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Prove that there is no permutation $\sigma$ such that $\sigma(123)\sigma^{-1}=(124)(567)$.

Cycle $(123),(124),$ and $(567)$ has order $3$ so if the equation $\sigma(123)\sigma^{-1}=(124)(567)$ is powered by 3, I get identity in both sides. I don't have idea to prove this.

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$\sigma\tau\sigma^{-1}$ is a permutation with the same cycle structure as $\tau$ that is obtained by applying $\sigma$ to the cycle notation of $\tau$. The result follows because $(123)$ and $(124)(567)$ have different cycle structures.

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  • $\begingroup$ What do you mean of applying $\sigma$ to the cycle notation of $\tau$ ? $\endgroup$ – james brown Mar 30 '16 at 16:01
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    $\begingroup$ @jamesbrown joriki means that you have the equality $\sigma (123)\sigma^{-1}=(\sigma(1)\sigma(2)\sigma(3))$. $\endgroup$ – Arnaud D. Mar 30 '16 at 16:04
  • $\begingroup$ Thank you joriki and arnaud $\endgroup$ – james brown Mar 30 '16 at 16:14
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Here's a representation-theoretic approach to this problem which uses somewhat of the same fundamental logic of joriki's answer (which is centered on the invariants of the permutations). Permutation matrices and permutations are effectively the same object: permutation matrices just permute the standard basis vectors. The permutation $(123)$ corresponds to the matrix

$$ P_{(123)} = \left(\begin{array}{ccccccc} 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right). $$

The permutation $(124)(567)$ corresponds to the matrix

$$ P_{(124)(567)} = \left(\begin{array}{ccccccc} 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right). $$

Suppose then that such a permutation (matrix) $\sigma$ existed, then we would have $\sigma P_{(123)} \sigma^{-1} = P_{(124)(567)}$. Taking a trace, we have that

$$ \operatorname{tr}(\sigma P_{123}\sigma^{-1}) = \operatorname{tr}(P_{(124)(567)}). $$

Since traces are cyclic, we have

$$ \operatorname{tr}(\sigma P_{(123)}\sigma^{-1}) = \operatorname{tr}(P_{(123)}\sigma^{-1}\sigma) = \operatorname{tr}(P_{(123)}).$$

But note that $\operatorname{tr}(P_{(123)}) = 4$ whereas $\operatorname{tr}(P_{(124)(567)}) = 1$, thus no such $\sigma$ can exist.

The reason I chose to use the trace is manyfold, but the main interpretation I like is that the trace (when it comes to permutations) counts the number of fixed points. If two permutations do not have the same number of fixed points, they cannot be conjugates of each other as the above shows.

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