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Thread: One last question for the night...

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

    Consider $\displaystyle S_5$, the symmetric group of degree five. Does it have a subgroup isomorphic to $\displaystyle C_5 \times C_5$ ? Does it have elements of order $\displaystyle 6$? Does it have a subgroup isomorphic to $\displaystyle D_5$ ? What about a subgroup isomorphic to $\displaystyle D_6 $? Let $\displaystyle C \subset S_5$ be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of $\displaystyle C$.
    Last edited by MichaelMath; Oct 27th 2010 at 09:17 AM.
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  2. #2
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    Quote Originally Posted by MichaelMath View Post
    Consider $\displaystyle S_5$, the symmetric group of degree five. Does it have a subgroup isomorphic to $\displaystyle C_5 \times C_5$ ?


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


    Does it have elements of order $\displaystyle 6$?

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


    Does it have a subgroup isomorphic to $\displaystyle D_5$ ?


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


    What about a subgroup isomorphic to $\displaystyle D_6 $?


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

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


    You try this yourself.

    Tonio

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

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

    where:

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

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

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

    $\displaystyle e \rightarrow 1$

    $\displaystyle r \rightarrow (12345)=\sigma$

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

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

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


    Does that make sense?
    Last edited by MichaelMath; Oct 28th 2010 at 08:12 PM.
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  4. #4
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    Quote Originally Posted by MichaelMath View Post
    For $\displaystyle S_5$ having a subgroup isomorphic to $\displaystyle D_6$, is the subgroup of $\displaystyle D_6$, the group of rotations?

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

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

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

    $\displaystyle e \rightarrow 1$

    $\displaystyle r \rightarrow \sigma$

    $\displaystyle r^2 \rightarrow \sigma^2$

    $\displaystyle r^3 \rightarrow \sigma^3$

    $\displaystyle r^4 \rightarrow \sigma^4$


    Does that make sense?

    Caution: Most people denote by $\displaystyle D_n$ 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; Oct 28th 2010 at 08:30 PM.
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