TITLE: $|G| + \frac{|G|}{\left|\langle a\rangle\right|} + \frac{|G|}{\left|\langle b\rangle\right|} + \frac{|G|}{\left|\langle ab\rangle\right|}$
QUESTION [16 upvotes]: Show that for every finite group $G$ and for every elements $a, b \in G$ the following expression
$$
|G| + \frac{|G|}{\left|\langle a\rangle\right|} + \frac{|G|}{\left|\langle b\rangle\right|} + \frac{|G|}{\left|\langle ab\rangle\right|}
$$ is even.

REPLY [5 votes]: I think I got it. Check it please.

If $|G|$ is odd. Obvious.
$|G|$ is even. Then let's assume embedding $f:G \rightarrow S_{|G|}$. According to Cayley theorem $g$ maps to the product of $\frac{|G|}{\left|\langle g\rangle\right|}$ independent cycles of length $\left|\langle g\rangle\right|$. In particular $g$ maps to the odd permutation iff $|G|$ is even and $\frac{|G|}{\left|\langle g\rangle\right|}$ is odd. 
There are either $2$ or $0$ odd permutations among $f(a), f(b), f(ab)$, i.e. there are zero or two odd summands in this sum ($|G|$ is even).