# Using Lattice Isomorphism Theorem

I am working on this for my algebra class and I am stuck at the very end.

$\textbf{QUESTION:}$ Let $p$ be a prime and let $G$ be a group of order $p^\alpha$. Prove that $G$ has a subgroup of order $p^\beta$, for every $\beta$ with $0\le\beta\le\alpha$.

$\textbf{SOLUTION:}$ We will proceed with induction on $\alpha$. When $\alpha=1$, we have $\lvert G\rvert = p$. So $1$ and $G$ are subgroups of order $1$ and $p$, respectively. Hence $G$ has a subgroup of order $p^\beta$ for all $0 \le \beta \leq 1$. Thus our basis is established.

Next, we assume that a group of order $p^\alpha$ has a subgroup of order $p^\beta$ for every $\beta$ with $0\le\beta\le\alpha$.

Now let $G$ be a group of order $\alpha+1$. By Theorem 8, page 125, we know that $G$ has a nontrivial center $Z(G)$. We have two cases to consider: if $Z(G)=G$, then $G$ is abelian and we let $H\le G$ with $\lvert H\rvert=p$ by Cauchy's Theorem; if $Z(G)\ne G$, then let $H=Z(G)$ which is nontrivial and normal in $G$.

In either case, we consider the two groups $H$ and $G/H$. By Lagrange's Theorem, both $H$ and $G/H$ are groups of order $p^\beta$ for some $0\le\beta\le\alpha$. So by the induction hypothesis, both $H$ and $G/H$ have subgroups of $p^\gamma$ for all $0\le\gamma\le\beta$.

Then by Fourth Isomorphism Theorem, $G$ has a subgroup of order $p^\beta$ for all $0\le\beta\le\alpha+1$?

Suppose that $|H|=p^\beta$, where $1\le\beta\le\alpha$. Every subgroup of $H$ is a subgroup of $G$, so by the induction hypothesis $G$ has subgroups of every order $p^\gamma$ with $0\le\gamma\le\beta$. Now suppose that $\beta<\gamma\le\alpha+1$, and let $\delta=\gamma-\beta$. $|G/H|=p^{(\alpha+1)-\beta}$, and $\delta\le(\alpha+1)-\beta\le\alpha$, so by the induction hypothesis $G/H$ has a subgroup $K$ of order $p^\delta$. The fourth isomorphism theorem says that $K=L/H$ for some subgroup $L$ of $G$ that contains $H$, and clearly $|L|=|K||H|=p^{\delta+\beta}=p^\gamma$. Thus, $G$ has subgroups of order $p^\gamma$ for all $\gamma\le\alpha+1$.