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Math Help - One last question for the night...

  1. #1
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    One last question for the night...

    Consider S_5, the symmetric group of degree five. Does it have a subgroup isomorphic to C_5 \times C_5 ? Does it have elements of order 6? Does it have a subgroup isomorphic to D_5 ? What about a subgroup isomorphic to D_6 ? Let C \subset S_5 be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of C.
    Last edited by MichaelMath; October 27th 2010 at 10:17 AM.
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  2. #2
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    Quote Originally Posted by MichaelMath View Post
    Consider S_5, the symmetric group of degree five. Does it have a subgroup isomorphic to C_5 \times C_5 ?


    Of course not: |C_5\times C_5|=25\nmid 120=5! , so by Lagrange it can't be


    Does it have elements of order 6?

    Lots of: any element of the form (ab)(cde) , with different a,b,c,d,e has order 6


    Does it have a subgroup isomorphic to D_5 ?


    Check the subgroup generated by (14)(23),\,(12345)


    What about a subgroup isomorphic to D_6 ?


    Check the subgroup <(34),\,(12)(345)>

    Let C \subset S_5 be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of C.


    You try this yourself.

    Tonio

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  3. #3
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    For S_5 having a subgroup isomorphic to D_6,

    Can I use the group of rotations in  D_6: r^n \, ( n =1,2,3,4,5), and \sigma = (1,2,3,4,5) in S_5

    where:

    X = \{ e,r,r^2,r^3,r^4 \}

    Y = (1,\sigma, \sigma^2,\sigma^3,\sigma^4) in S_5

    Then the isomorphism \phi: X \rightarrow Y is:

    e \rightarrow 1

    r \rightarrow (12345)=\sigma

    r^2 \rightarrow (13524)=\sigma^2

    r^3 \rightarrow (14253)=\sigma^3

    r^4 \rightarrow (15432)=\sigma^4


    Does that make sense?
    Last edited by MichaelMath; October 28th 2010 at 09:12 PM.
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  4. #4
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    Quote Originally Posted by MichaelMath View Post
    For S_5 having a subgroup isomorphic to D_6, is the subgroup of D_6, the group of rotations?

    So if the subgroup of rotations X = \{ e,r,r^2,r^3,r^4 \}

    and \sigma = (1,2,3,4,5) in S_5

    Then the isomorphism \phi: X \rightarrow S_5 is:

    e \rightarrow 1

    r \rightarrow \sigma

    r^2 \rightarrow \sigma^2

    r^3 \rightarrow \sigma^3

    r^4 \rightarrow \sigma^4


    Does that make sense?

    Caution: Most people denote by D_n the dihedral group of order 2n , not the one of order n.

    Tonio
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  5. #5
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    I made a few changes above...

    arghgh...

    So out of (12)(35), (13,45), (15)(24), (25)(34), and (14,23), the last one fixes 5... does this mean anything...
    Last edited by MichaelMath; October 28th 2010 at 09:30 PM.
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