Thread: S4 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.

(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?

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.

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.

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.