Another Cycle Decomposition Question

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Mar 14th 2010, 04:45 PM
craig

Another Cycle Decomposition Question

In $S_4$ let $\sigma$ be a cycle of length 4, and $\tau$ be a transposition.

Show $\sigma \tau$ is even, and briefly explain why $\sigma \tau\neq id$

Let $\sigma = (abcd)$ , then the decomposition would be $(ab)(ac)(ad)$ , and let $\tau = (ef)$ . This means that $\sigma \tau$ would be $(ab)(ac)(ad)(ef)$ , which is even. Think I've gone about this in the correct way?

For any transposition $(ij)$ , $(ij)^2 = id$ , this means that $(ij) = (ij)$ .

For our transpositions above, $(ab)\neq(ac)\neq(ad)$ , so even if $(ef) = (ab) or (ac) or (ad)$ , it would still not equal the identity transposition. Would this be enough to briefly explain?

I managed to attempt the first couple of sections of the question, but this last part has got me very confused:

Deduce that $\sigma \tau$ is either a cycle of length 3, or a product of 2 disjoint transpositions. Conclude that $\sigma = \gamma \lambda \tau$ for some not necessarily disjoint permutations $\gamma, \lambda \in S_4$

Would greatly appreciate some help in how to go about this problem.

Thanks in advance
Mar 14th 2010, 09:35 PM
Tinyboss

You're definitely on the right track. In general, an n-cycle is even for odd n, and vice-versa, so the first result is immediate. For the second part, I'd just note that the 4-cycle moves 4 elements, and any transposition can only restore at most 2 to their original place.
Mar 15th 2010, 01:26 AM
craig

Quote:

Originally Posted by Tinyboss

You're definitely on the right track. In general, an n-cycle is even for odd n, and vice-versa, so the first result is immediate. For the second part, I'd just note that the 4-cycle moves 4 elements, and any transposition can only restore at most 2 to their original place.

Thanks for the reply. That's a good idea about only restoring 2, didn't think of that.

Any idea how to approach the last section?
Mar 15th 2010, 12:34 PM
rubic

cycle decompositoions

we know there is 12 even permutations in s4, we also know that id is not one of the solutions. (as you have said). solution have to be one of the other 11 even permutation, which are 8x3-cycles and 2xproduct of two disjoint transpositions.

hope it will help
Mar 15th 2010, 03:53 PM
craig

Yes that helps, cheers for the reply!