# No permutation of $S_4$ satisfying the expression.

Q: Show that there does not exist a permutation $\sigma \in S_4$ satisfying $(1 2)(3 4)\sigma = \sigma(1 2 3 4)$.

I think, there must be an easy way around or do I have to show the results for every permutation of $S_4$?

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There is an easy way around. Hint: Think about odd and even permutations.

Or another way to do it would be to observe that if such $\sigma$ existed, you would have $\sigma^{-1} (12)(34) \sigma = (1234)$, and perhaps you know something about conjugation and cycle structure that tells you that can't happen. But the odd and even thing is simpler.

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+1: Always very nice, when somebody comes up with a simpler than expected solution! – Jyrki Lahtonen Apr 29 '12 at 11:02
@Jyrki: Thanks. I've been teaching this stuff recently, which helps. – Tara B Apr 29 '12 at 11:03
I agree with Jyrki as above. +1 – Babak S. Mar 30 '13 at 7:08

$(12)(34) = \sigma^{-1} (1234)\sigma$. You should know that $\sigma^{-1} (1234)\sigma=(\sigma(1)\sigma(2)\sigma(3)\sigma(4))$, so the right hand is cycle of length 4 but the left hand is 2 cycle of length 2. This is contradiction.

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Besides to theoretical formal @Tara's approach and by using GAP as follows, you can see that we achieve nothing in GAP's environment:

for alpha in SymmetricGroup(4) do if(alpha^(-1)(1,2)(3,4)alpha=(1,2,3,4)) then Print(alpha,"\n"); fi; od;

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$\ddot \smile \;\;+1\;\;$ – amWhy Mar 30 '13 at 18:43