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I have the three permutations $$a=(1\;3\;4\;8),\quad b=(2\;3\;5\;7),\quad c=(4\;3\;2\;8)$$ and I have to find all $x$ satisfying $$axb=c.$$ I have found one solution (I hope it's good): $$x=(7\;5\;1\;8\;3\;2)$$ Now my problem is, that the task says I have to find every solution.

Is there any other solution?

Spenser
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Seasharp
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1 Answers1

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Recall that a permutation is a bijective function, and is hence invertible. Therefore we can isolate $x$ by left multiplying each side of the equation by $a^{-1}$ and right-multiplying each side of the equation by $b^{-1}$ to get

So $$axb = c \iff xb = a^{-1} c \iff x = a^{-1} c b^{-1}$$

Now, we need the inverses of $a$, $b$:$$a^{-1} = (8431) = (1843),\quad b^{-1} = (7532) = (2753)$$

amWhy
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  • But the task says I have to give all the solutions for x in S8. So there isn't any other solution? – Seasharp Dec 16 '14 at 15:34
  • The solution is unique. The product (composition) of permutation is a permutation, and as such it is uniquely determined by the factors. – amWhy Dec 16 '14 at 15:38
  • No, it is the only solution. Every step above is if and only if. – mpv Dec 16 '14 at 15:40