elements of order 2 in a ring Let $m $ be a positive integer number and $(R,+,\cdot)$ be a ring that has $2m$ elements. Prove that the sum of all elements from $(R,+,\cdot)$ is equal with $1_{(R,\cdot)}+1_{(R,\cdot)}+1_{(R,\cdot)}+...+1_{(R,\cdot)}$ where we have in total $m$ summands and $$1_{(R,\cdot)}$$ represents the identity element of the monoid $(R,\cdot)$.
I have searched more about the sum of elements in a ring, yet I did not find anything helpful in this respect. What I found is that the product of elements of a multiplicative group is equal with the identity element of that group except for when there is exactly one element of order 2.
However, this didn't solve the problem unless I manage to prove that:

*

*If there is exactly one element of order 2 in the group $(R,+)$ then that element must be $1_{(R,\cdot)}+1_{(R,\cdot)}+1_{(R,\cdot)}+...+1_{(R,\cdot)}$, where we have in total $m$ summands

*If there is no element of order 2 or there is more than one element of order 2 then $1_{(R,\cdot)}+1_{(R,\cdot)}+1_{(R,\cdot)}+...+1_{(R,\cdot)}$
must be equal with $0$.

I do not know what to do with this. I would really appreciate any help, and I am looking forward to a complete and elementary proof.
 A: Note that there definitely is an element of order $2$, since the order of the additive group underlying $R$ is divisible by $2$.  We shall be more precise than this below.
Write $n$ for the additive order of $1_R$, so that $n$ divides $\lvert R\vert = 2m$.  As a finite Abelian group, the additive group underlying $R$ has a so called invariant-factor decomposition:  that is, we may write it as a direct sum of finite cyclic groups $C_1, \dotsc, C_k$ so that the order of $C_i$ divides that of $C_{i + 1}$ for all $i \in \{1, \dotsc, k - 1\}$.  In particular, the order $2m = \prod_{i = 1}^k \lvert C_i\rvert$ of $R$ divides $\lvert C_k\rvert^k$, so $\lvert C_k\rvert$ is even, so $C_k$ contains an element of order $2$.  Since $R$ contains an element $1_R$ of additive order $n$, we have that $n$ divides $\lvert C_k\rvert$; but, since the additive group underlying $R$ is $n$-torsion, we have that $\lvert C_k\rvert$ divides $n$.  Thus, $\lvert C_k\rvert$ equals $n$.
If $n$ does not divide $m$, then $2m/n = \lvert R/C_k\rvert$ is odd, so all even-order elements of $R$ are contained in $C_k$.  In particular, since $C_k$ is cyclic of even order, there is a unique such element.  Moreover, since the additive order $n$ of $1_R$ does not divide $m$, we have that $m 1_R$ is non-$0_R$ but $2m 1_R = (2m/n)(n 1_R)$ equals $0_R$, so $m 1_R$ is the unique element of $R$ of additive order $2$.  In this case, as you have observed, we are done.
If $n$ does divide $m$, then $m 1_R = (m/n)(n 1_R)$ equals $0_R$; and the order $\lvert R/C_k\rvert = 2(m/n)$ of $\bigoplus_{i = 1}^{k - 1} C_i$ is even.  It follows that $C_k$ and $\bigoplus_{i = 1}^{k - 1} C_i$ both contain (necessarily distinct) elements of order $2$.  Again, as you observed, we are done.
