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    Hi, I'm doing FP3 and my teacher told me to do the Groups option because its a good headstart for uni but I'm stuck on this question, just wondering if anyone could give a hand:

    G is a group and D(G) is the set of ordered pairs of elemts (g,\epsilon) where g is an element of G and \epsilon =\pm 1. Show that D(G) together with the binary operation ((g,\epsilon)\bullet(h,\delta)=(  gh^\epsilon, \epsilon\delta) is a group if and only if G is Abelian. (Here (h,\delta) denotes another element of D(G)).

    Find the condition on G which makes D(G) an Abelian group and show that, if this condition is satisfied, then D(D(G)) is also an Abelian group.

    Basically I managed to do the first part (showing that the operation produces a group only if G is Abelian). I did this by showing that in order for the operation to be associative (axiom for a group), then \epsilon\delta=\delta\epsilon (i.e. commutative). Since both \epsilon and \delta are members of G, then the operation on G must be commutative, i.e. G must be an Abelian group.

    However, I can't seem to do the second part (finding the condition). According to my teacher, you don't need to use the operation \bullet because thats another operation which takes place on D(G), apparently the operation is D (the assigning of the pairs). The answer is that all elements of G are self-inverse, but I don't understand how to arrive to this because these elements of G are not operated on themselves, so how does self-inverse affect them.

    Sorry if that's not that good of an explanation, I was just trying to explain my thought processes. I would appreciate if someone could give me a hand.
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    If what I think you mean is correct...

    if we define the binary operation as:

    (g,\epsilon)\bullet (h,\delta) = (g\cdot h,\epsilon\delta) (where  \cdot is the group operation in G), then what you are essentially defining D(G) as is  G \times C_2 (cf. direct product, http://en.wikipedia.org/wiki/Direct_product). This is because the group  C_2 is the same as the group of  \{\pm 1\} under multiplication.

    This is abelian if and only if  G is abelian.

    N.B.  D(D(G)) = G \times C_2 \times C_2 , and this also is abelian if and only if G is.

    If you're new to group theory I can give you a few things you might want to look into:

    Groups essentially are all about symmetry. The most familiar groups are the cyclic groups - the  C_n for an integer n. These groups represent the rotations of n-gons (e.g.  C_4 is essentially the group of the rotations of a square. There are four elements corresponding to the four rotations: 0, 90, 180 and 270 degrees.) If you want to go further check out other interesting groups like the dihedral 'D' groups, symmetric 'S' group, alternating 'A' groups. Group theory is a massive branch of maths.

    Random interesting group theory fact: 99% of the groups of size less than 2000 have size 1024.
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    (Original post by wellison999)
    If what I think you mean is correct...

    if we define the binary operation as:

    (g,\epsilon)\bullet (h,\delta) = (g\cdot h,\epsilon\delta) (where  \cdot is the group operation in G), then what you are essentially defining D(G) as is  G \times C_2 (cf. direct product, http://en.wikipedia.org/wiki/Direct_product). This is because the group  C_2 is the same as the group of  \{\pm 1\} under multiplication.

    This is abelian if and only if  G is abelian.

    N.B.  D(D(G)) = G \times C_2 \times C_2 , and this also is abelian if and only if G is.

    If you're new to group theory I can give you a few things you might want to look into:

    Groups essentially are all about symmetry. The most familiar groups are the cyclic groups - the  C_n for an integer n. These groups represent the rotations of n-gons (e.g.  C_4 is essentially the group of the rotations of a square. There are four elements corresponding to the four rotations: 0, 90, 180 and 270 degrees.) If you want to go further check out other interesting groups like the dihedral 'D' groups, symmetric 'S' group, alternating 'A' groups. Group theory is a massive branch of maths.

    Random interesting group theory fact: 99% of the groups of size less than 2000 have size 1024.
    oh my i'm sorry, you went to all that effort, i didn't type out the question properly. I apologise. I've edited the original post now but what the operation was supposed to represent was this:

    (g,\epsilon)\bullet(h,\delta)=(g  h^\epsilon, \epsilon\delta). again, sorry about that.

    also, its this bit that i'm stuck on: Find the condition on G which makes D(G) an Abelian group and show that, if this condition is satisfied, then D(D(G)) is also an Abelian group.
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    Yep I agree with your teacher - all elements must be self inverse. I will explain:

     D(G) abelian  \iff  (g,\epsilon) \bullet (h,\delta) = (h,\delta) \bullet (g, \epsilon) . Note that this must be true for ALL  g,h \in G and  \epsilon,\delta \in \{\pm 1\} because for a group to be abelian, we need ab=ba FOR EVERY a and b in the group.

    I.e.  (gh^{\epsilon},\epsilon \delta) = (hg^{\delta},\delta\epsilon) . BUT  \delta\epsilon = \epsilon\delta so this is the same as
     gh^{\epsilon} = hg^{\delta} .

    So we get that:

     gh = hg AND  gh^{-1}=hg \implies gh = hgh^2.

    So  h^2 = e . Since this must be true for every element  h of  G , every element must be self inverse (h^2 =e).

    The last bit isn't too hard, and I'll let you work it out.
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    OK i'm really not to sure about this D(D(G)) one because I end up with 2 pairs of elements next to each other with no operation between them??????!!. Could you check if this is right:

    Let 2 elements from D(D(G)) be ((g, \epsilon),\alpha) and ((h,\delta), \beta) where g,h are members of G and \alpha , \beta, \delta, \epsilon are members of {\pm1}.

    So

    ((g, \epsilon),\alpha) \bullet h,\delta), \beta) = ((g, \epsilon)(h, \delta)^\alpha , \alpha \beta) = ((gh^\alpha, \epsilon \delta^\alpha),\alpha \beta).

    Is it alright to "expand" it like this. I only need you to check this bit because if this is correct then I can answer the question. If it is not correct, then I have no clue what to do. Thanks.
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    (Original post by lilman91)
    ((g, \epsilon),\alpha) \bullet h,\delta), \beta) = ((g, \epsilon)(h, \delta)^\alpha , \alpha \beta) = ((gh^\alpha, \epsilon \delta^\alpha),\alpha \beta).
    The last bit isn't quite right. We 'multiply' as we have defined  \bullet originally (we just are not writing  \bullet here - this is quite a standard thing to do in group theory, by ab we mean  a\bullet b for the group operation of the group of which a and b are members of).

    Here,  (g, \epsilon) and  (h, \delta) are members of D(G), so by their 'product' we mean  (g, \epsilon)\bullet(h, \delta) .
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    Oh OK that makes sense, but then how do we deal with the \alpha power on (h, \delta)^\alpha
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    Well since  \alpha is either 1 or -1, you either get  (h,\delta) or  (h,\delta)^{-1} . To find the inverse, just see what you need to send  (h,\delta) to the identity.
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    http://www.cafepress.com/cp/moredeta...olorNo=23&pr=F
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    Indeed, it's the only group theory joke I know of. I always get odd looks if I mention it!
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    (Original post by wellison999)
    Indeed, it's the only group theory joke I know of. I always get odd looks if I mention it!
    It is almost as old as "why did the chicken cross the road?"

    But next time someone tells it to you, you can reply "been there, done that, got the T shirt!"
 
 
 
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