0

Let A be a finite abelian group. Let m be the smallest natural number such that ma=0 for every a in A. Prove that there is an element in A of the order m

user3395933
  • 97
  • Although this is an interesting question, you should show some of your efforts trying to solve this problem. So: what have you tried? – Crostul Jul 11 '15 at 07:46
  • We assumed negatively that there aren't any elements in A from the order m---> every element in A has an order smaller than m. Let k (<m) be the maximal order of every element in A. m=nk+r (n is natural number and r<k)---->ma=(nk+r)a=nka+ra=0---->r=0. m=nk. thats as far as I got. – user3395933 Jul 11 '15 at 07:53
  • @Crostul "you should show some of your efforts trying to solve this problem." Did you always follow your own advice? –  Jul 15 '15 at 04:50
  • 4
    @billford This is a standard question to ask in such a situation for lower rep users. It shows good faith on part of the poster. – Cameron Williams Jul 15 '15 at 04:56
  • @CameronWilliams So it's a rep problem. What if I had 100,000 reps? You would be a low rep user for me. Would it be OK for me to ask you to show your effort every time you ask a question? –  Jul 15 '15 at 05:09
  • 2
    @billford You're getting into an argument that is only going to end poorly for everyone involved. It is not discriminatory. There are lots of users who simply ask a few homework questions on this site without showing any effort and then never contribute again. Once a user has reputation in the several hundreds, there is an implicit trust developed. Even users who are as established as I am have to ask questions that don't lend themselves to much effort shown (see my recent post about involutions), however it is assumed that the poster has worked sufficiently hard on the problem. – Cameron Williams Jul 15 '15 at 05:19
  • @CameronWilliams You seem to think that a poster should pay some effort before asking a question. What's wrong with posting a question without paying much effort? Let me guess. You just don't like giving away an answer for free. I think it's not very broadminded. –  Jul 15 '15 at 06:09
  • @user3395933 You should add how you tried to solve the question in your question to avoid the closure. –  Jul 16 '15 at 22:23

2 Answers2

2

use the fundamental theorem of finitely generated Abelian groups: The invarint factor version. G is nothing but a k finite direct sum of Z_n's ordered in an ascending order of divisibility of the n's. Then certainly there is a summant Z_n. Now take a k-tuple with all entries 0 except one entry corresponding to Z_n in which it is 1.

Adelafif
  • 1,275
  • The fundamental theorem of finite Abelian groups will suffice :-) – joriki Jul 11 '15 at 16:16
  • @joriki What if $G$ is an infinite abelian group? I think the assertion still holds. –  Jul 15 '15 at 05:44
  • @billford: The original question was about finite Abelian groups. It would have to at least be modified even for finitely generated Abelian groups, since in this case there may not be $m$ such that $ma=0$ for all $a\in A$. I do suspect that if there is such an $m$, the assertion still holds even in non-finitely generated Abelian groups, but I don't know. In any case, my remark above was just for the sake of economy of proof :-) – joriki Jul 15 '15 at 05:53
  • @joriki I meant what if $G$ is an infinite abelian group under the condition that there is an integer $n\ge 1$ such that $n G = 0$. –  Jul 15 '15 at 05:56
0

Here is an elementary proof.

Suppose $m=p^{r_1}_{1}⋯p^{r_s}_{s}$, where $p_i$ are distinct primes and $r_i \geq 1$. Suppose for a moment for each $i$, there is an element $a_i$ of order $p^{r_i}_{i}$. I claim that $a = a_1\cdots a_s$ has the order $m$. Let $n\ge 1$ be an integer. Suppose $a^n = 0$. Since $a^m = 0$, it suffices to prove that $n$ is divisible by $m$. Since the claim is obviously true if $s = 1$, we assume that $s\ge 2$. Since $a^n = 0, {a_1}^n = {a_2}^{-n}\cdots {a_s}^{-n}$. Hence $({a_1}^n)^{p^{r_2}_{2}⋯p^{r_s}_{s}} = 0$. Hence $np^{r_2}_{2}⋯p^{r_s}_{s}$ is divisible by $p^{r_1}_{1}$. Thus $n$ is divisible by $p^{r_1}_{1}$. Similarly $n$ is divisible by $p^{r_i}_{i}$ for all $i$. Hence $n$ is divisible by $m$ as claimed.

It remains to prove that for each $i$, there is an element of order $p^{r_i}_{i}$. Since $m$ is the LCM of orders of all the elements of the group, there exists an element whose order is divisible by $p^{r_i}_{i}$. Hence there exists an element whose order is $p^{r_i}_{i}$ and we are done.