Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.14
Let $\sigma = (1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12)$. For each of the following integers $a$, compute $\sigma^a$: 13, 65, 626, 1195, -6, -81, -570, -1211.
Solution: First we prove a lemma.
Lemma: Let $G$ be a group and $x \in G$ and element with $|x| = n < \infty$. If $a = b \pmod n$, then $x^a = x^b$.
Proof: We have $a = b + nk$ for some integer $k$. Then $$x^a = x^{b+nk} = x^b (x^n)^k = x^b 1^k = x^b.\quad \square$$ We know that $|\sigma| = 12$.
$13 = 1 \pmod{12}$, so $\sigma^{13} = \sigma$.
$65 = 5 \pmod{12}$, so as we saw in a previous exercise, $$\sigma^{65} = \sigma^5 = (1\ 6\ 11\ 4\ 9\ 2\ 7\ 12\ 5\ 10\ 3\ 8).$$ $626 = 2 \pmod{12}$, so we have $$\sigma^{626} = \sigma^2 = (1\ 3\ 5\ 7\ 9\ 11)(2\ 4\ 6\ 8\ 12).$$ $1195 = 7 \pmod{12}$, so we have $$\sigma^{1195} = \sigma^7 = (1\ 8\ 3\ 10\ 5\ 12\ 7\ 2\ 9\ 4\ 11\ 6).$$ $-6 = 6 \pmod{12}$, so we have $$\sigma^{-6} = \sigma^6 = (1\ 7)(2\ 8)(3\ 9)(4\ 10)(5\ 11)(6\ 12).$$ $-81 = 3 \pmod{12}$, so we have $$\sigma^{-81} = \sigma^3 = (1\ 4\ 7\ 10)(2\ 5\ 8\ 11)(3\ 6\ 9\ 12).$$ $-570 = 6 \pmod{12}$, so we have $$\sigma^{-570} = \sigma^6 = (1\ 7)(2\ 8)(3\ 9)(4\ 10)(5\ 11)(6\ 12).$$ $-1211 = 1 \pmod{12}$, so we have $\sigma^{-1211} = \sigma$.
