Group Theory Subgroups Sylow Theory Let $G$ be a finite group, $p$ a prime and $e$ a nonnegative integer. If $p^e$
divides the order $|G|$ of $G$, show that $G$ has a subgroup of order $p^e$.
 A: Let $G$ be a group, then suppose $|G|=p^nm$, where $(p^n,m)=1$. By Syllow theorem, $G$ has a subgroup $H$ of order $p^n$. $H$ has a normal subgroup $H_1$ of order $p$ as $H$ is a p-group and p-groups has no-trivial center by class equation(which means you can find an element $a$ of order a power of $p$ in the center, say $a^{p^r}=1$ then the element $a^{p^{r-1}}$ is of order $p$ and then you let $H_1 := \langle a^{p^{r-1}} \rangle$). Then $H/H_1$ is of order $p^{n-1}$. $H/H_1$ has a normal subgroup $H_2$ of order $p$ by the same argument. So, $(H/H_1)/H_2$ is of order $p^{n-2}$. And notice that if we let $\phi_i :(..((H/H_1)/H_2)/...)/H_{i-1}) \rightarrow (..((H/H_1)/H_2)/...)/H_{i-1})/H_i$ be the canonical quotient map, for example,  $\phi_1: H \rightarrow H/H_1$ be the canonical map from $H$ to $H/H_1$. Then $\phi_1^{-1}(\phi_2^{-1}(...\phi_k^{-1}(H_k))...) \subset H$ is a subgroup of H and of $G$ of order $p^k$. 
Here is an image to illustrate a subgroup of order $2^3$ in a group of order $2^5$:
