Suppose $ \tau = (a_1, ..., a_k) $ it's a cycle in the group $ S_n $, and $ \sigma $ is any permutation from $ S_n $. Calculate the coupling $ \tau^\sigma $
Could you help me?
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Does coupling mean conjugation? – anon Aug 19 '13 at 00:01
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Yes, I guess so. I tried translate this word ;). – Mat Aug 19 '13 at 00:06
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Hint: what do we need to know about $a$ in order to find $\tau^\sigma a$? – Jonathan Y. Aug 19 '13 at 00:06
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I don't know. This $ \sigma $ looks weird. – Mat Aug 19 '13 at 00:09
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For example: $$ \sigma = {1\ 2\ 3\ 4\ 5\ 6\ 7 \choose 5\ 7\ 2\ 1\ 6\ 4\ 3} \ \ \tau = {1\ 2\ 3\ 4\ 5\ 6\ 7 \choose 6\ 7\ 5\ 3\ 4\ 2\ 1} $$ And we should find: $ \gamma $ in $ \sigma = \gamma^{-1}\circ \tau \circ \gamma $ – Mat Aug 19 '13 at 00:14
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But in this task, I haven't idea, how I must calulcate. – Mat Aug 19 '13 at 00:15
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I think looking at a specific $\sigma$ is what's holding you back. Try to find $\tau^\sigma a$ for arbitrary $\sigma,a$ and ${a_j}$ (you can solve by cases, if that helps). – Jonathan Y. Aug 19 '13 at 00:25
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Hint: what effect does $\sigma(a_1~a_2~\cdots~a_k)\sigma^{-1}$ have on $\sigma(a_1)$, $\sigma(a_2)$, etc.? – anon Aug 19 '13 at 00:25
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@JonathanY. what does mean "arbitrary $ \sigma $" – Mat Aug 19 '13 at 00:30
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@anon can you expand your method, because I started to learn it today, and I don't know all the concepts. – Mat Aug 19 '13 at 00:31
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It means, without assuming any particular structure to $\sigma$ (but still assuming knowledge of it, of course). But anyway, I think anon gave the game away. – Jonathan Y. Aug 19 '13 at 00:32
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Wait, no, your comment explaining that we're looking for $\gamma$ implies that we aren't dealing with calculating the conjugate $\tau^\sigma$ after all; please clarify. – Jonathan Y. Aug 19 '13 at 00:36
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It's only example how I understand it permutation. And in task I must calulate $ \tau^\sigma $ - it's other example. So I must solve problem in the topic. – Mat Aug 19 '13 at 00:40
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Maybe you have other hints? Because still I don't see anything :( – Mat Aug 19 '13 at 00:54
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As anon said, what does $\tau^\sigma(\sigma(a_j))$ equal? What about $\tau^\sigma(\sigma(b))$, where $b\not\in{a_1,\ldots,a_k}$? – Jonathan Y. Aug 19 '13 at 01:14
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hmm, i realy i don't know – Mat Aug 19 '13 at 01:21
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I've never solved this kind of task, so I'm learning now new thing. – Mat Aug 19 '13 at 01:27
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Could you look on the last comment below and answer on my question :) ? – Mat Aug 19 '13 at 22:56
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It will be easier to see the answer if you use cycle notation. For example instead of $\sigma$ and $\tau$ that you've written above, in cycle notation $\sigma = (1564)(273)$ and $\tau=(1627)(354)$. Neither of these are actually cycles in $S_n$, so instead let's pick $\sigma=(1564)$, $\tau = (1627)$. Then $\sigma^{-1} = (1465)$ and
$$\tau^\sigma = \sigma \circ \tau \circ \sigma^{-1} = (1564)(1627)(1465)=(2754).$$
Can you see the relation between $\tau^\sigma=(2754)$ on the one hand, and $\tau = (1627)$, $\sigma = (1564)$?
To make it even more apparent, let's write $\sigma$ in the old notation again:
$$\sigma = {1\ 2\ 3\ 4\ 5\ 6\ 7 \choose 5\ 2\ 3\ 1\ 6\ 4\ 7}. $$
Can you see the relation between $\tau=(1627)$ and $\tau^\sigma=(2754)$ using the above? If you still don't see it, how about if we rewrite $\tau^\sigma=(5427)$. How can we get this from $\tau=(1627)$ using the function $\sigma$ above?
Zavosh
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I know something :), but if I want understand whole the method, I must repeat a few things. And I will write comment for six hours, because in my country is 4 a.m., and I must go to sleep. – Mat Aug 19 '13 at 01:38
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So I must firsly calculate $$ {1 2 3 4 5 6 7 \choose 5 2 3 1 6 4 7} \circ {1 2 3 4 5 6 7 \choose 6 7 3 4 5 2 1} \circ {1 2 3 4 5 6 7 \choose 4 2 3 6 1 5 7}$$ Right? – Mat Aug 19 '13 at 13:08
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Ok, I calculate: $$ {1 2 3 4 5 6 7 \choose 5 2 3 1 6 4 7} \circ {1 2 3 4 5 6 7 \choose 6 7 3 4 5 2 1} = {1 2 3 4 5 6 7 \choose 4 7 3 1 6 2 5} $$ $$ {1 2 3 4 5 6 7 \choose 4 7 3 1 6 2 5} \circ {1 2 3 4 5 6 7 \choose 4 2 3 6 1 5 7} = {1 2 3 4 5 6 7 \choose 1 7 3 2 4 6 5 } $$ So we get $ (2754) $ but I can't see this, can you tell more something? – Mat Aug 19 '13 at 20:30
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In the big bracket for $\sigma$, if you look underneath the numbers $1$, $6$, $2$, $7$ for $\tau$, you will find exactly the numbers for $\tau^\sigma$: $5$, $4$, $2$, $7$. So when writing $\tau^\sigma$ in cycle notation, we just change the numbers in the cycle notation of $\tau$ using the rule given by $\sigma$. – Zavosh Aug 20 '13 at 02:15
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