Order of any element divides the largest order. Let $A$ be a finite Abelian group and let $k$ be the largest order of elements in A. Prove that the order of every element divides $k$. This is my attempt, I sense there is something wrong\incorrect in it, but I can't figure out what....Also I didn't use that fact that $A$ is abelian... I would appreciate you reply. 
$Attempt$: $|G|=n$. Suppose there is an element $a\in A$ of order $m$ that does not divide $k$. $k$ is the largest order of elements in $A$, therefore there exist $g\in A$ such that $o(g)=k$. $(ag)^m=g^m\ne 1$ and $(ag)^k=a^k\ne 1$ (that follows from the commutativity of A). The order of $ag$ has to be some $p$, where $p$| lcm$(m,k)$. If $lcm(m,k)=p$ then $(ag)^p=(ag)^{mt}=1$ where $mt=p,p\in \Bbb{N}$. But $(ag)^{mt}=((ag)^m)^t=(g^m)^t=g^{mt}=1$, but since $t$ is the smallest natural number that fulfills it, $mt=k$, i.e, $m$ divides $k$. A contradiction. 

Let $x\in A$ such that $x$ has largest order $m$ in $A$. Assume that $g\in A$ such that $|g|$ does not divide $m$. Wlog, assume that $|g|=p$ where $p$ is a prime. Then $gcd(p,m)=1$. Now, as $A$ is abelian, $g$ and $x$ commute and $|gx|=pm=lcm(p,m)$. Note that $lcm(p,m)=pm$ as $p$ is a prime and $p\nmid m$. Hence, $|gx|=pm > m$ which gives the contradiction as $x$ has largest order $m$. (I hope.)
