# 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.

• Maybe the fact that any finite Abelian group is isomorphic to $\prod_{finite} (\mathbb{Z}/ p_i \mathbb{Z})^{\alpha_i}$ could help? Jan 26 '16 at 20:16
• has an answer here as well: math.stackexchange.com/questions/191981/… Oct 21 '19 at 0:37

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 $a$....)

• I edited it... is that admissible? Jan 25 '15 at 2:53
• Not yet. First, you treat the case lcm$(m,k)=p$, but you don't consider the case $p<{}$lcm$(m,k)$. Second, your phrase "$t$ is the smallest natural number that fulfills it" is imprecise; and I think when you make it precise, you'll see that what you're claiming isn't necessarily true. Jan 25 '15 at 6:17
• As I understand, this zone is considered pretty hard and tricky by other question, so with this confusion and more than one time misunderstanding of me and other, I would have to quite. :( Jan 25 '15 at 11:38

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.)

• I only works for primes though, but that's a progress. Jan 25 '15 at 14:25
• But you know that if $|g|$ does not divide $m$, then it should have a prime divisor that does not divide $m$ also and by Cauchy's theorem, for every prime $p$ dividing the order of the group $G$, it has an element of order $p$. So, I think there no problem in taking the order of the element as a prime. If I'm wrong, pls correct.
– sez
Jan 26 '15 at 16:15