Calculate $\alpha \sigma \alpha ^{-1}$

Question

Exercise with permutations $$\alpha = (12)(135), \sigma = (1579)$$ Then $$\alpha \sigma \alpha ^{-1}= (12)(135)(1579)(531)(21)$$

In this case, I'm using permutation from right to left. I have the following steps $$1\to 2, 2\to2, 2\to2, 2\to2, 2\to 1$$ $$2\to1, 1\to3,3\to3, 3\to1, 1\to 2$$ $$3\to1, 1\to5,5\to3, 3\to 3$$ $4$ isn't here then $4\to 4$, also $6\to 6$, $8\to 8$ $$5\to5, 5\to1, 1\to5,5\to3, 3\to 3$$ $$7\to7, 7\to7,7\to7, 9\to 9, 9\to 9$$ $$9\to9, 9\to9,9\to1, 1\to 5, 5\to 5$$ $$5\to1, 1\to5,5\to3, 3\to 3$$ Then $$5\to 3, 7\to 9, 9\to 5$$ but this isn't a cycle.

Answer 1

You have $\alpha \sigma \alpha ^{-1} = (1\,\,2)(1\,\,3\,\,5)(1\,\,5\,\,7\,\,9)(5\,\,3\,\,1)(2\,\,1)$. Now $$ 1 \mapsto 2 \mapsto 2 \mapsto 2 \mapsto 2 \mapsto 1 $$ so $1 \mapsto 1$. Similarly $$ 2 \mapsto 1 \mapsto 5 \mapsto 7 \mapsto 7 \mapsto 7 $$ so $2 \mapsto 7$. By doing the rest of the permutations, we see that $$ 3 \mapsto 2, \,\,4 \mapsto 4, \,\,5 \mapsto 5, \,\,6 \mapsto 6, \,\,7 \mapsto 9, \,\,8 \mapsto 8, \,\,9 \mapsto 3, $$ so to summarise we have $$ (1)(2\,\,7\,\,9\,\,3)(4)(5)(6)(8) = (2\,\,7\,\,9\,\,3). $$

Answer 2

Hints:

• $\alpha(i_1\:i_2\:\dots i_r)\alpha^{-1}=\bigl(\alpha(i_1)\:\alpha(i_2)\:\dots\:\alpha(i_r)\bigr)$.
• $\alpha = (1\:2)(1\:3\:5)=(1\:3\:5\:2)$.