Permutations and Permutation Groups

Yesterday I had a meeting with Dr. Prudhom in which we discussed a few exercises from the chapter on permutations and permutation groups—the next important type of group after the dihedral groups and cyclic groups. One key topic that is necessary to understand permutation groups is cycle notation for permutations, which I’ve seen before but always struggled with. This time around, I spent a fair amount of time working specifically on cycle notation just to be sure that I understood it, because it’s the notation that is used whenever permutations are involved.

The exercises that we discussed in the meeting were the following:

For 50, we ended up trying out some examples and realized that whenever you conjugate a permutation by a 2-cycle, there are two possibilities. Either the two permutations are disjoint, in which case they commute and so you get that the 2-cycle just cancels with itself. If, however, the permutations are not disjoint, then we worked through the example as follows.

We can write the situation like (ab)(….a…b….)(ab), where the middle permutation is a t-cycle involving a and b at some point. When we work through this, we see that the a gets mapped to the element right after b in the t-cycle and b gets mapped to the element right after a in the t-cycle. Furthermore, the element right before a in the t-cycle gets mapped to b, and the element right before b in the t-cycle gets mapped to a. So this means that a and b have their positions swapped in the t-cycle when we conjugate by a 2-cycle, that is, (ab)(….a…b….)(ab)=(….b…a….). This is plainly still a t-cycle, so we’re done. (There’s technically also the case where a and b are adjacent in the t-cycle, but this doesn’t change anything. And in the case where only a or only b appears in the t-cycle, but not both, then the result is just replacing a with b in the t-cycle and holding the other one fixed. In both scenarios, the result is still just a t-cycle.)

51 follows fairly quickly from 50: we can assume that the cycles that form b are all disjoint, so they commute. Then we can write a as a product of (not necessarily disjoint, in this case) 2-cycles. These 2-cycles will commute with any cycles of b with which they are disjoint, so we can move them to be adjacent to the cycles of b with which they share elements. Then we can apply the previous exercise multiple times in order to collapse all of the 2-cycles into the cycle of b with which they share an element. In the case that there’s a 2-cycle from a that shares one element with one cycle of b and one element with another cycle of b, then we have something that looks like (ab)(….a…..)(….b….)(ab). This sends b to the element after a, a to the element right after b, the element right before a to the b and the one right before b to a. In other words, (ab)(….a….)(….b…..)(ab)=(….b….)(….a….). This doesn’t change the length of either of the cycles. Thus we can collapse all of the 2-cycles of a into the cycles of b without changing the length of the cycles of b, which is sufficient to complete the exercise.

And 52 follows from 51: we just showed that conjugating a permutation by another permutation doesn’t change the length of the disjoint cycles that make up the inner permutation. Therefore, conjugation doesn’t change the parity of the inner permutation either. (I also found an alternate proof that uses a slightly more abstract approach here—see Corollary 2.11 and the preceding theorem. I like this approach as a cool “other method” to approaching the problem.)

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