Finding permutation with a condition

Question

Let $a$ be a permutation in $S_6$. I'm asked whether there is an $a$ so that $a^2 = (123)(456)$
I'm quite confused about where to start. I do know the $a$ must consist of $3$ elements (right?).

How do I proceed?

Answer 1

Hints:

• What order can $a$ have?
• What kind of permutations have that order? (Think cycle structure.)
• Now figure out what squaring does to those permutations.

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More complete answer.

The order of $(123)(456)$ is $3$, that is $(a^2)^3=e$, or $a^6=e$. Therefore the order of $a$ must be one of the divisors of $6$. It can't be $1$, and it can't be $2$ (since then $a^2$ would be $e$). Can it be $3$ or $6$?

If it has order $3$ then it has cycle structure either $(xyz)$ or $(xyz)(pqr)$. In the former case, the square of a $3$-cycle is another $3$-cycle, so cannot be a product of disjoint $3$-cycles. That leaves the possibility $a=(xyz)(pqr)$. Squaring yields $a^2=(xzy)(prq)=(123)(456)$. See what to do?

If it has order $6$ then $a=(\color{Blue}{q}\color{Green}{r}\color{Blue}{s}\color{Green}{t}\color{Blue}{u}\color{Green}{v})$ and $a^2=(\color{Blue}{qsu})(\color{Green}{rtv})=(123)(456)$. See what to do here?