I'm fairly sure that my basic reasoning is correct, but I'm worried about a lack of rigor in my answers. Nitpicking would be highly appreciated. The exercises are from Aluffi's Algebra Chapter 0.
Problems.
1.8 Let $G$ be a finite abelian group, with exactly one element $f$ of order 2. Prove that $\prod_{g\in G}g=f$
1.9 Let $G$ be a finite group, of order n, and let m be the number of elements $g \in G$ of order exactly 2. Prove that n-m is odd. Deduce that if n is even, then G necessarily contains elements of order 2.
Solutions.
1.8 Every other element has a distinct inverse (or is the identity), and so gets cancelled out, but $f$, being its own inverse, remains uncancelled.
Consider a product of elements $g_1 g_2 .. f .. g_{n-1}$ where $n=|G|$. Now since $G$ is abelian, we may reorder this product as we choose. One of the $g_k$ is the identity, so we may ignore that. Now consider some other $g_k$ which is neither the identity nor $f$. Thus this $g_k$ has order greater than 2. This means that $g_k$ has an inverse in $G$ which is different than itself. (If it was its own inverse then its order would be 2). So we may place the elements $g_k$ next to their inverses in the product, so that the whole product becomes $eee..f = f$.
1.9 We will try to pair elements with their inverses. $n-m$ is the number of elements that have order not equal to $2$. We can split this into two types: elements with order greater than $2$, and the identity. So it suffices to show that there is an even number of elements with order greater than $2$, since there is only $1$ identity. Consider the distinct elements of the group with order greater than $2$. Since they have order greater than $2$, by the same logic as given in the previous exercise, they must have a inverse which is different than itself and shared by no other element. So we can pair each element with its inverse and obtain a list containing all the elements of the group with order greater than two in pairs. Thus there is an even number of elements in the group with order greater than two.