# Math Help - group theroy question

1. ## group theroy question

Let a be an element of maximum order from a finite Abelian group G. Prove that for any element b, |b| divides |a| (order of b divides order of a). Show by example that this need not be true for finite non-Abelian Groups.
I'am really stuck on this one. Any help would be great

2. Hi

Hint: In a group, let $g,f$ be two elements of finite order which commute, what is the order of $gf$ ?

For the non-abelian counterexample, you can take a look at $S_3$.

3. If the order of g=n and the order of f=m, then the order of gf would divide m*n. So if an element does not divide the element of maximum order in a finite Abelian group G, the order of fg>f when f is the maximal order element, this would result in a contradiction. Is this right?

4. also, are all finite abelian groups cyclic?

5. Originally Posted by wutang
also, are all finite abelian groups cyclic?

No: $\{e,a,b,ab\}$ is an abelian non-cyclic group, with $ab=ba$

Tonio

6. Originally Posted by clic-clac
Hint: In a group, let $g,f$ be two elements of finite order which commute, what is the order of $gf$ ?
i don't think anybody knows the answer! you probably had this in your mind that the answer is the lcm of the orders, but that is not necessarily true even for abelian groups. for example look at

the case $g=f^{-1}.$ anyway, what wutang needs here is this fact that for any two elements $a,b$ in a finite abelian group, there exists an element $c$ such that $|c|=\text{lcm}(|a|,|b|).$ see here.

7. Ouch that will teach me to think more before posting, thank you.

8. NonCommAlg could you help explain to me how the results in your link would result in a contradicition if |b| does not divide the order of |a| when a is an element of maximum oirder in a finite abelian group?

9. Originally Posted by wutang
NonCommAlg could you help explain to me how the results in your link would result in a contradicition if |b| does not divide the order of |a| when a is an element of maximum oirder in a finite abelian group?
choose $c \in G$ such that $|c|=\text{lcm}(|a|,|b|).$ now if $|b|$ does not divide $|a|,$ then $|c|=\text{lcm}(|a|,|b|) > |a|$ and so $a$ cannot have the maximum order. contradiction!

10. Originally Posted by clic-clac
Ouch that will teach me to think more before posting, thank you.
it's ok. this is one of the most notorious mistakes in (elementary) group theory that even some graduate students make! i don't know why?

11. Originally Posted by NonCommAlg
it's ok. this is one of the most notorious mistakes in (elementary) group theory that even some graduate students make! i don't know why?
Probably because it looks so convincing, especially to those smart people. When I was grading intro algebra homework (when I was the TA), the people who made this mistake were all top students in the class.