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.
Best Answer
Unfortunately, your attempt at a proof isn't valid. You define $n$ to be the order of $a$, then prove that $a^n=1$ (which doesn't need proof, actually, because that's what $n$ does), and then derive a contradiction because you had claimed $n<m$. Indeed, I don't think you ever used the hypothesis that $m\nmid k$, other than to (correctly) derive $a^k\ne1$ which was never used.
Here's a hint: if $\gcd(m,k)$ were equal to $1$, could you prove the statement? (Let $b$ be an element of order $k$; what's the order of $ab$?) Then, can you reduce the general case to the case where $\gcd(m,k)=1$? (Replace $a$ by an appropriate power of $a$....)