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Thread: Permutation groups

  1. #1
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    Permutation groups

    1a) Prove that for any two 3-cycles $\displaystyle \sigma_1, \sigma_2 \in S_n (n>4)$ there exists a $\displaystyle \tau \in A_n $such that $\displaystyle \tau\sigma_1\tau^{-1}=\sigma_2$
    1b) Let $\displaystyle Z(S_n) $ be the center of $\displaystyle S_n $. Prove that $\displaystyle Z(S_n)=Z(A_n)=\{e\}$ for all $\displaystyle n>3 $ and $\displaystyle Z(S_3)=\{e\}$
    1c) Prove that the only nontrivial normal subgroup of $\displaystyle S_n (n>4)$ is $\displaystyle A_n$
    1d) Give a nontrivial normal subgroup $\displaystyle N$ of $\displaystyle S_4$ such that $\displaystyle N \ne A_4$
    can anyone help? thanks in advance
    Last edited by tszhin8831; Nov 22nd 2008 at 11:36 PM.
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  2. #2
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    Quote Originally Posted by tszhin8831 View Post
    1a) Prove that for any two 3-cycles $\displaystyle \sigma_1, \sigma_2 \in S_n (n>4)$ there exists a $\displaystyle \tau \in A_n $such that $\displaystyle \tau\sigma_1\tau^{-1}=\sigma_2$
    If $\displaystyle \sigma_1 = (a,b,c)$ then $\displaystyle \tau \sigma \tau^{-1} = (\tau (a), \tau(b) , \tau (c))$.
    Therefore, if $\displaystyle \sigma_2 = (d,e,f)$ then let $\displaystyle \tau$ be defined as $\displaystyle a\mapsto d, b\mapsto e, c\mapsto f$ and fixes everything else.
    The problem is that $\displaystyle \tau = (a,d)(b,e)(c,f)\not \in A_n$.
    In order to fix that choose $\displaystyle a'\not \in \{a,b,c\}$ and $\displaystyle b'\not \in \{a',d,e,f\}$ and redefine $\displaystyle \tau$ to send $\displaystyle a'\mapsto b'$.
    This is possible since $\displaystyle n>4$. And we see that $\displaystyle \tau = (a,d)(b,e)(c,f)(a',b')\in A_n$.
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  3. #3
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    Quote Originally Posted by tszhin8831 View Post
    1b) Let $\displaystyle Z(S_n) $ be the center of $\displaystyle S_n $. Prove that $\displaystyle Z(S_n)=Z(A_n)=\{e\}$ for all $\displaystyle n>3 $ and $\displaystyle Z(S_3)=\{e\}$
    Let $\displaystyle \sigma\in Z(S_n)$ if $\displaystyle \sigma_1$ is conjugate to $\displaystyle \sigma$ then it means $\displaystyle \tau \sigma \tau^{-1} = \sigma_1 \implies \sigma = \sigma_1$.
    Therefore, elements in the center are only conjugate to themselves.
    We will show that any non-identity element in $\displaystyle S_n$ ($\displaystyle n>2$) are conjugate to some other element.
    Let $\displaystyle \theta \in S_n$ be a cycle, $\displaystyle \theta = (a_1,...,a_k)$, $\displaystyle k\geq 2$.
    Then $\displaystyle \tau \theta \tau^{-1} = (\tau(a_1), ... , \tau(a_k))$.
    Therefore if we pick $\displaystyle \tau = (a_2,...,a_k,a_1)$ then $\displaystyle \tau \theta \tau^{-1} \not = \theta$ if $\displaystyle k>2$.
    If $\displaystyle k=2$ then $\displaystyle (a_1,a_2)$ and $\displaystyle (a_2,a_1)$ are the same thing.
    But in this special case we can just let $\displaystyle \tau = (a_1,a_3)$ where $\displaystyle a_3 \not \in \{a_1,a_2\}$ and so $\displaystyle \tau \theta \tau^{-1} \not = \theta$.
    We have shown that no cycle can be in $\displaystyle Z(S_n)$ for $\displaystyle n>2$.

    Let $\displaystyle \sigma \in S_n$ be a non-identity permutation.
    Then we can write $\displaystyle \sigma = \theta_1 \theta_2 ... \theta_j$ into a product of disjoint cycles.
    Therefore, $\displaystyle \tau \sigma \tau^{-1} = ( \tau \theta_1 \tau^{-1})(\tau \theta_2 \tau^{-1}) ... (\tau \theta_j \tau^{-1})$.
    For the moment assume there is one of $\displaystyle \theta_1,...,\theta_j$ that is not a transposition, say, $\displaystyle \theta_1$ by relabing.
    Then it means, $\displaystyle \theta_1 = (x_1,...,x_r)$ where $\displaystyle r>2$.
    Let $\displaystyle \tau = (x_2,...,x_r,x_1)$ and we see that $\displaystyle \tau \theta_i \tau^{-1} = \theta_i$ for $\displaystyle i\not = 1$.
    Therefore, $\displaystyle \tau \sigma \tau^{-1} = \tau \theta_1 \tau^{-1} \theta_2 ... \theta_j$ while $\displaystyle \sigma = \theta_1 ... \theta_j$ so $\displaystyle \tau \sigma \tau^{-1} \not = \sigma$ since $\displaystyle \tau \theta_1 \tau^{-1} \not = \theta_1$.
    We have constructed an element conjugate and not equal to $\displaystyle \sigma$ thus $\displaystyle \sigma \not \in Z(S_n)$.
    But we left the case when each $\displaystyle \theta_i$ is a transposition as an exceptional case.
    Try to see what you can come up with to complete the proof for $\displaystyle Z(S_n) = \{ e \}$ for $\displaystyle n>2$.

    As for $\displaystyle Z(A_n)$ the proof should be similar but needs to be done with more care.
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  4. #4
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    Quote Originally Posted by tszhin8831 View Post
    1c) Prove that the only nontrivial normal subgroup of $\displaystyle S_n (n>4)$ is $\displaystyle A_n$
    If you happen to know that $\displaystyle A_n$ is simple for $\displaystyle n>4$ then that is really helpful here. Accepting this fact we see that $\displaystyle \{ e \} \subset A_n \subset S_n$ is a composition series. By Jordan-Holder theorem this is the unique composition series. Consequently, it must mean that $\displaystyle A_n$ is the only proper non-trivial normal subgroup of $\displaystyle S_n$.
    1d) Give a nontrivial normal subgroup $\displaystyle N$ of $\displaystyle S_4$ such that $\displaystyle N \ne A_4$
    can anyone help? thanks in advance
    Choose $\displaystyle N = \{ \text{id}, (12)(34),(13)(24),(14)(23)\}$.

    Here is another related problem can you show that $\displaystyle N$ and $\displaystyle A_4$ are the only proper non-trivial normal subgroups of $\displaystyle S_4$?
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    thanks a lot
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