Does Sylow's theorem assert the existence of subgroups of order $p^j$ for all $j=1,\dots,k$ ?

Question

Sylow's theorem says that there exists a subgroup of order $p^k$, where $p^k$ is the highest power of $p$ dividing the order of the group.

But for example if the group order is $24$, then we can write $24$ as $2^3 \cdot 3$. So there exists a subgroup of order $2^3 = 8$, but does Sylow's theorem also say that there exists subgroups of orders $2$ and $4$ as well (from $2^1$ and $2^2$)?

Answer 1

Yes, Sylow's first theorem can be widened to the following:

Claim Let $G$ be a finite group of order $p^nm$, $(m,p)=1$. Then for each $k=1,\ldots,n$ there exists a group of order $p^k$, and every group of order $p^k$ is normal in a group of order $p^{k+1}$.

This in particular gives that every finite $p$-group is solvable right away. In fact, finite $p$-groups are supersolvable.