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I'm trying to show that there is no $g$ such that $g^{-1}(1,2,3)g = (1,3)(5,7,8).$ I am having some general issues figuring out where to move with this problem, since it seems difficult to figure out how to make a contradictory statement to prove the lack of existence. If $g$ were to exist, I know that in $g$, $5$ must map to $1,2,$ or $3$ because if it did not, it would not be able to map to $7$ in $g^{-1}(1,2,3)g.$ However, I am having problems figuring out where this leads to a contradictory implication cycle.

Melinda
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1 Answers1

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Easy if you know the following result.

Theorem. Cycle type is invariant under conjugation.

Here you have a $3$-cycle $(1\ 2\ 3)$, and so $g^{-1}(1\ 2\ 3)g$ must also be a $3$-cycle. It can't be the product of a $3$-cycle and a $2$-cycle.

David
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