# Determine whether the following elements of $S_6$ are of the form $\sigma^2$.

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 $$\sigma^2$$.]

• (c) is odd, not even. – Theo Bendit Jan 10 at 1:55
• @TheoBendit Oh, yes. You're right. I'm not sure how I made that mistake. I'll edit to reflect this. – thisisourconcerndude Jan 10 at 1:57
• 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? – Theo Bendit Jan 10 at 1:59
• Any permutation of odd order is a square. – Lord Shark the Unknown Jan 10 at 4:14
• Hint: if $\tau^5=id$ then $\tau=\tau^6=(\tau^3)^2$. – bof Jan 10 at 13:59

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.
$$(12345)$$ 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.