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Find 8 elements that commute with (12)(34)(56). Do they form a subgroup of S6?

I actually have found 8 elements randomly, but I found ab is not in my 8 elements( for some a,b of my 8 elements), so I can conclude this is not a subgroup, right?

Evan
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  • If indeed you found that two elements when multiplied were not in the set, then it is not a subgroup. This is of course predicated on you properly calculating the elements. – Cameron Williams Feb 16 '15 at 02:33
  • And to add to @Cameron’s comment, there is definitely an $8$-element subgroup of $S_6$ consisting of elements that commute with $(12)(34)(56)$. – Brian M. Scott Feb 16 '15 at 02:35
  • so the answer to this question is flexible? What I list are: – Evan Feb 16 '15 at 02:53
  • (13)(24)(56), (1324)(56), but multiplication of them is (34), which is not in my list. Does it make sense? – Evan Feb 16 '15 at 02:54
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    I think you should check $(1324)(56)$ again... – Dustan Levenstein Feb 16 '15 at 03:08
  • @DustanLevenstein I checked it again (12)(34)(56) commutes with (1324)(56) such that I both get (1423). Right? – Evan Feb 16 '15 at 03:59
  • @BrianM.Scott can you help me show that (12)(34)(56) does not commute with (1324)(56)? Why I get (1423) in both ways? – Evan Feb 16 '15 at 16:26
  • @Matthew: There are $48$ permutations commuting with $(1,2)(3,4)(5,6)$, forming a subgroup. You only listed $8$ of them that happened not to form a subgroup. – orangeskid Feb 16 '15 at 17:54
  • @Matthew: You're right; it does. I was focussing so hard on the eight elements of the form $(12)^i(34)^j(56)^k$ with $i,j,k\in{0,1}$ that I made a logic error in my argument supposedly showing that they are the only ones and then didn't notice that you must have meant $(12)(34)(56)$ and not $(13)(24)(56)$ in your earlier comment. I apologize for the confusion. In fact $(1324)^2=(12)(34)$, so $(12)(34)$ is in the subgroup generated by $(1324)$, and the two have to commute. – Brian M. Scott Feb 16 '15 at 17:55

4 Answers4

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The permutation $\sigma= \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6\\ a_1 & a_2 & a_3 & a_4 & a_5& a_6 \end{array} \right) $ commutes with the permutation $(1,2)(3,4)(5,6)$ if and only if the permutation $(a_1 a_2)(a_3a_4)(a_5a_6)$ is the same as the permutation $(12)(34)(56)$, that is, if and only if $(a_1 a_2)(a_3a_4)(a_5a_6)$ is a rearrangement of $(12)(34)(56)$. There are $2^3 \times 3! = 48$ such permutations. For instance, $\ \sigma= \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 4 & 2 & 1 & 5 & 6 \end{array} \right) $ commutes with $(12)(34)(56)$ since $(34)(21)(56)= (12)(34)(56)$.

All these $48$ elements form a subgroup of $S_6$. Now, if you take just $8$ of them, they may not form a subgroup.

orangeskid
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Think as if it was the Klein group in $S_4$ but in $S_6$, then: $$(13)(24)(56) \\ (13)(26)(45) \\ (13)(25)(46)\\ (14)(23)(56)\\ (14)(25)(36)\\ (14)(26)(35)\\ (12)(35)(46)\\ (12)(36)(45)\\$$

Are the elements you are looking for. And the set: $$ A = \{ (13)(24)(56), (13)(26)(45), (13)(25)(46), (14)(23)(56), (14)(25)(36), (14)(26)(35), (12)(35)(46), (12)(36)(45), (12)(34)(56), Id \}.$$ Is in fact a subgroup of $S_6$.

Diego Robayo
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  • Here $A$ has nine elements, so is not a set of 8 which commute with $(12)(34)(56)$ and form a subgroup of $S_6.$ – coffeemath Mar 23 '15 at 10:49
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In any group $G$, for any specific element $g\in G$, the cyclic subgroup generated by $g$ always commutes with $g$.

Here the subgroup generated by three transpositions $(12), (34), (56)$ is in fact of order 8. (counting the identity element): they are $\{id, (12), (34), (56), (12)(34), (12)(56), (34)(56), (12)(34)(56)\}$.

P Vanchinathan
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In permutation group ,disjoint cycles commute and inverse pairs commute ( and a transposition is its own inverse ) So consider this set

{(5,6),(3,4),(1,2),(5,6)(1,2),(5,6)(3,4),(1,2)(3,4),(5,6)(3,4)(1,2),e=(1,2)(2,1)} It can be easily seen that this group is closed under composition and inverses . So it forms the subgroup of S6.

I cannot figure out how many of the elements of S6 will commute though.

Bluey
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