Exercise 4.3.29 (If $|G| = p^\alpha$, then $G$ has a subgroup of order $p^\beta$ ($0\leq \beta \leq \alpha$))

Proof. Let’s reason by induction on $\alpha$. The property holds if $\alpha =0$: the only possible value of $\beta$ is $0$, and $G=\{1\}$ admits the subgroup $\{1\}$ of order $p^0$.

Suppose that every group of order $p^{\alpha -1}$ admits subgroups of order $p^{\beta }$ for every $l$ such that $0\le \beta \le \alpha -1$. Let $G$ be a group of order $p^{\alpha }$, where $\alpha \ge 1$. Since $G$ is a nontrivial $p$-group, Theorem 8 shows that $G$ has a nontrivial center. Lagrange’s theorem then shows that $|Z|=p^s$, where $1\le s\le n$ (the case $s=\alpha$ corresponds to the case where $G$ is abelian).

There exists an element $h\ne e$ in $Z$. The order $d$ of $h$ divides $|Z|=p^s$, so $d$ is a power of $p$. Thus, $d=p^r$, where $1\le r\le s$. Then $g=h^{p^{r-1}}$ is of order $p$, since $g^p={(h^{p^{r-1}})}^p=h^{p^r}=1$, and $g\ne 1$.

The group $⟨g⟩\le Z$ is normal in $G$. Indeed, for all $x\in G$, since $g\in Z$, then $g^k\in Z$, $g^{k}x=x{g}^k$, so $x{g}^k{x}^{-1}=g^k$. Since every element of $⟨g⟩$ is a power of $g$, $⟨g⟩\subset Z$ is indeed normal in $G$, which allows us to consider the quotient group $G∕⟨g⟩$.

The group $G∕⟨g⟩$ has order $p^{\alpha }∕p=p^{\alpha -1}$. We can therefore apply the induction hypothesis to it: it admits a subgroup $H$ of order $p^{\beta -1}$ for every $\beta$ such that $1\le \beta \le \alpha$.

If $\pi :G\to G∕⟨g⟩$ is the natural projection $x\mapsto x⟨g⟩$, then ${\pi }^{-1}(H)$ is a subgroup of order $p|H|=p^{\beta }$. This shows that $G$ admits subgroups of order $p^1,...,p^{\alpha }$, and also the subgroup $\{1\}$ of order $p^0$. This completes the induction. □