Existence of group of order 12 (Sylow's theorem?)

[ephedyn]

Homework Statement

Is there a group of order 12 which contains one involution and ten elements of order 3? Give an example or otherwise prove that such a group cannot exist.

2. The attempt at a solution

Let G be a group of order 12 = (p^k)*m where p is a prime number, k is greater than or equal to 1, and p does not divide m. Let H be a Sylow 2-subgroup of G which has order 4, and K be a Sylow 3-subgroup of G which has order 3. By Sylow's third theorem, the number of such subgroups s is 1 mod p: s | m. Hence, there are either 1 or 3 Sylow 2-subgroups and 1 or 4 Sylow 3-subgroups. Clearly, a group which has 10 elements of order 3 cannot exist.

Did I get this right? Even if so, is there another (easier?) way to do this question which doesn't invoke Sylow's third theorem?

 

[jbunniii]

Well, you can simplify your proof a bit by ignoring the involution.

A group of order 12 must have 1 or 4 subgroups of order 3 (Sylow), so it must have 2 or 8 elements of order 3. Therefore it can't have 10 and we're done.

I'm not sure how you would prove it in a more elementary way without essentially proving a special-case version of the Sylow theorems for |G| = 12.

 

[jbunniii]

Here's another argument which uses Sylow's 1st theorem instead of the 3rd.

If |G| = 12, then G must have a subgroup H of order 4.

There are two isomorphism classes of order 4: H must be either cyclic or isomorphic to the Klein 4-group. The latter is ruled out because it has 3 involutions.

Therefore H is cyclic of order 4, so it contains an involution and two elements of order 4. This doesn't leave enough room in G for 10 elements of order 3.

 

[ephedyn]

Oh! Thanks for the quick response. One last question,

1 or 4 subgroups of order 3 (Sylow), so it must have 2 or 8 elements of order 3

This sounds like a silly question but how did you get "2 or 8"? (This question was meant to be under a few chapters before Sylow's theorems were introduced, so I haven't grasped the material fully yet; and I thought the desired answer would have something to do with conjugacy).

I like your second argument!

 

[jbunniii]

Oh! Thanks for the quick response. One last question,



This sounds like a silly question but how did you get "2 or 8"?

A subgroup of order 3 contains exactly two elements of order 3. (The third element is the identity.)

Also, any two distinct subgroups of order 3 (or any prime order) must intersect trivially, so their elements of order 3 are distinct.

Therefore N subgroups of order 3 implies 2N elements of order 3.

 

[jbunniii]

Here's another argument which uses Sylow's 1st theorem instead of the 3rd.

If |G| = 12, then G must have a subgroup H of order 4.

There are two isomorphism classes of order 4: H must be either cyclic or isomorphic to the Klein 4-group. The latter is ruled out because it has 3 involutions.

Therefore H is cyclic of order 4, so it contains an involution and two elements of order 4. This doesn't leave enough room in G for 10 elements of order 3.

This can be further simplified. It doesn't matter what the isomorphism classes are.

If |G| = 12, then G must have a subgroup H of order 4. By Lagrange's theorem, each of the three non-identity elements of H must have order 2 or 4. That means there can be at most 8 elements with order 3. Done.