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Group extensions

  1. Apr 10, 2006 #1
    On my exam review sheet, the prof said that we would have a question where we would have to find 4 extensions of A by B (A,B groups of order 4 or less). My friend said that by research on possible extensions ( plus the fact that this is an introductory algebra course), he/she has concluded that it is most probable that A would have be Z_4, and B would have to be Z_2 (Do you have any input as to whether this is "most probable", considering this is an intro. course?). Now, I have two questions

    1)Considering the groups mentioned above, I have come up with 4 extensions of A by B... (Z_4 X Z_2), Z_8, (The symmetries of a square), and Hamilton's Quaternions. What are other possibilities?

    I understand that I have asked about group extensions in another section, but I think this thread has somewhat of a difference. Thanks for the help.
    Last edited: Apr 10, 2006
  2. jcsd
  3. Apr 10, 2006 #2

    matt grime

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    It's a group of order 8, just work the possibilities. Try working out the possible extensions of Z_3 by Z_2, there are only two possible groups or order 6. That is an easy introduction.

    There really aren't that many groups of order 8, and even fewer that must contain an element of order 4.

    Seriously, just do it.

    Z_4 by Z_2, we know there's an element t of order 4, and an element s of order 2, so all we need to do is work out sts^{-1} and that's it. Since it is a finite group, what are the options? sts^{-1} can't be s or Id if it's t then you've got Z_4xZ_2, so what are the other options? (Hint the elements are Id,t,t^2,t^3,s,st,st^2,st^3 and that's it)
    Last edited: Apr 10, 2006
  4. Apr 10, 2006 #3
    Thanks. I have an exam right now, and so I'll work it out when I get home (By "work out sts^{-1}", you DID mean to plug in values and do the bit of scratchwork right?). As for the hint... how did you come up with those elements? I guess I have a lot more studying to do, If I can't even understand what you're saying (not completely, anyway). I'll be back in some hours with my response..

    thanks for the post
  5. Apr 11, 2006 #4

    matt grime

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    Actually I was being too hasty there.

    Let's think what group extension means here

    we have an extension of Z_4 by Z_2, that is we have G with Z_4 normal in G and G/Z_4=Z_2

    So, let t generate the subgroup Z_4, and there are two cosets for G/<t>, let s lie in the non-identity one.

    Thus every element in G is exactly one of:

    e,t,t^2,t^3 (the coset of Z_4 corresponding to the identity)
    s,st,st^2,st^3 (the coset of s)

    to specify all possible G we now need to work out some relations. In particular, we need to know the order of s, and what sts^{-1} is, and when we do that we have completely fixed G.

    Now, I made a mistake when I said s^2=e, that is not necessarily true. What is true is that in the quotient group the coset ^2 =[e], thus s^2 is one of e,t,t^2,t^3, and you need to consider all these possibilities too, and how s conjugates with t.

    Why? Every element in G is simply some 'word' in the letters 's' and 't', since they are generators for G. What is now important is knowing how to simplify these 'words' so you can identify how many different rules of composition there are.

    To do this with any group it suffices to know how to write the words in a special order. The best special order is to have all the s at the start and t at the end, ie so it is one of our list of elements in the cosets. Thus given two elements in the cosets I need to know the rule for obtaining their composition. Knowing what sts^{-1} is allows me to do this. Note, Z_4 is normal so sts^-1 is some power of t.

    Suppose that s^2=t^a and s^{-1}ts=t^b (which is the same as ts=st^b) , then I can compose any two elements, say st and st^3 using these rules:

    stst^3 = s(st^b)t^3 by rule two = s^2t^{3+b} = t^at^{3+b}=t^{a+3+b}

    so now I need to figure out all choices of a and b that are compatible.

    It was impossible to get the cyclic group of order 8 out of my first analysis written too close to getting home from the pub last night, and that corresponds to the choice of s^2=t, and st=ts: this gives an abelian group and s has order 8 which is hence cyclic generator.
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