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I know how to prove the following using parity ie breaking down the following into its transpositions and then showing it is impossible due to the parity but was wondering if anyone had any ideas on how to show it using order.

I was thinking it would be easy in the first case to show that $\sigma$ would have to be of order $8$ but I am not too sure on how to show that.

Prove there does not exist $\sigma^2=(1, 2, 3, 4)$ for $\sigma \in S_7$.

And more generally prove for all $n\ge 4$ there does not exist $\sigma \in S_n, \sigma^2=(1, 2, 3, 4)$.

Shaun
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user690808
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    Hint: What order would such an element $\sigma$ have to have? – ahulpke May 10 '21 at 22:06
  • I'm assuming it is 8 for the first case but not sure on how to actually prove that – user690808 May 10 '21 at 22:07
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    You know $\sigma$ would necessarily have order $8$, because $\sigma^2$ has order $4$. Thus, $\sigma^8 = (\sigma^2)^4 = e$, and the order divides $8$. Since $\sigma^4\neq e$, then the order must be $8$. – Arturo Magidin May 10 '21 at 22:09
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    Do you know the formula that if $g$ has order $n$, its $e$-th power has order $n/gcd(n,e)$ ? – ahulpke May 10 '21 at 22:09
  • How would you generalize this to the general case then? – user690808 May 10 '21 at 22:13
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    What "general case"? $n\geq 4$? What does an element of order $8$ in $S_n$ look like? – Arturo Magidin May 10 '21 at 22:14
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    It could be some 8 length permutation product any number of 2-length or 4-length permutations? – user690808 May 10 '21 at 22:15
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    And what happens when you square a cycle of length 8? – Arturo Magidin May 10 '21 at 22:17
  • Do you get two cycles of length 4? – user690808 May 10 '21 at 22:18
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    That's what happens when I do it. But I asked what happens when you do it. – Arturo Magidin May 10 '21 at 22:19
  • I would call this question a duplicate of this mother lode. I won't cast my vote as I have earned two applicable dupehammers, but don't want to use them without feedback from others. If anyone thinks this is, indeed, a close enough duplicate, please ping me. If you, like me, have the dupehammer, but are uncertain, feel free to act with my implied support. – Jyrki Lahtonen Aug 18 '24 at 07:39
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    There was a really nice answer posted here recently, which IMHO was the most elegant way to solve this: $(1234)$ is an odd permutation, so it can't be a square. This works in all $S_n$, not just the $n=7$ case. I'm not sure why the answer was deleted, but I wanted to record that nice argument here for posterity. – Steve D Aug 19 '24 at 04:44

2 Answers2

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Hint: if $\sigma^2=(1234)$, then $\sigma^4=(13)(24)$ and it follows that $\sigma$ has order $8$. But $S_7$ does not have any elements of order $8$.

Nicky Hekster
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Well, I'm not sure if this would be of interest to you, but we can show the following: If a permutation $\pi$ of order 4 in $S_n$ for any $n \in \mathbb{N}$ that has a 4-cycle $(x_1,x_2,x_3,x_4)$ can be written $\pi=\sigma^2$, then $\pi$ must have at least two 4 cycles. This may be of interest because it will show that a permutation $\pi$ with exactly one 4-cycle cannot be a square even for instances where $\pi$ is an even number of transpositions [i.e., a 4-cycle and then a disjoint 2-cycle].

Claim 1: Suppose that $\pi=\sigma^2$ for some permutation $\sigma$. The size of the orbit of element $x_1$ must be exactly 8 under $H\doteq \langle \sigma \rangle$, and so the only element that fixes $x_1$ is $e$.

Indeed, the orbit of $x_1$ must have exactly 4 or 8 elements under $H$. If the orbit of $x_1$ has exactly 4 elements then there is a subgroup of $H$ of order 2 that fixes $x_1$.However, the only subgroup of $H$ of order 2 is $\{1,\sigma^2\}$ and $\sigma^2$ does not fix $x_1$. $\surd$

In light of Claim 1, the number of elements in the orbit $O_H(x_1)$ of $x_1$ under $H$ must be $|H|$, and for each $x_j \in O_H(x_1)$, the only element that fixes $x_j$ is $e$. So for each of the 8 elements $x_j \in O_H(x_1)$ and for any subgroup $H'$ of $H$, the number of elements in the orbit $O_{H'}(x_j)$ of $x_j$ under $H'$ must be exactly $|H'|$. So if $\pi=\sigma^2$ for some $\sigma$, then taking $H'=\langle \sigma^2 \rangle$ $=\langle \pi \rangle$; then $|H'|$ has exactly 4 elements, so for each of the 8 such elements $x_j$, the number of elements in the orbit $O_{H'}(x_j)$ of $x_j$ must have exactly 4 elements.

So it follows from this that each of the 8 elements $x_j \in O_H(x_1)$ is in a 4-cycle of $\pi$, so $\pi$ must have 2 4-cycles.

One can extend this line of reasoning to show that in fact $\pi$ must have precisely an even number of 4 cycles to be a square $\pi=\sigma^2$, where the elements in a 4-cycle of $\pi$ are precisely the elements in an 8-cycle of $\sigma$.

In fact, you could use this line of reasoning to show that a permutation $\pi$ in $S_n$ is a square iff for each $k \in \mathbb{N}$ there are an even number of $2k$-cycles.

Mike
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