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Math Help - Group Theory Tough Problems

  1. #1
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    Unhappy Group Theory Tough Problems

    Howdy all, having issues with some group theory problems. Hints or suggestions are very welcomed.

    (1) Prove that for n>=3 the subgroup generated by the three cycles is An.

    (2) Prove that the smallest subgroup of Sn containing (1, 2) and (1, 2, . . . , n) is Sn

    This is what I have so far:

    Sn is generated by 2-cycles. Any element in An is a product of an even number of transpositions, so we only need to show the product of any two transpositions can be written in terms of 3-cycles.
    (12)(13) = (123), while
    (12)(34) = (123)(143).

    Is this sufficient enough, I feel as if this is not proofy. HELP!!

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RoboMyster5 View Post
    Howdy all, having issues with some group theory problems. Hints or suggestions are very welcomed.

    (1) Prove that for n>=3 the subgroup generated by the three cycles is An.


    Note that any permutation writable as a 3-cycle is an even permutation thus clearly the subgroup generated by all three cycles is a subgroup of A_n. To show the other inclusion we first note that any \sigma\in A_n may be written as the product of an even number of transpositions. It suffices then to show that the product of two distinct transpositions can be written as a 3-cycle. To do this we break it into two cases:

    1. The transpositions aren't disjoin, in other words we have (a,b)(a,c). But a little work shows that the above is actually equal to (a,c,b)

    2. The transpositions are disjoint, or we have (a,b)(c,d). Again, with a little work, we can show that this is equivalent to (a,b,d)(c,b,d)

    This finishes the proof.


    What have you done for the second one?
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  3. #3
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    See I'm not sure where to begin. I Know the symmetric group is generated bythe set of all transpositions. Since every permutation can be written as a product of disjoint cycles and each cycle is a product of transpositions. And I beleive that those two cycles are generators of the group. Thats all I can fathom.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RoboMyster5 View Post
    See I'm not sure where to begin. I Know the symmetric group is generated bythe set of all transpositions. Since every permutation can be written as a product of disjoint cycles and each cycle is a product of transpositions. And I beleive that those two cycles are generators of the group. Thats all I can fathom.
    But that is all you need! Think about it is like this, suppose you have some group G such that (1,2),(1,\cdots,n)\in G, if you can prove that (1,2),\left(1,\cdots,n\right) generates S_{n} then clearly S_n\le G but remeber here that we are only considering G\le S_n. Put the two together and we get G=S_n!
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