Consider the additive abelian group $(\mathbb{R},+)$. Does there exists a binary operation $\circ:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ such that the following holds

  • $(\Bbb{R},\circ)$ is a group.

  • For all $S\subseteq \mathbb{R}$ the subgroup generated by $S$ in $(\mathbb{R},+)$ is equal (as a set) to the subgroup generated by $S$ in $(\mathbb{R},\circ)$.

  • $(\mathbb{R},+)$ is not isomorphic to $(\mathbb{R},\circ)$.

This question is a follow up of this MSE post of mine and due to it I guess that the answer to both the questions are negative. But I can't construct such a group (or prove the existence of such one).

Thanks for any help.

  • 2
    $\begingroup$ The second condition is equivalent to the requirement that subgroups are the same for both group laws. $\endgroup$
    – YCor
    Aug 25, 2018 at 12:36
  • $\begingroup$ @DerekHolt this is my expectation and I can check it when $\circ$ is assumed to be abelian. $\endgroup$
    – YCor
    Aug 25, 2018 at 12:39
  • $\begingroup$ I deleted the comment because I was assuming that the other group was abelian. $\endgroup$
    – Derek Holt
    Aug 25, 2018 at 12:40
  • 2
    $\begingroup$ the $\circ$ group generated by each element is isomorphic to the + group generated by it, as Z is the unique group having a unique subgroup of each index. A similar argument for Z^2 solves the problem. $\endgroup$
    – Uri Bader
    Aug 25, 2018 at 12:41
  • 1
    $\begingroup$ Related: mathoverflow.net/questions/312177/… $\endgroup$
    – YCor
    Oct 7, 2018 at 21:52

1 Answer 1


The answer is no: every such group law is isomorphic to the standard law.

Let me prove something stronger: the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{R})$ characterizes $\mathbf{R}$ up to group isomorphism.

Let us first check it among abelian groups. Let $G$ be a group with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{R})$ as posets. The absence of nontrivial minimal elements characterizes torsion-freeness, so $G$ is torsion-free. Since $G$ is abelian, the absence of nontrivial maximal element characterizes the absence of proper subgroups of finite index, and hence $G$ is divisible. So $G$ is isomorphic to $\mathbf{Q}^{(I)}$ for some cardinal $I$. Cyclic subgroups can be characterized, among abelian groups, as those groups with infinitely many subgroups, and in which every nontrivial subgroup is contained in only finitely many subgroups. The number of cyclic subgroups of $\mathbf{Q}^{(I)}$ for nonempty $I$ is equal to $\max(\aleph_0,|I|)$. So here we have $|I|=c$.

This concludes the abelian case.

In the nonabelian case, we need two lemmas (of independent interest, and probably proved somewhere in the literature).

(1) the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{Z}^2)$ characterizes $\mathbf{Z}^2$ up to group isomorphism.

In a lattice $X$ with maximal element $q$, let us say that $x$ satisfies Property $A_p$ ($p$ a given prime) if the interval $[x,q]$ consists of $x$, $q$, and exactly $p+1$ pairwise non-comparable elements. Then in $\mathrm{Sub}(\mathbf{Z}^2)$, it is easy to check that the only subgroup satisfying $A_p$ is $p\mathbf{Z}^2$.

Consider $G$ with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{Z}^2)$. Then $G$ has a unique subgroup $G_p$ satisfying $A_p$ for every $p$, and $\bigcap_p G_p=\{1\}$ (since the corresponding property holds in $\mathbf{Z}^2$). In particular, $G_p$ is a normal subgroup. The quotient $G/G_p$ has finitely many subgroups and hence is a finite group. So $G$ is residually finite.

If by contradiction $G$ were non-abelian, there would exist $p$ such that $G/G_p$ is non-abelian; take $p$ minimal for this. The finite group $G/G_p$ is nonabelian and has all its proper subgroups cyclic of prime order. The only possible groups having this property have the form $C_q\ltimes C_{q'}$ with $q,q'$ prime and $q$ divides $q'-1$. It has $q'+3$ subgroups ($q'+1$ is we exclude the trivial and whole subgroups), so $q'=p$. Now $G/(G_p\cap G_q)$ inherits from the corresponding group in $\mathbf{Z}^2$, the property of having exactly $(p+3)(q+3)$ subgroups, including $p+q+2$ maximal ones. On the other hand, in $G$, the uniqueness of a subgroup with Property $A_q$ implies that $C_q^3$ is not a quotient of $G$; it follows that $G/(G_p\cap G_q)$ has the form $C_q^2\ltimes C_p$. This group has $q+1$ maximal subgroups that are normal, and $p$ maximal subgroups that are non-normal (the $q$-Sylow subgroups), so the total is $p+q+1$, which does not match: contradiction.

