A following theorem of Cauchy's theorem

Question

I'm reading a theorem as follows:

Let $G$ be a group of order $pq$, where $p, q$ are primes and $p>q$. If $a \in G$ is of order $p$ and $A$ is the subgroup of $G$ generated by $a$. Then $A \lhd G$.

The proof is to suppose $B$ is another group of order $p$ and consider $AB$, so as to give a contradiction. Thus $A$ is the only subgroup of order $ p$.

What I don't understand is that the proof says:

If $x \in G$, $B=x^{-1}Ax$ is a subgroup of $G $ of order $p$, in consequence of which we conclude that $x^{-1}Ax=A$ ; hence $A \lhd G$.

I don't understand why it makes this so simple. From Lagrange's theorem, I know a subgroup of $G$ must be of order $p$, $q$, or $pq$. If $A$ is a subgroup of $G$, then $x^{-1}Ax$ is also a subgroup of $G$. $x^{-1}Ax$ is isomorphic to $A$, therefore $x^{-1}Ax$ and $A$ have same order $p$. But the proof says $x^{-1}Ax=A$, why?

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$I suppose you haven't covered Sylow's theorems yet.

Then, what is missing in the argument that has been presented to you is that if $C = x^{-1} A x \ne A$, then the subset $A C$ of $G$ has order $p^2 > pq = \Size{G}$, a contradiction.

This follows from the following fact

If $G$ is a finite group, and $A, C \le G$, then $$\Size{A C} = \dfrac{\Size{A} \cdot \Size{C}}{\Size{A \cap C}}.$$

Here $AC = \Set{a c : a \in A, c \in C }$ is just a subset of $G$.