I need to prove that S3 is a solvable group by providing an actual series, and demonstrating it via cycle notation. I am using S3, A3, e as my solvable series, but I do not know how to show it in cycle notation.

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- May 5th 2011, 06:24 PMgrace412S4 solvable
I need to prove that S3 is a solvable group by providing an actual series, and demonstrating it via cycle notation. I am using S3, A3, e as my solvable series, but I do not know how to show it in cycle notation.

- May 5th 2011, 07:07 PMTinyboss
(I assume the subject should read "S3 solvable"?)

What exactly do you need to demonstrate with cycle notation? The normality of A3<S3? The abelian-ness of the quotient? Those are trivial by order considerations, as I'm sure you know. Is it just an exercise in composing permutations? - May 5th 2011, 07:27 PMgrace412
well. the exact question is to prove that s3 is solvable by specifying an actual solvable series (which I specified above), and to justify the solution completely via cycle notation. That probably does not answer your question at all though.

- May 5th 2011, 07:36 PMTinyboss
The only relevance of cycle notation that I can see is if you're expected to go through element-by-element and show that A3 is closed under conjugation by elements of S3, and that S3/A3 is abelian. ...which seems like a terrible waste of time since both results are immediate from [S3:A3]=2 and |A3|=3.

- May 5th 2011, 11:46 PMDeveno
well, conjugation preserves cycle decomposition.

the elements of A3 must be either: a) the identity, or b) a 3-cycle.

since conjugation preserves cycle decomposition, this immediately shows that A3 is normal.

for example, σ(a b c)σ^-1 = (σ(a) σ(b) σ(c)).

so we can start our composition series like so: S3 > A3, and S3/A3 has order 2, so it is necessarily abelian.

but |A3| = 3, since the 3 elements of A3 are: {e, (1 2 3), (1 3 2)}, and 3 is prime, so A3 is cyclic, and thus abelian.

therefore S3 > A3 > {e} is a composition series where every factor is abelian, so S3 is solvable.