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Math Help - Diagonal elements of a permutation group

  1. #1
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    Diagonal elements of a permutation group

    Let
     <br />
\alpha_1=\left(\begin{array}{ccc}1&2&3\\1&2&3\end{  array}\right)\:\\\alpha_2=\left(\begin{array}{ccc}  1&2&3\\1&3&2\end{array}\right)\;\alpha_3=\left(\be  gin{array}{ccc}1&2&3\\3&2&1\end{array}\right)<br />

     <br />
\alpha_4=\left(\begin{array}{ccc}1&2&3\\2&1&3\end{  array}\right)\:\alpha_5=\left(\begin{array}{ccc}1&  2&3\\2&3&1\end{array}\right)\;\alpha_6=\left(\begi  n{array}{ccc}1&2&3\\3&1&2\end{array}\right)<br />

    The following is the group table:
     <br />
\begin{array}{c|c|c|c|c|c|c|}\circ &\alpha_1&\alpha_2&\alpha_3&\alpha_4&\alpha_5&\alp  ha_6\\<br />
\hline<br />
\alpha_1 &\alpha_1&\alpha_2&\alpha_3&\alpha_4&\alpha_5&\alp  ha_6\\<br />
\hline<br />
\alpha_2 &\alpha_2&\alpha_1&\alpha_5&\alpha_6&\alpha_3&\alp  ha_4\\<br />
\hline<br />
\alpha_3 &\alpha_3&\alpha_6&\alpha_1&\alpha_5&\alpha_4&\alp  ha_2\\<br />
\hline<br />
\alpha_4 &\alpha_4&\alpha_5&\alpha_6&\alpha_1&\alpha_2&\alp  ha_3\\<br />
\hline<br />
\alpha_5 &\alpha_5&\alpha_4&\alpha_2&\alpha_3&\alpha_6&\alp  ha_1\\<br />
\hline<br />
\alpha_6 &\alpha_6&\alpha_3&\alpha_4&\alpha_2&\alpha_1&\alp  ha_5\end{array}\end{array}<br />

    In this case along the diagonal, there two non-identity elements.

    Question:
    1. Is it possible at all to have all identity elements along the diagonal?
    2. Is it possible at all to have only one identity along the diagonal?
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  2. #2
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    Answer to question 1.
    If you have all identities along the diagonal, this is what I got.

    ab = c for some element c \in G.
    abab = cc = c^2 = e since the square of every element is the identity.

    However, since a = a^{-1} and b = b^{-1},
    <br />
abab = aba^{-1}b^{-1} = e

    Multiply both sides by ba to get

    aba^{-1}b^{-1}ba = ab = ba, so the group is Abelian.

    Thus if your group is non-Abelian, you cannot have all identity along the diagonal. Here is an example:


    \begin{array}{c|c|c|c|c|}<br />
\circ & e & a & b & c\\<br />
\hline<br />
e & e & a & b & c \\<br />
\hline<br />
a & a & e & c & \b \\<br />
\hline<br />
b & b & c & e & a \\<br />
\hline<br />
c & c & b & a & e<br /> <br />
\end{array}
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