Groups of Order 12 aren't Simple

Question

Suppose $G$ is a group of order $12=2^2*3$. Let $n_p$ denote the number of Sylow p subgroups. Then $n_2$ is 1 or 3 and $n_3$ is 1 or 4. I want to show that one of them is one since if that is the case I can conclude that we have a normal subgroup, in which case $G$ is not simple. For the purpose of arriving at a contradiction, suppose $n_2=3,n_3=4$. I am not sure how to proceed. I was told that this implies that $G$ has 9 elements of order 2 or 4 and 8 elements of order 3, but how does one conclude that? I would just conclude that $G$ has 3 elements of order 2 and 4 elements of order 3.

Answer 1

If $n_3=4$ you have $8$ elements of order $3$. so there are $4$ elements of other orders. Those are exactly enough to have a subgroup of order $4$, so it is unique. Hence normal.

Answer 2

Another contradiction is that there are 8 elements of order 3. And we find more that $|G|$ element in the group using the other information. We can do this by considering $n_2=4$. So let the 4 Sylow 2 subgroups called $H_1,\ldots, H_4$. Then $|H_1\cap H_2|<2^2$ So therefore $|H_1\cup H_2| =1$ or $2$ by Lagrange’s Thrm. Then compute $|H_1\cup H_2|$ to be at least 6 by the inclusion exclusion principle (8+8-2=6). Then $|G|\geq 6+8$ so contradicting $|G|=12$.