Show that there is no permutation $\sigma$ in $S_4$ that satisfies $(1\,2\,3\,4) \circ \sigma = \sigma \circ (1\,2) (3\,4) $

Question

My attempt is as below.

$(1\,2\,3\,4)=(1\,4)(1\,3)(1\,2)$ is an odd permutation.

$(1\,2)(3\,4)$ is an even permutation.

$\sigma$ can be either even or odd but not both.

Suppose $\sigma$ is even then $(1\,2\,3\,4) \circ \sigma$ is odd and $\sigma \circ (1\,2) (3\,4)$ is even. Which implies that $(1\,2\,3\,4) \circ \sigma \neq \sigma \circ (1\,2) (3\,4) $.

On the other hand suppose $\sigma$ is odd then $(1\,2\,3\,4) \circ \sigma$ is even and $\sigma \circ (1\,2) (3\,4)$ is odd. Which again means that $(1\,2\,3\,4) \circ \sigma \neq \sigma \circ (1\,2) (3\,4) $.

Please let me know if this is the correct way to tackle the mentioned problem. Thanks

Answer 1

Your argument looks valid. Whether $\sigma$ is even or odd, $\sigma^{-1}(1234)\sigma$ is odd, but $(12)(34)$ is even.

Another argument is that conjugation preserves cycle structure,

so we know that $\sigma^{-1} (1234) \sigma\ne(12)(34)$,

because $(1234)$ and $(12)(34)$ have different cycle structure.

Answer 2

Yet another argument is that conjugation preserves the order of the element in question: $(1234)$ has order $4$, while $(12)(34)$ has order $2$.