# Permutations in Abstract Algebra

Can an odd permutation be conjugate to an even one? I want to say no, because odd/even permutations have defined cycle types, and conjugation always preserves the cycle type. Is this the correct approach? If not, could someone give a counterexample? And if so, how would I make this rigorous?

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no, because conjugated permutations have the same cycle type, and the cycle type determines whether the permutation is even or odd –  Camilo Arosemena Dec 16 '12 at 2:55
Even simpler, look at the sign: conjugate permutations have the same sign, but even and odd permutations do not. –  KCd Dec 16 '12 at 2:55

Given a permutation $\sigma$, you can write $\sigma$ as a product of transpositions. This product is not unique, but the fact is that given $\sigma$, the number of factors in the product will always be either even or odd. We say that the sign of the permutation $\text{sgn}(\sigma)$ is $1$ if the number of factors is even and $-1$ if the number of factors is odd.
The next fact is that $\text{sgn}(\sigma) = \text{sgn}(\sigma^{-1})$
The last fact is that $\text{sgn}(\sigma\tau) = \text{sgn}(\sigma)\text{sgn}(\tau)$.
So if two permutations are conjugates of each other, say: $\sigma_1 = \tau\sigma_2\tau^{-1}$, then \begin{align} \text{sgn}(\sigma_1) &= \text{sgn}(\tau\sigma_2\tau^{-1})\\ &= \text{sgn}(\tau)\text{sgn}(\sigma_2)\text{sgn}(\tau^{-1}) \\ &= \text{sgn}(\sigma_2)\text{sgn}(\tau)\text{sgn}(\tau) \\ &= \text{sgn}(\sigma_2)\text{sgn}(\tau)^2 \\ &= \text{sgn}(\sigma_2). \end{align} (So this agrees with what was said in the comments to the question above).