1. ## Sylow's Counting Argument

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.

I will attempt myself once I have a good idea of what to do . Thnx

2. Originally Posted by joanne_q
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.
The group has a Sylow 3-subgroup of order 3. It also has a Sylow 2-subgroup of order 4. Suppose that x is the number of Sylow 3-subgroups then x|4 and x = 1(mod 3) so x=1,2,4 it cannot be 2 so x = 1 or 4. Let y is the number of Sylow 2-subgroups then y|2 and y = 1(mod 2) so y = 1 or 2. Our goal is to show that either x=1 or y=1, because that would imply that it has a Sylow subgroup which is left fixed by conjugation and is therefore a normal subgroup. Suppose for contradiction that x=4 and y=2. But if it has 4 distinct subgroups of order 3 then the intersection of any two must be a subgroup and hence a proper divisor of 3, i.e. it must be 1. Thus, together these 4 subgroups have exactly 9 distinct elements. While the Sylow 2-subgroup has a trivial intersection with any of the Sylow 3-subgroups and hence it has an additional 3 distinct elements, thus we have so far 12 elements. But we still have a remaining order 4 subgroup that adds at least one more element so in total we have at least 13 distinct elements. Which is a contradiction because the group has only 12 elements. Which means either x=1 or y=1. And therefore it has a normal subgroup.