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For some reason I'm finding permutation cycles to be strange and hard to deal with.

Let $\alpha$ and $\beta$ be cycles of odd length (not disjoint). Prove that if $\alpha^2 = \beta^2$, then $\alpha = \beta$.

I don't see what the odd lengths of $\alpha$ and $\beta$ have to do with anything. I see that they are even permutations. I see that $\alpha^2 = \beta^2 \implies \alpha^2\beta^{-2} = \varepsilon$. I don't think that helps. I don't know. The book went over the proof that a permutation can't be both odd and even, but I don't see anything in the proof that helps.

If $\alpha = (a_1, a_2, \dots, a_s)$, then $\alpha^2 = (a_1, a_3, \dots, a_s, a_2, \dots, a_{s-1})$. If $\beta = (b_1, b_2, \dots, b_r)$, then $\beta^2 = (b_1, b_3, \dots, b_r, b_2, \dots, b_{r-1})$. I don't see how to match up the elements of $\alpha$ and $\beta$ if I'm using different letters, but I can't use the same letters because I don't know which ones are the same or different. And maybe they are the same, but what if one of the cycles is rotated and starts on a different element? How do you deal with things like this?


Edit: Is the following proof sound?

The only way for $\alpha^2$ to be equal to $\beta^2$ while $\alpha \neq \beta$ is for $\alpha^2$ and $\beta^2$ to both be equal to $\varepsilon$. But the only way for that to happen is if the exponent, 2, is equal to a multiple of the length of the cycle. But since the lengths of $\alpha$ and $\beta$ are odd, their lengths have to be at least 3, since if they were length 1, they would be the same as length 0. So $\alpha^2$ and $\beta^2$ cannot be equal to $\varepsilon$, so $\alpha$ must be equal to $\beta$.

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Hint: If $\alpha$ is an $n$-cycle and $k$ is relatively prime to $n$, then $\alpha^k$ determines $\alpha$. Prove this in general, then set $k=2$.

If you didn't have an assumption of odd length, two different transpositions would be a counterexample -- then $\alpha^2$ and $\beta^2$ are both the identity permutation.

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  • $\begingroup$ Is the following proof sound? The only way for $\alpha^2$ to be equal to $\beta^2$ while $\alpha \neq \beta$ is for $\alpha^2$ and $\beta^2$ to both be equal to $\varepsilon$. But the only way for that to happen is if the exponent, 2, is equal to a multiple of the length of the cycle. But since the lengths of $\alpha$ and $\beta$ are odd, their lengths have to be at least 3, since if they were length 1, they would be the same as length 0. So $\alpha^2$ and $\beta^2$ cannot be equal to $\varepsilon$, so $\alpha$ must be equal to $\beta$. $\endgroup$ – Matt Gregory Feb 8 '12 at 15:27
  • $\begingroup$ @Matt: The first sentence lacks a justification -- and is actually not true for even $n$; consider (1 2 3 4) and (1 4 3 2). Was my hint not clear? It would lead to a proof of the form: "Given any odd $n$ there is an $m$ such that $(\alpha^2)^m=\alpha$ for every $n$-cycle $\alpha$, namely blah blah blah. Therefore, if $\alpha^2=\beta^2$, then $\alpha=(\alpha^2)^m=(\beta^2)^m=\beta$." $\endgroup$ – hmakholm left over Monica Feb 8 '12 at 15:37
  • $\begingroup$ But $(1 2 3 4)^2 = \varepsilon$ and $(1 4 3 2)^2 = \varepsilon$? No, I'm sorry, I didn't understand your hint. I'm not sure why. I need to think about what you just wrote in order to understand it. This stuff is not clear to me at all. $\endgroup$ – Matt Gregory Feb 8 '12 at 15:48
  • $\begingroup$ @Matt, um no. $(1\;2\;3\;4)^2=(1\;3)(2\;4)=(1\;4\;3\;2)^2$. Therefore they are a counterexample to your claim that $\alpha^2=\beta^2$ only when $\alpha^2=\varepsilon$. (I'm assuming that $\varepsilon$ is your notation for the identity permutation). $\endgroup$ – hmakholm left over Monica Feb 8 '12 at 15:50
  • $\begingroup$ Oh sorry! You're right, I was reading that wrong. Yeah, I meant $\varepsilon$ as the identity permutation. $\endgroup$ – Matt Gregory Feb 8 '12 at 16:02
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Note that they are not disjoint. Pick one of the common elements, call it $x$. Now see what happens to it under both $\alpha^2$ and $\beta^2$

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  • $\begingroup$ It is still true if they are disjoint, though. The parenthesis in the exercise probably ought to have said "not necessarily disjoint". $\endgroup$ – hmakholm left over Monica Feb 7 '12 at 19:22

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