Find the center of S3 and the centralizer of each element of S3

Find the center of S₃ and the centralizer of each element of S₃

Expert Solution

The centralizer of every element of S3 is a subgroup of S3, so by Lagrange’s theorem, it must
have order 1, 2, 3, or 6. Of course the centralizer of the identity element is all of S3, but notice
that, if a, b, and c are distinct, we have

(a b)(a b c) = (b c) while (a b c)(a b) = (a c). Hence since
all non-identity elements of S3 have the form (a b) or (a b c), we conclude the the centralizer of any
non-identity element of S3 is not all of S3. But certainly the centralizer of any element contains
the cyclic subgroup generated by that element, so this observation and Lagrange’s theorem imply
that the centralizer of any nontrivial element of S3 is precisely the cyclic subgroup generated by
that element. The observations above also imply that the center of S3 is trivial as the center of S3 is

Cyclic group generated by two cycles and cyclic group generated by three cycles has trivial intersection.

Colby Messinger answered 2 years ago

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