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I have given the following permutation $$\sigma=\left(\begin{array}{rrr} 1&2&3&4&5&6&7&8&9 \\ 4&5&6&7&8&9&1&2&3 \end{array}\right)$$ and $I_1=\{1,4,7\},I_2=\{2,5,8\}, I_3=\{3,6,9\}$. I just know that $\sigma(I_i)=I_i$ for all $i$. Now we have given $\tau\in C_{S_9}(\sigma)$ the centralizer. I need to show that $\tau(I_i)\in \{I_1,I_2,I_3\}$.

We just know that $$\tau(I_1)=\tau(\sigma(I_1))=\sigma(\tau(I_1))$$. Now I only need to show that $\tau(I_1)\in \{I_1,I_2,I_3\}$. But I don't see why this needs to be true. Could someone help me?

angryavian
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user123234
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  • In these situations, cycle notation is very helpful. Do you know it? That's because cycle notation allows you very easily to compute $\tau^{-1} \sigma \tau$. – Robert Shore Nov 20 '21 at 22:17
  • but I don't know anything about $\tau$. So I don't see why this is helpful – user123234 Nov 20 '21 at 22:18
  • You know a lot about $\tau$. You know that $\tau^{-1} \sigma \tau = \sigma$. When you write the expression in cycle notation that gives you a strong constraint on $\tau$. It's easy to see this way, for instance, that there are exactly $162$ elements of the centralizer of $\sigma$. – Robert Shore Nov 20 '21 at 22:21
  • Sorry I don't see your point, and neither how to write $\tau$ in a cyclic notation – user123234 Nov 20 '21 at 22:24
  • Can you write $\sigma$ in cycle notation? And do you know how to calculate $\tau^{-1} \sigma \tau$ in general when $\sigma$ is written in cycle notation? – Robert Shore Nov 20 '21 at 22:26
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    Start with writing $\sigma$ rather than $\tau$ in cycle notation, and figure out how conjugation generally works on a permutation written in cycle notation. (In contrast to $\sigma$, it doesn't seem to be particularly illuminating to think of $\tau$ in terms of its cycle decomposition here). – Troposphere Nov 20 '21 at 22:26
  • Yes so I know that $\sigma=(147)(258)(369)$ and I know how in general one can calculate $\tau^{-1}\sigma \tau$ – user123234 Nov 20 '21 at 22:29
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    So how must $\tau^{-1}$ look if $$(1,4,7)(2,5,8)(3,6,9) = (\tau^{-1}(1);\tau^{-1}(4);\tau^{-1}(7))(\tau^{-1}(2);\tau^{-1}(5);\tau^{-1}(8))(\tau^{-1}(3);\tau^{-1}(6);\tau^{-1}(9)) $$ ? (It would probably be simpler to start out with $\tau\sigma\tau^{-1}=\sigma$ instead). – Troposphere Nov 20 '21 at 22:34
  • Sorry I don't see how you get to the left side, since for example I need to compute $\tau^{-1}(\sigma(\tau(1)))$ but I don't know $\tau(1)$ – user123234 Nov 20 '21 at 22:36
  • You said you know how to calculate $\tau^{-1}\sigma\tau$? If what I wrote doesn't match that knowledge, could you describe the procedure you do know? – Troposphere Nov 20 '21 at 22:37
  • So we learned that for example to compute $\tau^{-1}(\sigma(\tau(1))$ we take the element on the right of $\tau$ then the one on the right of $\sigma(\tau(1))$ and so on – user123234 Nov 20 '21 at 22:39
  • I'm not speaking about how to apply $\tau^{-1}\sigma\tau$ to a particular element, but how to write down the entire cycle decomposition for $\tau^{-1}\sigma\tau$ when you know the cycle decomposition of $\sigma$. Do you know, for example, that for a single cycle we have $$ \alpha(x_1,x_2,\ldots,x_n)\alpha^{-1} = (\alpha(x_1);\alpha(x_2);\ldots;\alpha(x_n))$$ – Troposphere Nov 20 '21 at 22:44
  • Yes we saw this in the lecture. But here we have three cycles? – user123234 Nov 20 '21 at 22:46
  • So I have looked at it and your computation from above makes sense. So I would say that $\tau^{-1}=\sigma$ is this correct? and thus $$\tau=(741)(852)(369)$$ and thus $\tau(I_i)\in {I_1,I_2,I_3}$ – user123234 Nov 20 '21 at 23:02
  • Hmm, perhaps you're misunderstanding the exercise. Your comments sound like you think there is one particular permutation $\tau$ that the exercise wants a proof about -- but it actually asks for a proof that such-and-such is true about every permutation in the centralizer (all 162 of them, as Robert computed). – Troposphere Nov 21 '21 at 00:01
  • Hmm okei so I‘m really missunderstanding this part. Could you maybe explain more sbout it, because I don‘t understand how it should look like – user123234 Nov 21 '21 at 07:08
  • because I don't see how to get from this $\tau^{-1}(1)...$ term to tau since this therm is not tau it is equal to sigma – user123234 Nov 21 '21 at 07:22

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