Now $G$ is abelian, torsion-free. The (at most) uniqueness of subgroups with Property $A_p$ therefore applies to subgroups of $G$ and hence $G$ has $\mathbf{Q}$-rank $\le 2$. Since $G/G_p\simeq C_p^2$ for all $p$, we deduce $G\simeq\mathbf{Z}^2$. $\Box$

(2) the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{Z})$ characterizes $\mathbf{Z}$ up to group isomorphism.

Let $G$ be such a group. Then $G$ is torsion-free. Say that a subgroup has Property B if it is contained in finitely many subgroups. From the property in $\mathbf{Z}$, we deduce that every nontrivial subgroup has Property B, and that the intersection of any two nontrivial subgroups has Property B. We deduce that $G$ has a commensurated subgroup $N$ isomorphic to $\mathbf{Z}$. The germ at infinity of the action on $N$ yields a homomorphism $G\to\mathbf{Q}^\times$. The image has to be contained in $\{\pm 1\}$: otherwise some element of $G$ acts by some element $\neq\pm 1$, and hence generates a cyclic subgroup with trivial intersection with $N$. Also, since $N$ has Property B, we see that $G$ is finitely generated. We deduce that some finite index subgroup of $N$ is normal, say $N$ itself (up to rename). Since $G/N$ has finitely many subgroups, it is finite, and hence $G$ is virtually cyclic and torsion-free (and infinite). This is known to imply that $G$ is infinite cyclic. $\Box$

Now let us conclude. Consider $G$ with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{R})$. So $G$ is torsion-free. By the lemma, we can recognize subgroups of $G$ isomorphic to $\mathbf{Z}$ and $\mathbf{Z}^2$; in particular, we inherit from $\mathbf{R}$ the property that any two infinite cyclic subgroups is contained in a subgroup isomorphic to $\mathbf{Z}^2$. Hence $G$ is abelian. Since the abelian case is done, this ends the proof.

Edit: Here's a reference: the book by M. Suzuki, Structure of a Group and the Structure of its Lattice of Subgroups, Ergebnisse der Mathematik und Ihrer Grenzgebiete 10, Springer-Verlag Berlin Heidelberg, 1956

Theorem 3, page 35 (attributed to R. Baer, 1939), says that for any abelian group $G$ of $\mathbf{Q}$-rank $\ge 2$ (i.e., containing a subgroup isomorphic to $\mathbf{Z}^2$), any "projectivity" of $G$ (that is, an isomorphism $\mathrm{Sub}(G)\to\mathrm{Sub}(H)$ for another group $H$), is induced by a unique (up to pre-composition by $x\mapsto -x$) isomorphism $G\to H$. In particular, $H$ is isomorphic to $G$.

(While the "in particular" assertion holds for $G=\mathbf{Z}$ by Lemma (2) above, the stronger assertion is false for $G=H=\mathbf{Z}$, as the lattice of subgroups has an uncountable automorphism group then.)

  • 1
    $\begingroup$ Can we conclude that $x\circ y=x+y$ for all $x,y\in\mathbb R$? $\endgroup$ Aug 25, 2018 at 14:35
  • 8
    $\begingroup$ @TarasBanakh No: conjugate the standard law by any permutation exchanging $t$ and $-t$ for some given subset of positive reals $t$. $\endgroup$
    – YCor
    Aug 25, 2018 at 14:35
  • $\begingroup$ If in the problem we replace $(\mathbb{R},+)$ by any group $(G,\circ_G)$ then what is the reason for which the generalized problem fails to be true? $\endgroup$
    – user57432
    Aug 26, 2018 at 17:00
  • $\begingroup$ @user170039 this depends on what you want to call "the generalized problem". $\endgroup$
    – YCor
    Aug 26, 2018 at 17:21
  • 1
    $\begingroup$ @user170039 OK; I thought it could be related to the notions of my post since you ask it here rather than as a comment to the question. Consider a Tarski monster on some prime $p$. The subgroups are: $\{1\}$, the all group, and others, minus 1, form a partition of $G-1$ into countably many subsets of cardinal $p-1$. Since there are several (uncountably) many pairwise non-isomorphic Tarski monsters on $p$ (for large $p$), this yields a answer to your question. I don't have simpler examples in mind for the moment. $\endgroup$
    – YCor
    Aug 27, 2018 at 14:10

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