Okay, I'm trying to list all the Sylow 2-subgroups of $\displaystyle D_{12}$. I have that n2 = 1 or n2 = 3 and I know that $\displaystyle | Syl_2 ( D_{12} ) | = 4$. It happens that there is only one subgroup of $\displaystyle D_{12}$ of order 4 so n2 = 1 is correct. However I'd like a more general argument to rule out n2 = 3 if possible.

If n2 = 3 then we must have 3 subgroups of order 4 that are all conjugate to each other and I'm trying to use this and the class equation to rule it out but I'm missing something simple. The conjugacy classes of $\displaystyle D_{12}$ are $\displaystyle \{ 1 \}, ~ \{ r^3 \}, \{ r, ~ r^5 \}, ~ \{ r^2 , ~ r^4 \}, ~ \{ s, ~sr^3 \}, ~ \{sr^2 , ~ sr^4 \} $. I'm thinking that this can't be if we have 3 conjugate order 4 subgroups... I think there is a factor of 1/3 somewhere here but I can't figure out how to finish the argument.

Any thoughts?

Thanks!

-Dan