What are the possible forms and classifications of groups of order 12?

[mathusers]

Hey there guys.

Let G be a group of order 12. Show by a Sylow counting argument that if G does not have a normal subgroup of order 3 then it must have a normal subgroup of order 4.

Deduce that G has one of the following forms:
(i) $C_3 \rtimes C_4$
(ii) $C_3 \rtimes (C_2 \times C_2)$
(iii) $C_4 \rtimes C_3$ or
(iv) $(C_2 \times C_2) \rtimes C_3$

Hence, classify all groups of order 12 up to isomorphism.

Any suggestions on how to go about doing this one please?
I will attempt myself once I have a good idea of what to do . Thnx :)

 

[NateTG]

http://en.wikipedia.org/wiki/Sylow_theorems

There is a finite number of possible cases for the number of Sylow 2 and 3 subgroups. You basically have to go through case by case.

 

[mathusers]

ok here is what i have for the first part.

Please verify whether this is correct or not.

Let G be a group of order 12. Show by a Sylow counting argument that if G does not have a normal subgroup of order 3 then it must have a normal subgroup of order 4.

suppose G has sylow 3-subgroup: order 3 and sylow 2-subgroup: order 4.

Let x be number of sylow 3-subgroups then x|3 and x=1(mod3). so x = 1 or 4.
Let y be number of sylow 2-subgroups then y|2 and y=1(mod2). so y= 1 or 2.

We have to show either x=1 or y=1 as that would imply it has a sylow subgroup which is fixed by conjugation and is therefore a normal subgroup.

Let's contradict:
------------------
let x=4, y=2. If x has 4 distinct subgroups of order 3 then the intersection of any 2 must be a subgroup and hence a proper divisor of 3. i.e. it's 1.
Together, the 4 subgroups have 9 distinct elements.

On the other hand, the sylow 2-subgroup has a trivial intersection with any of the sylow 3-subgroups and hence has 3 additional distinct elements.
So altogether so far, we have 12 elements.

However, we still have a remaining order 4 subgroup that adds atleast 1 more element. So in total, we have atleast 13 distinct elements. This is a contradiction because the group has only 12 elements. This means either x=1 or y=1. So there is a normal subgroup of order 4.

any ideas on how to solve the rest of the question. i.e. as below?.

Deduce that G has one of the following forms:
(i) $C_3 \rtimes C_4$
(ii) $C_3 \rtimes (C_2 \times C_2)$
(iii) $C_4 \rtimes C_3$ or
(iv) $(C_2 \times C_2) \rtimes C_3$

Hence, classify all groups of order 12 up to isomorphism.

 

[NateTG]

$$2 \equiv 0 (\rm{mod} 2)$$
(The choices for $y$ are 1 and 3.)

The rest of the proof for the first part is OK.

You can do the second part as cases again.
(What are the possibilities for $G/H$ for each of the normal subgroups.)