In Pinter's "A Book of Abstract Algebra", Chapter 8 Exercise E2 asks the following:
Prove that any two cycles of the same length are conjugates of each other.
Pinter puts forth the the beginning of the solution:
"If $\alpha$ and $\beta$ are cycles of the same length, where $\alpha = (a_1 … a_s)$ and $\beta = (b_1 … b_s)$, let $\pi$ be the following permutation: $\pi (a_i)= b_i$ for $i = 1,…,s$ and $\pi(k) = k$ for $k\neq a_1,…,a_s, b_1,…,b_s$. Finally, let $\pi$ map distinct elements of $\{b_1,…,b_s\} – \{a_1,…,a_s\}$ to distinct elements of $\{a_1,…,a_s\} – \{b_1,…,b_s\}$".
The portion that I have bolded is the part that confuses…as I cannot see why this is necessary.
It looks to me like Pinter is specifying a property of $\pi$ that accounts for possible overlap in the two different cycles…so that "non-overlapped elements" only map to "non-overlapped elements", but I cannot see why this must be done.
From the first two properties of $\pi$, namely:
- $\pi (a_i)= b_i$ for $i = 1,…,s$
and
- $\pi(k) = k$ for $k\neq a_1,…,a_s, b_1,…,b_s$
it seems like all relevant features of $\pi$ have been sufficiently described. Why must the last property be included as well?
Best Answer
All relevant features of $\pi$ have yet to be defined: one hasn't yet said what $\pi$ does to elements of $\{b_1,\dots, b_s\}-\{a_1,\dots, a_s\}$.
Furthermore, the elements of $\{a_1,\dots, a_s\}-\{b_1,\dots, b_s\}$ haven't been "hit", or mapped to yet.
Thus the extra condition.