Math Is Fun Forum

laipou

Member

Registered: 2009-09-19

Posts: 24

help me plz~ property of S4

Prove:
For any positive integer d|24 and d != 4,
the subgroups of order d in S4 are isomorphic.
What about subgroups of order 4?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: help me plz~ property of S4

Subgroups of

have orders 1, 2, 3, 4, 6, 8, 12, or 24. Obviously there is only one subgroup (up to isomorphism) of each of the orders 1, 2 and 3. There is only one subgroup of

of order 24, namely

, and one subgroup of

of order 12, namely

.

Which leaves 4, 6 and 8. As

does not have any cyclic subgroup of order 6 (if it did it would have to have either a 6-cycle or a disjoint product of a 2-cycle and a 3-cycle, neither of which is possible) all subgroups of order 6 are thus isomorphic (to

). As for 8, notice that

. Hence every subgroup of order 8 is a Sylow 2-subgroup; therefore all subgroups of order 8 are conjugate and therefore isomorphic (to the dihedral group

).

has two nonisomorphic subgroups of order 4, namely

(cyclic) and

(the Klein Viergruppe).

Last edited by JaneFairfax (2009-09-19 04:15:56)

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: help me plz~ property of S4

Thank you for answering.
Could you explain "As S4 does not have any cyclic subgroup of order 6 (if it did it would have to have either a 6-cycle or a  disjoint product of a 2-cycle and a 3-cycle, neither of which is possible) all subgroups of order 6 are thus isomorphic (to S3 )."in detail(Why S4 does not have any cyclic subgroup of order 6=>all subgroups of order 6 are thus isomorphic (to S3 ))?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: help me plz~ property of S4

There are only two groups of order 6 up to isomorphism: S3 and Z6 (note D3 is isomorphic to S3).

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: help me plz~ property of S4

"There are only two groups of order 6 up to isomorphism" Why?
Is there any simple explanation?
I'm just a beginner in basic group theory.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: help me plz~ property of S4

Let G be a group of order 6.  An easy way is to use Sylow's theorem which proves that G is a semi-direct product of Z2 and Z3, which can give rise to at most 2 groups.  And here they are: S3 and Z6.

But this uses a bit of machinery.  If you know group actions, then you can let G and on H, a nonnormal subgroup of G of order 2.  Prove that this action is faithful, and then of course it goes into S3, and you're done.

It is also possible to prove that G = {1, x, x^2, y, z, w} where x is of order 3, y, z, and w all have order 2.  Then using the fact that G is a subgroup of S4, you should be able to prove that y, z, and w are all transpositions.  Given this, it should only take a few calculations to show that x, y, z, and w are permutations of only 3 letters (numbers).

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: help me plz~ property of S4

Thanks a lot!!
Now I can finish my homework:D