My Math Forum Permutations of a finite set
 User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

 June 9th, 2014, 08:46 AM #1 Member   Joined: May 2014 From: None of your business Posts: 31 Thanks: 0 Permutations of a finite set I'm trying to figure out this problem: Let alpha and Beta be cycles of odd length (not disjoint). Prove that if alpha squared equals beta squared, then alpha equals beta. What I know thus far is that... since l is an odd length, alpha squared and beta squared are cycles, and not a product of disjoint cycles. Both alpha and beta can be expressed as an even number of transpositions. I also know that I can express alpha as a product of the cycles alpha1 gamma alpha2, and that this product is associative. Since the problem states that alpha and beta are not disjoint, then I can reason that there is ai in the alpha cycle that is equal to some bj in the beta cycle. Now I tried expressing beta squared as a product of beta1 (bj bj+2) beta2, and since it is associative, rewrote it as (bj bj+2) beta1 beta2, then multipied by (bj+2 bj) on the left, also multiplying alpha squared by (bj+2 bj) on the left. With examples, I continued in this fashion until I had the identity permuation on the right hand side of the equation. Now I'm stuck. I'm thinking that I only showed that the product of beta squared inverse and alpha squared is the identity (not really illuminating). Please help.
 June 9th, 2014, 04:41 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Since $\alpha,\beta$ are not disjoint, they share at least one element in their respective cycles. Hence we may write: $\alpha = (a_1\ a_2\ \dots\ a_n)$ $\beta = (a_1\ b_2\ \dots\ b_m)$ Thus: $\alpha^2 = (a_1\ a_3\ \dots a_n\ a_2\ a_4\ \dots a_{n-1})$ (since $n$ is odd) $\beta^2 = (a_1\ b_3\ \dots b_m\ b_2\ b_4\ \dots b_{m-1})$ (since $m$ is odd). Since $\alpha^2 = \beta^2$ it follows that $m = n$, and $b_j = a_j$. To see why we have the stipulation that $n,m$ be odd, consider: (1 2) and (1 3). Then (1 2)(1 2) = (1 3)(1 3) = e, but (1 2) does not equal (1 3). ********** The trick here is to see it does not matter which element of a cycle we start it with: (1 2 3) = (2 3 1) = (3 1 2), and that two functions $f,g$ are equal iff their domains are equal, their co-domains are equal, and they take the same values in the co-domain for every value in the domain. (Recall that the cycle $(a\ b\ c)$ is the function: $f:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ with: $f(a) = b$ $f(b) = c$ $f(c) = a$ $f(k) = k$, for all other elements besides $a,b,c$). So once we prove that $a_3 = \alpha^2(a_1) = \beta^2(a_1) = b_3$, we compare $a_5 = \alpha^2(a_3) = \beta^2(a_3) = \beta^2(b_3) = b_5$, and so on, until we come "full cycle". Thanks from Awesomo

 Tags finite, permutations, set

,

,

,

### express alpha beta as product of two cycles

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post princevisram Advanced Statistics 6 April 17th, 2012 07:41 AM Hooman Real Analysis 0 February 20th, 2012 09:34 PM koolkidx45 Advanced Statistics 1 February 20th, 2012 11:09 AM bureddogs44 Advanced Statistics 3 September 18th, 2011 12:58 PM sinanbasar Advanced Statistics 0 April 16th, 2010 10:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top