# Diagonal elements of a permutation group

• Aug 16th 2010, 02:27 PM
novice
Diagonal elements of a permutation group
Let
$
\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)
$

$
\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:
$
\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, 04:53 PM
Vlasev
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 $b = b^{-1}$,
$
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|}
\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}$