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.
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?