How do I show every abelian group whose order is square free is cyclic without using the fundamental theorem of finite abelian groups?
I tried something like this
Let $|G| = p_1p_2…p_n$ By Cauchy's theorem, there's an element $x_i$ of order of $p_i$. Now, I want to show that $x_1x_2…x_n$ generates G.
Is this approach correct?
Best Answer
I presume you mean square-free order.
Hint: prove that if $C_m$ and $C_n$ are cyclic groups of orders $m$ and $n$ respectively with $\gcd(m, n)=1$ then $C_n\times C_m$ is cyclic.
(Your generator should be $(1, 1)$ - try some small examples to see how your proof should work, like $C_2\times C_3$.)