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The Problem:

Let $S_6$ be the symmetric group on six letters. Determine whether the following elements of $S_6$ are squares (i.e., of the form $\sigma^2$).

(a) $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$.

(b) $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$.

(c) $(1 \hspace{1mm} 2 \hspace{1mm} 3)(4 \hspace{1mm} 5)$.

My Progress:

(a) is clear. $\sigma^2$ must be an even permutation, and $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$ is an odd permutation. Therefore there can be no $\sigma \in S_6$ such that $\sigma^2 = (1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4)$.

(b) is not so clear to me. $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$ is, indeed, an even permutation, so the parity doesn't help me here. I've pretty much exhausted all of my tools. I'm basically at the trial-and-error point now (e.g., trying to write $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$ as the square of different permutations) which is getting me nowhere fast. Basically, I'm not sure what else can be said about an element of the form $\sigma^2$ other than the fact that it's even.

(c) is giving me the same issues as (b). [EDIT: Nevermind. $(1 \hspace{1mm} 2 \hspace{1mm} 3)(4 \hspace{1mm} 5)$ is odd since it has an odd number of cycles of even length; so it cannot be of the form $\sigma^2$.]

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    $\begingroup$ (c) is odd, not even. $\endgroup$ – Theo Bendit Jan 10 at 1:55
  • $\begingroup$ @TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this. $\endgroup$ – thisisourconcerndude Jan 10 at 1:57
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    $\begingroup$ For (b), notice that the permutation is essentially adding $1$ modulo $5$. Maybe look at adding $1/2$ (by which I mean, the multiplicative inverse of $2$) modulo $5$, and see if that becomes a valid square root? $\endgroup$ – Theo Bendit Jan 10 at 1:59
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    $\begingroup$ Any permutation of odd order is a square. $\endgroup$ – Lord Shark the Unknown Jan 10 at 4:14
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    $\begingroup$ Hint: if $\tau^5=id$ then $\tau=\tau^6=(\tau^3)^2$. $\endgroup$ – bof Jan 10 at 13:59
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Hint: If $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5) = \sigma^2$ , then $\sigma^{10}$ is the identity, so its order must divide 10. Also, $\sigma$ can be written as a product of disjoint cycles. What would its order be as a function of the orders of those cycles? What would the orders of those cycles have to be? Is it possible for a product of disjoint cycles of those orders to have its square equal to $(1 \hspace{1mm} 2 \hspace{1mm} 3 \hspace{1mm} 4 \hspace{1mm} 5)$ ?

Edit: In light of the observation of Lord shark the unknown's in his first comment above (which follows from the fact that if $p^{2n+1} = I$ then $p = p^{2n+2} = \left(p^{n+1}\right)^2$ ), trying to answer the questions I posed above turns out to be a rather clumsy way of tackling the problem.

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$(12345)$ generates a cyclic subgroup $H$ of order $5$. Since the squares of an abelian group are a subgroup, the squares $S$ in $H$ are a subgroup of $H$. Since $(12345)^2 \in S$, we have $S \neq \{id\}$. Since $5$ is prime, it follows by lagrange’s theorem that $|S|=5$, so $H=S$, and we are done.

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