$|D_6|=12=2^23.$ I started with $3$. I know that the number of $3$-Sylow subgroups, denoted $n_3$, is: $1,4,7...$ and I also know that $n_3|2^2$. e.g, $n_3=1, 4$. How can I show that it can't be $4$? After all, Sylow subgroups are not disjoint. Suppose I showed it is $1$. How exactly do I find the $3$-sylow subgroup itself? I would really appreciate you help.
2026-03-26 10:59:59.1774522799
Find all the $p$-Sylow subgroups of $D_6$.
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Find a $3$ sylow subgroup, and check if it's normal. if it is - it's the only one. otherwise, there are 4 of them.
to find the $3$ sylow subgroup, just find a subgroup of three elements, giving the structure of $D_6$