All non-abelian groups of order $pq$ are unique up to isomorphisms?

Question

A non-abelian group $G$ of order $pq$, where $p$ and $q$ are primes and $q>p$, can be represented as the product of its unique $q$-Sylow subgroup $A$ and one of the $q$ $\:p$-Sylow subgroups $B$. So $G=AB=\{ a^ib^j \mid a^q=e, b^p=e\}$. Now, the "composition law" must be $\: bab^{-1}=a^k$, $k \not = 0,1$.

I have two questions. I should perhaps say that I'm not familiar with direct or semidirect products.

1. Why does this show that all such groups are isomorphic? I.e. what is the isomorphism $\phi : G_1 \to G_2$ between $G_1$, in which $bab^{-1}=a^x$, and $G_2$, in which $bab^{-1}=a^y$? Could it be the isomorphism $\phi$ such that $\phi(a^x)=a^y$ and $\phi(b)=b$?

2. What values can $k$ take on? Certainly $k^p \equiv 1 \pmod q$, which implies $p \mid (q-1)$, but we already knew that. It is certainly not true that $k \in \Bbb U_q \implies k^p=e$, so I was wondering what $k$'s are allowed.

Answer 1

We can view the $q$-group as the additive group of $\Bbb F_q$, so instead of raising to $k$th power, we really multiply with an integer $k$. Repeating the multiplication $p$ times must give the identity, i.e., $k^p\equiv 1\pmod q$, or $k$ (viewed as element of $\Bbb F_q$) must be and can be any root of the polynomial $X^p-1\in\Bbb F_q[X]$. As we disregard $k=1$, we are only concerned with the case that $k$ is one of the $p-1$ primitive $p$th root of unity $\bmod q$ (which exist because $q\equiv 1\pmod p$).

Recall that we obtained $k$ from a chosen generator $b$ of the chosen $p$-group. Had we taken $b^r$ as generator instead, $1\le r<p$, we would have ended up with $k^r$ instead of $k$. But as $k$ cannot be a root of $X^r-X$ unless $r=1$, we conclude that by varying $r$, we loop over all primitive $p$th roots. Vice versa, no matter what $p$th root of unity we pick, we always get the same semi-direct product, just with a different choice of generator of the $p$-group-