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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 ?

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    $\begingroup$ Did you want to restrict to finite abelian groups? If not, then I suspect that this questions gets a lot more complicated. $\endgroup$
    – Joe Silverman
    Commented Oct 8, 2020 at 11:54
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    $\begingroup$ If anything I am interested in the case of infinity abelian groups. $\endgroup$
    – M masa
    Commented Oct 9, 2020 at 9:03

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I am going to assume that by "algebra" you simply mean a ring.

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 $n$.

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  • $\begingroup$ Why does $-u \in R^\times$ force $1 = -1$? $\endgroup$
    – LSpice
    Commented Oct 8, 2020 at 12:48
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    $\begingroup$ @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. $\endgroup$
    – Alex B.
    Commented Oct 8, 2020 at 13:10

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