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I'm trying to write this proof, but can't seem to finish it. I've tried to suppose that $a,b \in H$ and

$$ |a| = 2k+1 $$ $$ |b| = 2n+1 $$

with $n,k \in 2\mathbb{N}$. I also know that $|b^{-1}| = |b|$ so I tried

$$ (a)^{2k+1}(b^{-1})^{2n+1} = e $$ Suppose WLOG that $k < n$, and use the fact that $G$ is abelian. Then

$$ (ab^{-1})^{2k+1}(b^{-1})^{2(n-k)} = e $$

At which point I am stuck. I want to have

$$ (ab^{-1})^{2k+1} = e $$

But I have an even number of left-over $b^{-1}$ to deal with.

gen-ℤ ready to perish
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Anthony P
  • 609

1 Answers1

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The order of a product $ab$ in an abelian group is a divisor of the least common multiple of the orders of $a$ and $b$, i.e., ${\rm ord}(ab)\mid {\rm lcm}(({\rm ord}(a),{\rm ord}(b))$, see here:

Order of product of two elements in a group

Let $m={\rm ord}(a), n={\rm ord}(b)$, and $d={\rm ord}(ab)$. Then $$d|\frac{mn}{\gcd(m,n)}=\text{lcm}(m,n),\quad\frac{mn}{\gcd(m,n)^2}|d.$$

Furthermore the least common multiple of two odd numbers is odd:

Is the lowest common multiple of two odd numbers, an odd number?

Dietrich Burde
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  • Unfortunately, the claim about the lcm only holds if the subgroups generated by the elements intersect trivially, otherwise it may happen that one just gets a divisor in that (which still suffices for this purpose, as it will still be odd). – Tobias Kildetoft Jan 29 '18 at 19:15
  • @TobiasKildetoft Yes, you are right, thank you.I replaced the incorrect link; I will be more careful next time. – Dietrich Burde Jan 29 '18 at 20:03