Cardinality of a power set? Or is "all subsets of a set" $\neq$ power set? Again as usual, group theory is muddling me up. A proof of the Sylow $I$ theorem starts as follows > Let $X$ be the set of all subsets of $G$ with $|A|=p^m$. where, the setting I have for Sylow $I$ is > Let $G$ be a group $|G|=n$ where $p$ is a prime that divides $n$. Then we can write $n=p^mr$ where $p$ does not divide $r$....(and the theorem is stated). The issue is in the above notation, shouldn't $|A|=2^n$? The set of **all** subsets of $G$ that has cardinality $n$ is the power set, right? Then isn't its cardinality $2^n$? Why is it $p^m$? We can't just pick and choose that for convenience when we're discussing the power set. Can someone explain why??

You're misreading the English. The statement is defining $X$ as the set of all subsets $A$ of $G$ _such that_ the cardinality of $A$ is $p^m$. Namely, $$X=\\{A\subseteq G : |A|=p^m\\}.$$

Here $A$ is a free variable, not the power set of $G$.