I am doing exercise which is a problem from the book Algebra by Hungerford (exercise II.5.4).
Given an infinite $p$-group $G$, where $p$ is a prime, I need to show that either $G$ has a subgroup of order $p^n$ for every $n\geq1$ or there exists $m\in\mathbb{N}$ such that every finite subgroup of $G$ has order less than or equal to $p^m$.
Correct me if I'm wrong, but I see that this statement is an "either $P$ or $Q$" statement, meaning that I have to show "($P$ and $\neg Q$) or ($\neg P$ and $Q$)".
Assuming $P$: "For each $n\geq 1$, there is a subgroup of order $p^n$", clearly we have $\neg Q$: "For each $m\in \mathbb{N}$, there is a subgroup of order $> p^m$", so now if I assume $\neg Q$, how do I show $P$?
My feeling is that I need to construct a new larger subgroup somehow from smaller ones and then use induction, or maybe utilize Zorn lemma, but I am honestly stuck as I am working in infinite case.
Any help or hint is appreciated. Thanks a lot.
My attempt:
Claim: If $H$ is a finite group of order $p^k$ where $p$ is prime, then for each $i=0,1,2,...,k$, $H$ has a subgroup of order $p^i$.
Proof of claim: For the case $|H|=p^1$, it is trivial. Assume that every finite group of order $p^r$ has subgroups of order $1,p^1,...,p^r$. Let $H$ be a finite group of order $p^{r+1}$. It is known that the center $Z(H)$ is nontrivial by class equation and of order $p^l$ by Lagrange's theorem, and hence there is an $x\in Z(H)$ of order $p$ by Cauchy's theorem. It can be shown that $\langle x \rangle$ is normal in $H$, and hence we consider the quotient group $H/\langle x\rangle$ which is of order $p^r$. By induction hypothesis, there are subgroups of $H/\langle x\rangle$ of order $1,p^1,...,p^r$, which must be, respectively, in the form $H_0/\langle x\rangle$, $H_1/\langle x\rangle$, ..., $H_r/\langle x\rangle$, where $H_0,H_1,...,H_r$ are subgroups of $H$ containing $\langle x\rangle$, by correspondence theorem. The order of each $H_j$ is $p^{j+1}$, and thus the inductive step is complete. End proof.
Now, if $n\geq 1$, then by $\neg Q$ there is a subgroup $H$ of $G$ of order $>p^n$. By the above claim, there is also a subgroup of order $p^n$.