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I have found this topic here: How to find all the permutations which commute with a certain one, but I still could not understand it. Could anyone please explain how to think about doing it?

If I have (12)(34), working in the symmetric group 4 for instance, how would I find every possible permutation that permutes with it?

Thanks.

Valentin
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  • Look for conjugates of a permutation. A link that can help is https://math.stackexchange.com/questions/1532745/if-g-13524-how-can-we-find-some-permutation-k-such-that-kgk-1-246/1532749#1532749 For your problem, you want $k(12)(34)k^{-1}=(12)(34)$. – Anurag A Jan 12 '19 at 22:08
  • I am not really sure about what "(k(1)k(3)k(5))(k(2)k(4))" is meaning and how would I use that. It's from the post you send. – Valentin Jan 12 '19 at 22:26
  • You should work with the so useful decomposition of a parmutation into a product of disjoint cycles. – Jean Marie Jan 12 '19 at 22:49
  • @BurLeXyOOnuTz if you read the answer carefully you will see $k(1)$ is the image of $1$ under the (permutation) function $k$. – Anurag A Jan 13 '19 at 01:44

1 Answers1

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You want all the permutations $k \in S_4$ such that $$k(12)(34)k^{-1}=(12)(34).$$ Since $$ k(12)(34)k^{-1} =k(12)\color{red}k^{-1} \, \color{\red}k(34)k^{-1} = (k(1) \,\, k(2)) \,\,\,\, (k(3) \,\, k(4)) $$ Where $k(i)$ means the image of $i$ under permutation $k$.

You want this to be $$(k(1) \,\, k(2)) \,\,\,\, (k(3) \,\, k(4)) = (1 \,\, 2) \,\,\, (3\,\, 4)$$

So our permutation $k$ can be such that $$(k(1) \,\, k(2))=(1 \,\, 2) \quad \text{ and } \quad (k(3) \,\, k(4))=(3\,\, 4)$$ OR $$(k(1) \,\, k(2))=(3 \,\, 4) \quad \text{ and } \quad (k(3) \,\, k(4))=(1\,\, 2)$$

From the first scenario, we get either $k(1)=1,k(2)=2,k(3)=3,k(4)=4$ (identity permutation) or $k(1)=1,k(2)=2,k(3)=4,k(4)=3$ (this refers to the permutation $(34)$) or.....

Hope you can take it from here.

Added response:

I am adding this to answer the question you have raised in your comment.

Observe from the second scenario I have mentioned in my answer above, we can have $k(1)=3, k(2)=4$ and $k(3)=2, k(4)=1$. So ask yourself, what will be the cycle notation for this permutation?

we have ($1 \xrightarrow{k} 3 \xrightarrow{k} 2 \xrightarrow{k} 4 \xrightarrow{k} 1$) and this corresponds to $(1324)$.

Anurag A
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  • I saw in the answers that (1324) is a permutation that satisfies commutativity too, but I don't understand how would you get that using this technique. Could you please explain how would you get (1324) please? I would really appreciate it. – Valentin Jan 13 '19 at 02:18
  • @BurLeXyOOnuTz I have added some remarks to my answer, so please take a look. – Anurag A Jan 13 '19 at 02:48
  • excuse me I do not feel like I understand k. As a question, can k(1) be 2,3,4 then k(2)=1,3,4 then k(3)=1,2,4 and k(4)=1,2,3 and we are just rearranging these maps in all possible ways? – Valentin Jan 13 '19 at 18:45