Let us assume that we have a group G with the following property : for every 5 discrete elements of G , at least 2 of them can be commutated.
Is G an abelian group?
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"can be commutated" means that a.b=b.a – user3697730 Nov 15 '16 at 19:18
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No, there are easy example of nonabelian groups with this property (though for obvious reasons, only quite small ones). – Tobias Kildetoft Nov 15 '16 at 19:20
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Could you please provide an example of a nonabelian group with this property? – user3697730 Nov 15 '16 at 19:24
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I think it would be better if you just sat down and actually tried to come up with one (just try the nonabelian groups you know one by one, you should fairly quickly see an example). – Tobias Kildetoft Nov 15 '16 at 19:25
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Actually ,it took me hours to solve the same problem with 4 elements instead of 5. My experiance with abelian groups is limited unfortunatelly... – user3697730 Nov 15 '16 at 19:29
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You only need to actually start looking at examples. – Tobias Kildetoft Nov 15 '16 at 19:29
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For $3$ elements it is true, see here. – Dietrich Burde Nov 15 '16 at 19:50
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@DietrichBurde I think it should hold for $4$ as well, and for any fixed number, it will hold as long as the group is sufficiently large. – Tobias Kildetoft Nov 15 '16 at 19:52
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@DietrichBurde I didn't actually check if $Q_8$ worked, as a smaller example did the trick (for this specific number). – Tobias Kildetoft Nov 15 '16 at 19:57
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@DietrichBurde Sure, the two $3$-cycles commute. – Tobias Kildetoft Nov 15 '16 at 20:00
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Up to dimension $12$ there are $8$ non-abelian groups, namely $$S_3,D_4,Q_8,D_5,C_3\rtimes C_4, C_2\times S_3, A_4,D_6. $$ Their Cayley tables are available, so we can easily check whether or not for each $5$ distinct elements, there are two which commute. Certainly this is true for $S_3$ and $Q_8$, since for $S_3$, choosing $5$ elements means choosing all but one element. If the identity is among the $5$ elements, it commutes with everything and we are done, if not, then the two $3$-cycles commute. For $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$, choosing $5$ elements means, that we have $a$ and $-a$ among them, which do commute. I leave it to you to check the other groups.
So the answer is: no, the group need not be abelian in general. However, it is true if the order of $G$ is big enough.
Dietrich Burde
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