Permutation cycles in Jacobson's Basic Algebra I.

Question

Nathan Jacobson's Basic Algebra I Second Edition, Section 1.6 Cycle Decompositions of Permutations, page 51, exercise 4 says:

Show that if $\alpha$ is any permutation then

$$\alpha (i_1 i_2 \cdots i_r) \alpha^{-1} = (\alpha(i_1)\alpha(i_2) \cdots \alpha(i_r)).$$

This makes no sense. Suppose $\alpha = (12)$ and $(i_1 i_2 \cdots i_r) = (12345)$. Then

$$\alpha (i_1 i_2 \cdots i_r) \alpha^{-1} = (12)(12345)(12) = (31452)$$

but

$$(\alpha(i_1)\alpha(i_2) \cdots \alpha(i_r)) = (\alpha(1)\alpha(2)\cdots \alpha(5)) = (21345).$$

I really dislike this text, because it seems he's constantly abusing notation. Can anyone see what he's trying to say?

Answer 1

You computed the product incorrectly. In general, note that $$\alpha^{-1}(\alpha(i_1))=i_1,\;\;(i_1\cdots i_r)i_1=i_2,\;\;\alpha(i_2)=\alpha(i_2)$$ Therefore the image of the element $\alpha(i_1)$ under the permutation $\alpha (i_1\cdots i_n)\alpha^{-1}$ is the element $\alpha(i_2)$. Similarly, we have $$\begin{align}\alpha(i_2)\mapsto &\alpha(i_3) \\ \alpha(i_3)\mapsto &\alpha(i_4) \\ \vdots \\ \alpha(i_r)\mapsto &\alpha(i_1)\end{align}$$ Therefore one of the cycles for the permutation is $$(\alpha(i_1)\alpha(i_2) \cdots \alpha(i_r))$$ Consider any element $x$ not in that cycle. Then $\alpha^{-1}(x)\ne i_j$ for any $j$, because that would contradict the fact that permutations are bijections and hence one-one. Thus the image of $x$ will be $$\alpha(\alpha^{-1}(x))=x$$ All other elements are therefore fixed, and the equality follows. Note that this is a conjugation.