# Math Help - Diagonal elements of a permutation group

1. ## 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?

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}$