I'm trying to figure out how many cyclic groups there are of order $p^3q^2$ (up to isomorphism), where p and q are different primes.
My attempt at a solution here follows.
The number of abelian groups of order $p^3q^2$ (up to isomorphism) is 6 by 'The Fundamental Theorem of Finitely Generated Abelian Groups'.
There being more cyclic groups of a certain order than there are abelian groups of that order is silly, since being cyclic also means being abelian. So there is no more than 6 cyclic groups of order $p^3q^2$.
So my wish is to demonstrate whether or not there exists an element of order $p^3q^2$ in these six groups. So far I'm not having any success. Help would be greatly appreciated!
Best Answer
It seems rather silly to be invoking the Fundamental Theorem for Finite(ly generated) Abelian groups to attack this problem, given that that theorem implicitly assumes that one can talk about “the” cyclic group of order $n$ for any integer $n$...
Thinking about abelian groups (and their structure) is total overkill. For each $n$, $0\leq n\lt\infty$, there is one and only one cyclic group of order $n$ up to isomorphism (where “cyclic group of order $0$” should be interpreted as the infinite cyclic group).
Indeed, if $G$ and $H$ are both cyclic groups of order $n$, let $g\in G$ and $h\in H$ be such that $G=\langle g\rangle$ and $H=\langle h\rangle$. Then the map $f\colon G\to H$ that sends $g^k$ to $h^k$ is well-defined (since $g^k=g^{\ell}$ if and only if $k\equiv \ell\pmod{n}$, if and only if $h^k=h^{\ell}$); is a group homomorphism (since $g^rg^s = g^{r+s}\mapsto h^{r+s}=h^rh^s$); is one-to-one (since $h^k=e$ if and only if $n|k$ if and only if $g^k=e$); and is surjective (since the image contains the generator of $H$); and hence is an isomorphism.
(For $n=0$, note that $a\equiv b\pmod{0}$ if and only if $0|a-b$, if and only if $a-b=0$, if and only if $a=b$; that is, congruence modulo $0$ is the same as equality.)
Thus, whatever $n$ is, there is one and only one cyclic group of order $n$.