tex].

    Now take [tex](a,b,c)+H[/tex] we want an element [tex](0,0,d) \in (a,b,c)+H[/tex] and reasoning as before we have [tex]a=b=0[/tex], [tex]d=c[/tex] which clearly isn't satisfied by all elements [tex](a,b,c)[/tex] so not ALL cosets have an element of the form [tex](0,0,d)[/tex]. On the other hand if you want an element [tex](d,e,f)[/tex] we get [tex]a-d=b-e=c-f[/tex] and we loose the fact that in each coset the element [tex](d,e,f)[/tex] is unique.

    Another way to tackle this problems is to note that [tex]<(1,1,1)> \cong \mathbb{Z} \times 0 \times 0[/tex] , [tex]<(2,2,2)> \cong 2\mathbb{Z} \times 0[/tex] and in general [tex]<(k,...,k)> \cong k\mathbb{Z} \times 0 \times ... \times 0[/tex] (where there are n k's and (n-1) 0's ) and with a little knowledge of direct products and quotient groups you get:

    [tex]\frac{ \mathbb{Z} ^n }{ k_1\mathbb{Z} \times ... \times k_n\mathbb{Z} } \cong \mathbb{Z} _{k_1} \times ... \times \mathbb{Z} _{k_n}[/tex]
  5. Nov 19, 2009 #4
    Thanks, this was quite helpful.
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