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Given the following exchange in $S_8$: $a = \left(\begin{array}{cccccccc} 1& 2& 3& 4& 5& 6& 7& 8\\ 2& 5& 8& 3& 1& 7& 6& 4 \end{array}\right)$

I need to find what is $a^{14}$ .

First, I found $a$, in the following sequence : $a = (1\ 2\ 5) (3\ 8\ 4) (6\ 7).$

But how can I calculate $a^{14}$ using the above ?

Regards,

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    Then (6 7)^2=(6)(7) , and (1 2 5) (1 2 5) = (5 1 2) ? and the last one (5 1 2) (1 2 5) = (2 1 5) ? thanks :)2012-01-27

1 Answers 1

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If $\sigma$ and $\tau$ are disjoint cycles, then $\sigma\tau=\tau\sigma$; in particular, for every integer $k$, we have $(\sigma\tau)^k = \sigma^k\tau^k$.

So the first observation is that you can compute $a^{14}$ by computing $(1\ 2\ 5)^{14}$, $(3\ 8\ 4)^{14}$, and $(6\ 7)^{14}$ separately.

The second observation is that if $\sigma^n$ is the identity, and $a\equiv \pmod{n}$, then $\sigma^a = \sigma^b$.

Since $(1\ 2\ 5)^{3}$, $(3\ 8\ 4)^3$, and $(6\ 7)^2$ are the identity, you only need to compute $(1\ 2\ 5)^k$ where $k$ is any number congruent to $14$ modulo $3$; $(3\ 8\ 4)^s$ where $s$ is any number congruent to $14$ modulo $3$; and $(6\ 7)^m$ where $m$ is any number congruent to $14$ modulo $2$.

I'll leave that to you.

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    @ron: By definition, if $n$ is$a$positive integer, then $a$ and $b$ are "congruent modulo $n$", written $a\equiv b \pmod{n}$, if and only if $a-b$ is a multiple of $n$. This happens if and only if $a$ and $b$ have the same remainder when you divide them by $n$. So "any number that is congruent to $14$ modulo $2$" means "any $b$ such that $14-b$ is a multiple of $2$", which means "any $b$ that leaves a remainder of $0$ when divided by $2$", which means "any even number number". I'm not using the "modulo" **operation** that CS-types like, I'm using the modulo **relation**.2012-01-27