This is what I have. Not sure if this is right. $${4\times3\times2\over3}=8 $$ I think this is too easy to be true
Find all permutations of $S_4$ that commutes with (123)
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$\begingroup$
abstract-algebra
permutations
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2$8$ is not a permutation, is it? – 2017-02-01
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08 is the number of permutations – 2017-02-01
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1http://math.stackexchange.com/questions/2118144/prove-disprove-find-a-in-s-4-that-is-not-a-power-of-b-such-that-ab-ba-whi – 2017-02-01
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1@behold: But you are not asked for a number of permutations, you are asked for the permutations. – 2017-02-01
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1"Find all permutations..." doesn't mean just find out how many there are; it means find out which ones they are. – 2017-02-01
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0The identity and the inverse, of course. Any others? nope. – 2017-02-01
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0You've forgotten $(123)$ itself, lol. – 2017-02-01
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0doh... yes and itself. – 2017-02-01
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0I am not understanding @MichaelHardy. I do not see another way to do it – 2017-02-01
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0@Arthur This is different from the one that is claimed to be a duplicate. My question states exactly what I type – 2017-02-01
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0@behold You want all permutations that commute with $(123)$, while the linked question asks for all permutations that commute with $(123)$ and at the same time is not a power of $(123)$. I know your question is not a verbatim copy, but it's close enough in my book. – 2017-02-01
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0To commute with (123) in $S_4$ how am I suppose to go about it? Like (123) commutes with (13) right? @Arthur – 2017-02-01
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0To commute with (123) in $S_4$ how am I suppose to go about it? Like (123) commutes with (13) right? @DougM – 2017-02-01
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0$(13)(123) = (12)\cdots (123)(13) = (23)$ does not commute. – 2017-02-01