# If the number of units of a ring is odd, then the ring has cardinality as a power of two [duplicate]

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.

• Try proving the contrapositive. – Qiaochu Yuan Jul 16 '20 at 8:28
• @QiaochuYuan so, if a ring is not a power of $2$, then it still can have an odd number of units right? why should always have an even number of units? Should we assume commutativity of rings here? – vidyarthi Jul 16 '20 at 8:46
• No, commutativity is unnecessary. It is in fact true that if a finite ring has cardinality that's not a power of $2$ then it has an even number of units, and as a hint, the proof is short. – Qiaochu Yuan Jul 16 '20 at 8:51
• @Bernard well, here I assume the ring is finite. Edited the poat. – vidyarthi Jul 16 '20 at 9:48
• The early parts of this answer also answer your question. Not sure whether call this a duplicate or not. – Jyrki Lahtonen Jul 16 '20 at 18:20

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 $$2$$.
• @vidyarthi: by hypothesis, the order of $R^{\times}$ is odd, so it cannot contain a subgroup of order $2$ (namely, the injective image of $\mathbb{Z}^{\times}$). – Alex Wertheim Jul 16 '20 at 9:18
• @vidyarthi: For any ring homomorphism $\alpha \colon S \to T$, the canonical induced map $S/\mathrm{ker}(\alpha) \to T$ is always injective. (But as a cheap aside: a (unital) ring morphism from a field to any ring is always injective.) – Alex Wertheim Jul 16 '20 at 9:50
• Stahl: haha, yes, careless of me! @vidyarthi: yes, $\varphi(-1) = \varphi(1)$. But $1$ and $-1$ are identified in the quotient $\mathbb{F}_{2} = \mathbb{Z}/2\mathbb{Z}$, so the induced map $\mathbb{F}_{2} \to R$ is injective. – Alex Wertheim Jul 16 '20 at 9:58