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I'm reading Beardon's Algebra and Geometry, I'm a little confused about this exercise:

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If $n$ is not divisible by $3$, then $\rho$ fixes some $k$. Suppose $n=5$, then we have the set $\{1,2,3,4,5\}$. As $\rho$ is a $3-$ cycle, we write it as - for example:

$$\begin{pmatrix} {1}&{2}&{3}\\ {2}&{3}&{1} \end{pmatrix}$$

Or

$$\begin{pmatrix} {1}&{2}&{3}&{4}&{5}\\ {2}&{3}&{1}&{4}&{5} \end{pmatrix}$$

Which amounts to the same thing. Now suppose we apply it to $12345$, then we will have $23145$ then it obviously will have $4,5$ fixed. But suppose $n=6$, we apply the same permutation to $123456$ and obtain $231456$ and it also happens that some integers are fixed, also if the permutation is a $3-$cycle, it obviously can't reach all the integers in the permutation. I may have understood something wrong.

Red Banana
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  • when $n=6$, $\rho$ could equal $(1,2,3)(4,5,6)$. – Angina Seng Sep 17 '17 at 10:25
  • Yes, but the exercise seems to ask me to prove that whenever $n$ is not divisible by $3$, then some $k$ is fixed and when it's a multiple of $3$, it is not. Is that it? – Red Banana Sep 17 '17 at 10:35
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    No. The question only asks about non-multiples of $3$. – Angina Seng Sep 17 '17 at 10:37
  • Yes, but wouldn't it be equivalent? I mean, doesn't it says that: $n$ not divisible by $3$ $\implies$ $\rho$ fixes some $k$? – Red Banana Sep 17 '17 at 10:40
  • The two statements are not equivalent. If $n$ is not divisible by $3$, then some point surely is fixed. If $n$ is divisible by $3$, then anything could happen. – Crostul Sep 17 '17 at 11:19
  • Your very first example shows that if $n$ is divisible by $3$, a permutation $\rho$ can satisfy $\rho^3=I$ while fixing no element: $(123)$ is such a cycle for $n=3$. Don't read in a statement what it isn't saying. – egreg Sep 17 '17 at 14:28
  • That is weird. If we have $a \implies b$, then isn't $¬b \implies ¬ a$? Or that statement isn't actually an implication? I'm used to seeing "if ... then..." as implications, but I'm aware that sometimes, it is possible that they aren't. So why it isn't in this case? – Red Banana Sep 17 '17 at 20:16

1 Answers1

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Write $\rho$ as a product of disjoint cycles: $\rho=\sigma_1\sigma_2\dots\sigma_k$. Suppose that $\rho^s=I$. Then, by the properties of disjoint cycles, $$ \sigma_1^s \sigma_2^s \dots \sigma_k^s = I $$ If $\sigma_i^s\ne I$, then there is $p\in\{1,\dots,n\}$ such that $q=\sigma_i^s(p)\ne p$. On the other hand, $q$ is not moved by any $\sigma_j^s$, because of disjointness. This is a contradiction to the product being the identity.

Therefore $\sigma_i^s=I$, for $i=1,\dots,k$. For $s=3$ this implies that the length of $\sigma_i$ is either $1$ or $3$.

Without loss of generality, we can assume $\sigma_i$ has length $3$ for $1\le i\le h$ and length $1$ for $h+1\le i\le k$. Then $$ n=3h+(k-h) $$ If $n$ is not divisible by $3$, we conclude that $k-h\ne0$, which means that $\sigma_k$ has length $1$ and at least one element is fixed by $\rho$: indeed, $k-h$ is the number of length $1$ cycles, that is, of fixed elements.

Nothing can be said about fixed elements in case $n$ is divisible by $3$. There can be fixed elements or not. For instance, if $n=6$, we can consider $$ \rho_1=(123)(4)(5)(6) \qquad\text{and}\qquad \rho_2=(123)(456) $$ The former has fixed elements, the latter hasn't. What one can say is that the number of fixed elements is also divisible by $3$, because, with the same notation as before, $k-h=n-3h$.

egreg
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  • There is also an interesting take given with a corollary in the book: If $d$ is the LCM of $q_1, \dots , q_m$ with $q_j$ being the length of the cycle $\rho_j$, then $\rho^d=I$. Now if $d=3$, the length of the cycles can only be $1$ or $3$. – Red Banana Sep 18 '17 at 08:50
  • @AskYourself You're using the implication the wrong way. – egreg Sep 18 '17 at 08:56
  • What you mean? $$ – Red Banana Sep 18 '17 at 11:15
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    @AskYourself The corollary says that if $d$ is the LCM, then $\rho^d=I$; proving that if $\rho^m=I$ then $d\mid m$ is not obvious. – egreg Sep 18 '17 at 11:58