# Diagonal elements of a permutation group

• Aug 16th 2010, 01:27 PM
novice
Diagonal elements of a permutation group
Let
$\displaystyle \alpha_1=\left(\begin{array}{ccc}1&2&3\\1&2&3\end{ array}\right)\:\\\alpha_2=\left(\begin{array}{ccc} 1&2&3\\1&3&2\end{array}\right)\;\alpha_3=\left(\be gin{array}{ccc}1&2&3\\3&2&1\end{array}\right)$

$\displaystyle \alpha_4=\left(\begin{array}{ccc}1&2&3\\2&1&3\end{ array}\right)\:\alpha_5=\left(\begin{array}{ccc}1& 2&3\\2&3&1\end{array}\right)\;\alpha_6=\left(\begi n{array}{ccc}1&2&3\\3&1&2\end{array}\right)$

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

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?
• Aug 16th 2010, 03:53 PM
Vlasev
If you have all identities along the diagonal, this is what I got.

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

However, since $\displaystyle a = a^{-1}$ and $\displaystyle b = b^{-1}$,
$\displaystyle abab = aba^{-1}b^{-1} = e$

Multiply both sides by ba to get

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

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

$\displaystyle \begin{array}{c|c|c|c|c|} \circ & e & a & b & c\\ \hline e & e & a & b & c \\ \hline a & a & e & c & \b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \end{array}$