On the number of Sylow p-subgroups I am working on an exercise where, so far, I have constructed a group $G = C_7 C_6 = \big\{n'h'\;|\;n' \in C_7,\;h' \in C_6\big\}$ defined by the relations
$$\big\{n^7 = h^6 = 1,\;hnh^{-1} = n^2\big\}$$
Indeed, $G$ is the extension of $C_6$ by $C_7$ such that the generator of $C_6$ acts by conjugation on $C_7$ as an automorphism of order 3. Now, since $|G| = 2\cdot 3\cdot 7$, we know by the Sylow theorems that there is a unique Sylow 7-subgroup, 1 or 7 Sylow 2-subgroups, and 1 or 7 Sylow 3-subgroups. It's not clear to me how we can determine that there is exactly 1 Sylow 2-subgroup and 7 Sylow 3-subgroups. I have tried a proof by contradiction, but I can't get things to work out properly.
Update: I have proven that there is exactly one Sylow 2-subgroup and 7 Sylow 3-subgroups. However, my method was sort of by brute force. That being said, I'm still curious if there is a quick way to determine this result.
 A: The group elements can be written as $n^a h^b$ with $a\in\{0,\ldots,6\}$ and $b\in\{0,\ldots,5\}$. To do multiplications in that representation, we need to know how we can interchange powers of $h$ with powers of $n$.
So let us investigate that first. By $hnh^{-1} = n^2$ we have $hn = n^2 h$.
Iteratively, we get that $h n^a = n^{2a} h$ for all integers $a$. (For negative integers $a$, use that inverting the equality $hnh^{-1} = n^2$ yields $hn^{-1} = n^{-2} h$.)
Similarly, iterating over $b$ gives $h^b n^a = n^{2^b a} h^b$ for all integers $a, b$.
Now let us examine the obvious $3$-Sylow subgroup $P_3 = \langle h^2\rangle = \{1, h^2, h^4\}$ and the obvious $2$-Sylow subgroup $P_2 = \langle h^3\rangle = \{1,h^3\}$.
We compute
$$n h^2 n^{-1} = n (h^2 n^{-1}) = n (n^{2^2 \cdot (-1)} h^2) = n^{-3} h^2 \notin P_3.$$
Therefore, $P_3$ is not normal in $G$. The only other possibility admitted by Sylow is that there are seven $3$-Sylow groups.
Obviously, $P_2$ is normalized by $h$, and by
$$n h^3 n^{-1} = n n^{2^3 \cdot (-1)} h^3 = n^{-7} h^3 = h^3\in P_2$$ it is normalized by $n$, too.
Therefore, $P_2$ is normal in $\langle n,h\rangle = G$.
Hence $P_2$ is the unique $2$-Sylow group.
