# Given an permutation $a$ in $S_8$ how to find $a^{14}$?

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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Hint: start with $(1,2,5)^3$, or very easy $(6,7)^2$. –  draks ... Jan 27 '12 at 18:17
That's the thing : I don't exactly understand how to calculate the following : (125) (125) , or (125) (125) (125) , can you please explain ? –  ron Jan 27 '12 at 18:19
setup 8 things in front of you and permute element 6 and 7, this is $(6,7)$, permute twice for $(6,7)^2$. Or more difficult: Put thing 1 on place 2, thing 2 on place 5 and thing 5 on 1. This is $(1,2,5)$. Happy shuffling! –  draks ... Jan 27 '12 at 18:21
So if I'm not mistaken , (6,7)^2 equals (6) ? –  ron Jan 27 '12 at 18:25
Hmm, in fact I think $(6)$ might count as a representation of the identity, but I won't write it like that. Maybe $(6)(7)$ suits better... –  draks ... Jan 27 '12 at 18:29

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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(1 2 5)^3 = (1 5 2) , (3 8 4)^3 = (3 4 8) , (6 7)=(6)(7) , is this right ? if so , what now ? thanks –  ron Jan 27 '12 at 19:18
@ron: No; $(1\ 2\ 5)^3$ is the identity. $(3\ 8\ 4)^3$ is the identity. $(6\ 7) = (6\ 7)$. I think you didn't write what you meant. Were you trying to write the result of $(1\ 2\ 5)^{14}$, $(3\ 8 \ 4)^{14}$, and $(6\ 7)^{14}$? If so, then yes. Then $$a^{14} = \Bigl( (1\ 2\ 5)(3\ 8\ 4)(6\ 7)\Bigr)^{14} = (1\ 2\ 5)^{14}(3\ 8\ 4)^{14}(6\ 7)^{14};$$and since you know the far right hand side, you know what $a^{14}$ is. –  Arturo Magidin Jan 27 '12 at 19:31
Then I didn't get what you mean ,probably.When you say "the identity" then you mean "id" , right ? so a few questions: 1. what is the identity of (1 2 5)^3 ? 2. if what I wrote in my previous comment isn't correct , then how can I calculate what a^14 exact value ? thanks a lot! –  ron Jan 27 '12 at 20:08
@ron: $(1\ 2\ 5)^3$ is the identity of $S_8$*. It's the permutation that sends every number to itself: just compute it! That's why what you wrote in your previous comment is incorrect. You wrote "$(1\ 2\ 5)^3 = (1\ 5\ 2)$." That's *wrong. If you had written "$(1\ 2\ 5)^{14}=(1\ 5\ 2)$" instead, then that would have been correct. I suspect you made that mistake when you wrote down your first comment for each of the three permutations. And I already told you how to compute $a^{14}$ exactly, twice (once in the post, again in the previous comment). –  Arturo Magidin Jan 27 '12 at 20:15
Okay ,then : (3 8 4)^14 = (3 8 4)^2 = (3 4 8) ? but what about (6 7)^14 ? since 14 modulo 2 equals 0 , I get (6 7)^14=(6 7)^0. –  ron Jan 27 '12 at 21:52