Square free finite abelian group is cyclic 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?
 A: 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 $C_2\times C_3$.)
A: Your approach can be made to work, but you may more or less end up redoing the structure theorem of f.g. abelian groups in this simpler case. It is easier to consider, like user1729, only two coprime integers at a time.
So I might recommend a contrapositive route. Assume that the group is not cyclic. Let $x$ be of order $t$, where you make $t$ as large as possible. As the group is not cyclic, $t$ cannot be the product of all the primes dividing the order of $G$, so there is a prime $p$ that is a factor of $|G|$ but is not a factor of $t$. We know that there is an element $y\in G$ of order $p$. We also know (from Lagrange's theorem) that the subgroups $H=\langle x\rangle$ and $K=\langle y \rangle$ intersect trivially. Using all this it is easy to show that $xy$ has order $tp$. Spoilerized hint:

 If $(xy)^i=x^iy^i=1_G$, then $x^i=y^{-i}$ is in $H\cap K$, so $x^i=y^i=1_G$.

This contradicts the maximality of $t$ and proves your claim.
A: Yes, the approach works. All you need is a lemma:
Lemma. Let $G$ be a group, and let $x$ and $y$ be elements of $G$ such that $xy=yx$. If $\gcd(|x|,|y|) = 1$, then $|xy|=|x||y|$, where $|g|$ is the order of $g$.
Proof. Let $|x|=r$ and $|y|=s$. Note that if $x^a = y^b$ for some integers $a$ and $b$, then $x^a=y^b=1$: for the order of $x^a$ is $r/\gcd(r,a)$ and the order of $y^b$ is $s/\gcd(s,b)$. Sincee $x^a=y^b$, then $r/\gcd(r,a) = |x^a| = |y^b| = s/\gcd(s,a)$. Thus, the order of $x^a$ divides $s$ and divides $r$, hence divides $\gcd(r,s)=1$; so $x^a=1=y^b$, as claimed.
Now:
$$\begin{align*}
(xy)^k = 1 &\Longleftrightarrow x^kyk=1\\
&\Longleftrightarrow x^k= y^{-k}\\
&\Longleftrightarrow x^k = y^k = 1\\
&\Longleftrightarrow r|k\text{ and }s|k\\
&\Longleftrightarrow \mathrm{lcm}(r,s)|k\\
&\Longleftrightarrow rs|k.
\end{align*}$$
Thus, $|xy|=rs$, as claimed. $\Box$
Now use induction to show that $|x_1\cdots x_n|=|x_1|\cdots|x_n|$, and get your conclusion.
By the way, there are some nice generalizations of the Lemma above. See for example this question and this one. On the other hand, if $x$ and $y$ don't commute, then the orders of $x$, $y$, and $xy$ may be completely independent: given any three positive integers $r,s,t\gt 1$, there is a finite group $G$ with elements $x$ and $y$ such that $|x|=r$, $|y|=s$, and $|xy|=t$. In fact, this came up recently in MathOverflow.
