TITLE: Is there a case "infinite" p-group is meaningful?
QUESTION [5 upvotes]: I'm hesitating to choose my definition for "p-subgroup".
If i only consider finite cases, i would define it as a group with the order which is a power of $p$.
If i should consider infinite cases too, i would define it as a group such that every order of an element is a power of $p$.
Is there a case in which the concept of infinite p-group is useful?

REPLY [2 votes]: Another way both to form infinite examples where Sylows theorems can fail, or behave as expected, is to use amalgams (or free products with amalgamation). For example, if we take the free product of two cyclic groups of prime order $p,$ we get an infinite group in which any finite subgroup has order dividing $p,$ but the two generating subgroups of order $p$ are not conjugate. The infinite dihedral group $\langle t,u,: t^{2} = u^{2} = 1 \rangle$ is the "easiest" example, with $p=2.$ On the other hand, if $A,B$ are (possibly infinite) groups which each have a unique conjugacy class of maximal finite $p$-subgroups, and have isomorphic maximal finite $p$-subgroups, say $P$, then $A*_{H}B$ still has a unique conjugacy class of maximal finite $p$-subgroups, where $H$ is any common subgroup of $A$ and $B$ containing an isomorphic copy of $P.$
I suppose that for infinite groups, you have to be careful by what you might mean by a Sylow $p$-subgroup. One definition might be a maximal $p$-subgroup, which contains every other $p$-subgroup up to conjugacy. Another definition might be a maximal finite $p$-subgroup, which contains every other finite $p$-subgroup up to conjugacy. It is possible to have infinite groups which have Sylow $p$-subgroups in the second sense, but not in the first. There may be easier ways to do this, but you would (for sufficiently large $p$), take two Tarski monsters $S$ and $T$ such that all subgroups of order $p$ are conjugate in $S$, and likewise in $T$ (these exist)- it might be that $S$ and $T$ are isomorphic, I don't care. For the amalgam $A = S*_{C}T$, where $C$ is a common cyclic subgroup of order $p$ of $S$ and $T.$ Then all finite $p$-subgroups of $A$ are conjugate to subgroups of $C$, but $S$ and $T$ are maximal (infinite) $p$-subgroups of $A$ which are not conjugate in $A$- and furthermore, every infinite $p$-subgroup of $A$ is conjugate to either $S$ or $T.$