How to show that $2$ is invertible in a ring with odd cardinality?

Question

Let $R$ be a commutative ring with unity that has an odd number of elements. Show that $2$ is invertible in $R$.

Attempt

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I've found that $2 \ne 0$ from Lagrange, since the order of $1$ in the additive group of $R$ must be odd.

Answer 1

Because $R$ is finite, $2^m = 2^n$ for some $m \lt n \in \Bbb Z^+$. Thus, $2^m(2^{n-m}-1)=0$, so either (a) $2^{n-m}-1=0$ or (b) it has even order in the additive group of $R$. But option (b) is impossible since $R$ has odd order, so option (a) tells us $2^{n-m}=1$ with $n-m \geq 1$. Moreover, since $2 \neq 1$, in fact we know $n-m \gt 1$ and $2$ is invertible because $2 \cdot 2^{n-m-1} = 1$.

Answer 2

Let $R$ be a commutative ring with $|R|$ odd. The image of the unique map $\Bbb{Z}\ \longrightarrow\ R$ is a subring of $R$ containing the element $2$. This subring is isomorphic to $\Bbb{Z}/n\Bbb{Z}$ for some $n$, where $n$ is odd because $|R|$ is odd. Now it suffices to show that $2$ is invertible in $\Bbb{Z}/n\Bbb{Z}$.