TITLE: If the number of units of a ring is odd, then the ring has cardinality as a power of two
QUESTION [9 upvotes]: If the number of units of a finite ring is odd, then  does the ring has cardinality as a power of $2$?
I think yes. For fields, it is trivial. For non-fields, it is a hard question for me. I saw a paper here that sates that an odd number is the cardinality of the group of units of a ring if it is of the form $\prod_i (2^{n_i}-1)$. But, that proof is quite lengthy, and still the ring need not be a power of $2$. Any short proof? Thanks beforehand.

REPLY [16 votes]: Consider the canonical ring morphism $\varphi \colon \mathbb{Z} \to R$. Since $\mathbb{Z}^{\times} = \{-1, 1\}$, the induced group morphism $\mathbb{Z}^{\times} \to R^{\times}$ must be trivial by Lagrange, so $\varphi(1) = \varphi(-1) = 1$. In particular, $\varphi$ factors through an injective morphism $\mathbb{F}_{2} \to R$, so $R$ is an $\mathbb{F}_{2}$-vector space, and thus must have cardinality a power of $2$.