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Math Help - The center of the group T

  1. #1
    Forum Admin topsquark's Avatar
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    The center of the group T

    This is probably going to be a tedious job for someone. I apologize in advance.

    I have posted a multiplication table for this group. The group is generated by two elements a, and b, both cyclic and of order 4 and 3 respectively. That is to say that all the group elements are in the form a^mb^n where a^4 = e and b^3 = e. The only other piece of information needed to create the table is that I have set ba = a^3b.

    The problem I have is finding the center. By my definition, the center of a group G is the set of elements \{ g \in G |gx = xg ~\forall x \in G \}. I have determined the center of this group to be \{ e, a^2, b^2, a^2b^2 \}. The center is an Abelian subgroup, so the set I have labeled must be a group.

    But it is not closed. In particular (b^2)^2 = b and (a^2b^2)^2 = a^3.

    What have I done wrong?

    -Dan

    Edit: That attachment is huge! I'll have to figure out how to reduce the size at some point.
    Attached Thumbnails Attached Thumbnails The center of the group T-scan0001.jpg  
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  2. #2
    Forum Admin topsquark's Avatar
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    I have another thread in the chatroom on this group and it appears I shouldn't have put the thread in there after all. Referring to the comment that Opalg made (down near the bottom) he commented that the group elements are not distinct and the group is really the order 6 Abelian group \{ e, a, b, ab, b^2, ab^2 \}. Could this be why my calculation of the center of this group is having problems?

    -Dan
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by topsquark View Post
    I have another thread in the chatroom on this group and it appears I shouldn't have put the thread in there after all. Referring to the comment that Opalg made (down near the bottom) he commented that the group elements are not distinct and the group is really the order 6 Abelian group \{ e, a, b, ab, b^2, ab^2 \}. Could this be why my calculation of the center of this group is having problems?

    -Dan
    The whole thing is messed up--where particularly did this come from? Trying to construct groups from generators and relations is a horrific task. In fact, as I already think another member pointed out in the chat room there are some relatively simple presentations for which it is still uknown whether the generated group is even non-trivial!
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  4. #4
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    something is terribly wrong, here. check the definition of your relations.

    i think what you are trying to do is define the semi-direct product Z3 x| Z4.

    but the multiplication of a semi-direct product "twists" things.

    let id:Z3-->Z3 be the identity automorphism of Z3, and

    inv:Z3-->Z3 be the inverse automorphism of Z3. we can define a homomorphism of

    Z4 into Aut(Z3) by 1,3-->inv, 0,2-->id

    if we set b = (1,0), and a = (0,1) then ba = (1,0)(0,1) = (1+id(0),0+1) = (1,1)

    while ab^2 = (0,1)(2,0) = (0+inv(2),1+0) = (-2,1) = (1,1), so i think your third relation should be:

    ba = ab^2.

    see: http://mathworld.wolfram.com/FiniteGroupT.html
    Last edited by Deveno; April 5th 2011 at 09:17 PM.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    The whole thing is messed up--where particularly did this come from? Trying to construct groups from generators and relations is a horrific task. In fact, as I already think another member pointed out in the chat room there are some relatively simple presentations for which it is still uknown whether the generated group is even non-trivial!
    Not quite...it is the identity element. There are some presentations such that you cannot tell if a given element is the identity or not.

    What is related to your comment though is something similarly hideous, called the Andrews-Curtis conjecture, which states that `there exist some presentations of the trivial group which cannot be transformed into a trivial presentation using only Tietze transformations'. Unlike the word problem, no counter-examples have been found. But it is similarly hideous - it basically claims that Tietze transformations are rubbish! Which is, quite frankly, annoying...
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  6. #6
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    well, seeing as how the word problem can be insoluable, it doesn't surprise me that Teitze transformations might not transform a trivial group into a trivial presentation. it seems to me to be similar to deciding if a knot is the unknot.
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  7. #7
    Forum Admin topsquark's Avatar
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    I am beginning to feel like I'm seriously out of my depth here. All I was trying to do was to find a relatively large (non-Abelian) group with variously sized subgroups to test some of the problems in my textbook! Maybe I should have followed my first instinct and looked up T_h!

    I lack the necessary knowledge to follow the discussion so I'm going to strategically retreat, lick my wounds, do some more studying and come back to it later.

    Thank you all and I'll see you around.

    -Dan
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  8. #8
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    i recommend investigating S4. it has only 24 elements, and a healthy variety of interesting subgroups.
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