Let $(G,+)$ be a finite abelian group. I want to prove that $|G/2G|$ is equal to the number of the elements of $G$ of order $2$ plus the identity element.
I tried to count the number of elements of $G/2G$. I aslo tried to invoke an application of the first isomorphism theorem, but in both cases my attempts were not fruitful.
Do you have any hint, reference or proof for this algebraic fact? Thank you in advance for your valuable help.
Consider the map $f:G\to G$ defined by $f(a)=2a$. As $G$ is Abelian, this is a homomorphism. Its kernel is $G_2=\{a\in G:2a=0\}$. Its image is $2G$. By the First Isomorphism Theorem, $2G\cong G/G_2$. Therefore $|2G|=|G|/|G_2|$, equivalently $|G_2|=|G|/|2G|$.