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Wikipedia says that the center of the symmetric group n>=3 is trivial.

https://en.wikipedia.org/wiki/Center_(group_theory)#Examples

But IIRC, S4 is the same group as the rotations of a cube, and the group of rotations of a cube has an abelian subgroup Z2×Z2 consisting of the rotations by 180 degrees, which is nontrivial.

So ... what's going on?

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    $\begingroup$ Abelian subgroup means that the elements commute among all each other. There may be well none of them (except the identity) which commutes with all the elements of the parent group. $\endgroup$
    – Kan't
    Commented Oct 4, 2022 at 3:36
  • $\begingroup$ While the center is an abelian subgroup, not every abelian subgroup is central. An element is central in $S_4$ if it commutes with every element of $S_4$. And $S_4$ is not the rotations of the cube. That's smaller. It is the permutations of a set of four elements. $\endgroup$ Commented Oct 4, 2022 at 3:36
  • $\begingroup$ Rotations of a cube has 24 elements - same as s4. Each one of six faces can be on top, and that top face can be in 4 possible orientations. Doesn't mean it is s4, but apparently it is. Not sure I see the connection, though. $\endgroup$ Commented Oct 4, 2022 at 4:15
  • $\begingroup$ @PaulMurrayCbr Oh, I see. I think the connection is through the four diagonals that connect opposite pairs of vertices. Think of rotations as permuting the diagonals, verify any nontrivial rotation yields a nontrivial permutation of the diagonals. $\endgroup$ Commented Oct 4, 2022 at 4:23
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    $\begingroup$ Ignore the orientation. Just consider the diagonal (equivalently, unordered pairs of pairs of opposite vertices). See here. $\endgroup$ Commented Oct 4, 2022 at 4:50

2 Answers 2

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The statement $S_4$ has an abelian subgroup $H=\mathbb{Z}_2\times \mathbb{Z}_2$ means that every element in $H$ commutes with other element in $H$. This does not mean that every element in $H$ commutes with all the elements in $S_4$. So this does not imply $H\leq Z(S_4)$.

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  • $\begingroup$ It's not necessary that every (nontrivial) element in $H$ commute with all the elements of $S_4$ to solve OP's conflict. Some suffices, which yields a nontrivial center either. $\endgroup$
    – Kan't
    Commented Oct 4, 2022 at 3:59
  • $\begingroup$ So, rather, $H$ abelian does not imply $|H\cap Z(S_4)|>1$. $\endgroup$
    – Kan't
    Commented Oct 4, 2022 at 4:06
  • $\begingroup$ Yep - rotating my pen left and then flipping vertically does not produce the same result as flipping vertically and rotating it left. And there it is. $\endgroup$ Commented Oct 4, 2022 at 4:11
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Not every abelian subgroup of a group is central. Examples abound.

Let's see, to take a few, any finite group divisible by a prime $p$ has a (cyclic) subgroup of order $p$, by Cauchy's theorem (which is abelian). But that subgroup needn't be central. The $S_n'$s ($n\ge3$) indeed provide infinitely many examples.

So you're confusing the definitions. Central elements commute with every element of the parent group.

And, this may be a little like beating a dead horse now but, by Cayley's theorem, any finite abelian group embeds in an $S_n$.

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