# kth power of permutation equals identity element

• November 14th 2009, 02:33 PM
galactus
kth power of permutation equals identity element
I have been looking over this cycle and permutation business in an abstract algebra book I have. One of the problems says to prove that the kth power of a permutation equals the identity.

${\sigma}=(a_{1} \;\ a_{2} \;\ .... \;\ a_{k})\in S_{n}$.

Prove that $(a_{1} \;\ a_{2} \;\ ..... \;\ a_{k})^{k}=e$

How do we find the power of a permutation?. i.e $(1 \;\ 2 \;\ 3 \;\ 4 \;\ 5)^{3}$.

I can't seem to find it explained anywhere.
• November 14th 2009, 02:57 PM
Plato
Quote:

Originally Posted by galactus
I have been looking over this cycle and permutation business in an abstract algebra book I have. One of the problems says to prove that the kth power of a permutation equals the identity.
${\sigma}=(a_{1} \;\ a_{2} \;\ .... \;\ a_{k})\in S_{n}$.
Prove that $(a_{1} \;\ a_{2} \;\ ..... \;\ a_{k})^{k}=e$
How do we find the power of a permutation?. i.e $(1 \;\ 2 \;\ 3 \;\ 4 \;\ 5)^{3}$.

As you well know, notation depends on the author.
I like the function notation for permutations.
Say $\sigma = \left( {\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 1 & 2 \\\end{array} } \right)$ and we want to find $\sigma^3$
We do three compositions.
$\begin{gathered}
1\xrightarrow{1}3\xrightarrow{2}5\xrightarrow{3}2 \hfill \\
2\xrightarrow{1}4\xrightarrow{2}1\xrightarrow{3}3 \hfill \\
\vdots \hfill \\
5\xrightarrow{1}2\xrightarrow{2}4\xrightarrow{3}1 \hfill \\
\end{gathered}$

So $\sigma ^3 = \left( {\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ \end{array} } \right)$
• November 14th 2009, 03:41 PM
galactus
Oh, OK. that's how you do it. Thanks, Plato.
• November 14th 2009, 03:52 PM
galactus
DUH, I see what is going on now. I was misreading the problem.