# Algebra with a certain abelian group as the multiplicative group

Let $$A$$ be an abelian group. Are there an algebra $$\mathfrak{X}(A)$$ s.t. the multiplication group is isomorphic to A ? i.e. $$\mathfrak{X}(A)^{\times} \simeq A.$$

For example, for $$A=\mathbb{Z}/4\mathbb{Z}$$, $$\mathfrak{X}(A)=\mathbb{F}_{5}$$. I want to know the sufficient conditions of $$A$$ for existence of $$\mathfrak{X}(A)$$. Is it well-known ?

• Did you want to restrict to finite abelian groups? If not, then I suspect that this questions gets a lot more complicated. – Joe Silverman Oct 8 at 11:54
• If anything I am interested in the case of infinity abelian groups. – M masa Oct 9 at 9:03

The answer is "no", in general. For example $$\mathbb{Z}/5\mathbb{Z}$$ is not the unit group of a ring. Indeed, suppose it was the unit group of a ring $$R$$, let $$u\in R$$ denote a generator of the unit group. Now $$-1$$ is always a unit, and has order dividing $$2$$. But it is supposed to live in a cyclic group of order $$5$$, which forces its order to be $$1$$, i.e. $$1=-1$$ in $$R$$, so that $$R$$ has characteristic $$2$$. But then $$u$$ generates a subring of $$R$$ that is a quotient of $$\mathbb{F}_2[\mathbb{Z}/5\mathbb{Z}]\cong \mathbb{F}_2[x]/(x^5-1)\cong \mathbb{F}_2\times \mathbb{F}_{16}$$. Since that quotient has at least $$5$$ distinct elements, it must be at least all of $$\mathbb{F}_{16}$$, so that $$R^\times$$ contains $$\mathbb{F}_{16}^\times\cong \mathbb{Z}/15\mathbb{Z}$$ — contradiction.
In general, this argument shows that a group of prime order $$p$$ is isomorphic to the unit group of a ring if and only if $$p=2$$ or of the form $$2^n-1$$ for some $$n$$.
• Why does $-u \in R^\times$ force $1 = -1$? – LSpice Oct 8 at 12:48
• @LSpice: we are assuming that $R^\times=\{1,u,\ldots,u^4\}$, so we must have $-u=u^i$ for some $i$. Since $u$ is a unit, this means that $-1=u^{i-1}$ for some $i$. But $-1$ has order dividing $2$, and we are in a cyclic group of order $5$, so $-1$ must have order $1$, i.e. be equal to $1$. I guess, I could have bypassed the whole $-u$ business. Let me edit. – Alex B. Oct 8 at 13:10