Symmetric group (direct product and decomposition)

[Symmetryholic]

I am looking for a mechanism to find a decomposition of symmetric groups. For finitely generated abelian group G, there is a mechanism to decompose G such that G is isomorphic to a direct sum of cyclic groups.

For symmetric groups, it seems a bit complex for me to find it.

For example,


(1) Is it possible for S_{n} to be decomposed as a direct product of groups?


(2) Is there any mechanism to find an isomorphic group of a direct product of symmetric groups? Let's say,

G = S_{n-2} \times S_{n-1} \times S_{n}

Any isomorphic group of the above G?

 

[rodigee]

(1) No, you can't decompose the symmetric group into the direct product of two (or more!) of it's proper subgroups.

(2) Sorry, but I don't understand the question.

 

[Symmetryholic]

If we cannot decompose symmetric groups, question (2) might reduce to find isomorphic groups of S_{n}.

Is there any known group, let's say P, isomorphic to S_{n}? If so, by Caley's theorem, any group G is isomorphic to the subgroup of P. Is that right?

 

[adriank]

I'm not understanding you.

If we cannot decompose symmetric groups, question (2) might reduce to find isomorphic groups of S_{n}.

What do you mean by "isomorphic groups of S_n"? Groups isomorphic to S_n? You can take any set you like of size n! and define a group operation on it to make it isomorphic to S_n.

Is there any known group, let's say P, isomorphic to S_{n}?

Yes, put P = S_n. Or do you mean something different?

If so, by Caley's theorem, any group G is isomorphic to the subgroup of P. Is that right?

No, just pick a group G larger than P. Cayley's theorem states that any finite group G is isomorphic to a subgroup of S_n, where n = |G| (it depends on G!).

 

[Symmetryholic]

What do you mean by "isomorphic groups of S_n"? Groups isomorphic to S_n? You can take any set you like of size n! and define a group operation on it to make it isomorphic to S_n.

For instance,http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2006;task=show_msg;msg=0044.0002"

Like A_{n+2}, is there any known group that has a subgroup isomorphic to S_{n}? A_{n+2} is the only group (different from S_n ) I have found so far.

Is there any dihedral group(or variants) of order m > n that has a subgroup isomorphic to S_{n}?

 

[rodigee]

Right, A_{2n} is trivially isomorphic to S_n. Even more trivially, S_{n+k} has plenty of subgroups isomorphic to S_n for all k\geq 1. Continuing, so does S_n \times G for any group G... etc. You can construct as many as you like which is why I (and probably adriank as well) am confused by this "any known group" phrase.

I don't know then answer to your Dihedral group question off the top of my head. I'll think about it. Assuming you haven't already, see if you can figure out why the symmetric group can't be decomposed

 

[adriank]

Right, A_{2n} is trivially isomorphic to S_n.

Not true: |A_2n| = (2n)!/2, while |S_n| = n!. However, A_n+2 has a subgroup isomorphic to S_n, at least according to the post he linked.

As for the dihedral group question: I write D_n to mean the dihedral group of order 2n. Any subgroup of D_n is either cyclic or isomorphic to D_m, where m divides n.

Every dihedral group has its trivial subgroup isomorphic to S₁. Every dihedral group also has a subgroup isomorphic to S₂, which is a cyclic group of order 2. For S₃, note that S₃ is isomorphic to D₃, so the dihedral groups that have a subgroup isomorphic to S₃ are exactly the ones of the form D_3n. Thereafter, no symmetric group is isomorphic to a dihedral group, so there are no more.

 

[rodigee]

I meant to say " A_{2n} has a subgroup that is trivially isomorphic to S_n"

 

[adriank]

Alright, I guess that's true. :)

 

[mathwonk]

the analog for non abelian groups, of the direct decomposition of abelian ones, is the filtration by subnormal towers, used in galois theory. see the jordan-holder theorem. or the krull schmidt theorem.

 

[adriank]

I was going to list some composition series for the symmetric groups, but I had a feeling that those wouldn't be what he was looking for. I may as well do it anyway...

Here are composition series with the composition factors for the symmetric groups.
S_1 = \{\varepsilon\}
S_2 \rhd \{\varepsilon\}; factors C_2
S_3 \rhd A_3 \rhd \{\varepsilon\}; factors C_2, C_3
S_4 \rhd A_4 \rhd K \rhd H \rhd \{\varepsilon\}, where K = \{\varepsilon, (12)(34), (13)(24), (14)(23)\} and H = \{\varepsilon, (12)(34)\}; factors C_2, C_3, C_2, C_2
S_n \rhd A_n \rhd \{\varepsilon\}, n \ge 5; factors C_2, A_n

 

[Symmetryholic]

Thanks for all replies.

I have some further questions.

(1) Is it possible for S_{n} (n>=5) to be a product of its proper subgroups such that S_{n} = HK. H and K are proper subgroups of S_{n}.
(2) Is it possible for S_{n} to be a finite product of its proper subgroups such that S_{n} = H_{1}H_{2}..H_{k}, where H_{1}H_{2}..H_{k} are proper subgroups of S_{n}, where n >=5.

----------------------------------------------------------------------
Lemma 1. S_{n} is generated by a=(1 2) and b=(1 2 3 ... n).(http://mathforum.org/library/drmath/view/51685.html")
----------------------------------------------------------------------
My assumption is
If b \in A_{n} for n>=5, it is possible for (1). (Is this correct?)
(It's because a \in S_{2} and b \in A_{n}, and both S_{2}, A_{n} are proper subgroups of S_{n} ).
If b \notin A_{n}, I have no clue.

 

[adriank]

It's possible for any n > 2; take H = {e, (12)} and K = A_n. In that case, HK = A_n ∪ (12)A_n = S_n; this is the union of the even permutations of S_n and the odd permutations of S_n.

 

[Symmetryholic]

Thanks, adriank.

Is it possible for A_{n} to be a product of its proper subgroups as well such that A_{n} = H_{1}H_{2}..H_{k}, where H_{1}H_{2}..H_{k} are proper subgroups of A_{n}, where n >=5 ?

This problem is hard for me as well because I could not find any pattern on proper subgroups of A_{n}.