Take a group $G$, and consider the set of all its subgroups. Is there a natural way to define multiplication of subgroups, in such a manner that the set forms a group? If so, how is the operation constricted, and what is this group called?

The reason I came up with the question and why it might seem natural is this. A consequence of the isomorphism theorems is that $HH'/H\cong H'$ whenever $H,H'$ are disjoint subgroups and $H$ is a normal subgroup of $G$, and furthermore, if $H,H'$ both happen to be normal subgroups in $G$ with $H\leq H'\leq G$, then $(G/H)/(H'/H)\cong G/H'$. So in some twisted sense, it seems like for at least a certain situations and for certain types of subgroups of $G$, there may be a natural method of defining multiplication that allows us to actually "do arithmetic" over the subgroups of $G$? This is in the sense that, over this proposed group, we can actually just "cancel $H$" in $(G/H)/(H'/H)\cong G/H'$ as a consequence of how multiplication is defined. It seems very interesting (to me, at least) what the consequences of defining such a group might be.

Sorry if the question is vague, but here's an idea of what I'm not looking for: Take any group, consider all its $n$ subgroups, and define multiplication in some arbitrary way, such as the multiplication over $\mathbb Z/n\mathbb Z$. If we do this, we completely lose the fact that these elements started out as subgroups of a group, and the interesting algebraic structure is lost! Instead, I'm interested in whether there is such a way of defining this group of subgroups that preserves the algebraic properties of the original group $G$.

A naïve construction I tried that doesn't work: for two subgroups $H,H'$ of an abelian group $G$, define $HH'=\{hh':h\in H,h'\in H'\}$. It's quite easy to verify this always gives a subgroup of $G$, so we have closure, and associativity and such are easily checked too. The problem is that most subgroups of $G$ have no inverse in this "group" of subgroups for a general group $G$, because for $HH'=1$, we require $hh'=1$ for every $h\in H$, $h'\in H'$ which is not possible in all but the most trivial cases. As noted in the comments, if we removed the hypothesis that $G$ is abelian, then its subgroups need not be normal, and this construction fails even more badly since we don't even have closure anymore. So in general, this construction would not go anywhere close to working.

Any thoughts, or links to something that has already been done, are greatly appreciated!

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    $\begingroup$ Why is $HH’$ always a subgroup? $\endgroup$
    – Randall
    Jan 5, 2019 at 0:27
  • $\begingroup$ @Randall sorry, I missed the hypothesis that $H'$ is normal in $G$. (Actually, I started out writing the post assuming $G$ was abelian, but forgot to clean up when I removed that hypothesis.) Edited to fix that issue. So that's one more problem with that construction! $\endgroup$
    – YiFan Tey
    Jan 5, 2019 at 0:33
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    $\begingroup$ At any rate the answer is no for reasons you already pointed out. You can look up the Groethendieckization of a monoid, which is an extension of what you want (sort of). $\endgroup$
    – Randall
    Jan 5, 2019 at 0:35
  • $\begingroup$ @Randall what reason did I point out? As far as I know it seems like I only proved a particular construction doesn't work; is there a reason no other constructions can work? (Again, sorry if "work" sounds vague. I'm honestly not sure how to more precisely phrase the question.) $\endgroup$
    – YiFan Tey
    Jan 5, 2019 at 0:37
  • 1
    $\begingroup$ Try googling "Groethendieck group" $\endgroup$ Jan 5, 2019 at 1:13

2 Answers 2


I think for such a construction to be called "natural", it would need to have at least the following the property:

The induced action of $\mathrm{Aut}(G)$ on the set of subgroups of $G$ induces a subgroup of the automorphism group of the group of subgroups.

(For example, subgroups of $G$ that are "the same" with respect to the structure of $G$ should become elements of the group of subgroups that are also "the same" with respect to its structure, i.e. subgroups of $G$ that are conjugate under $\mathrm{Aut}(G)$ should also be conjugate under the automorphism group of the group of subgroups.)

But this is not always possible. For example, let $G$ be the Klein group. It has $5$ subgroups, so its group of subgroups should be cyclic of order $5$, but the automorphism group of the Klein group is $S_3$ which does not embed into the automorphism group of $C_5$.

  • $\begingroup$ Nice +1. Just to add a little: There is a natural map $Aut(G) \to Bij(SG)$ where $SG$ is the set of subgroups, $\phi\mapsto [H\mapsto \phi(H)]$. This map is not always injective (consider $C_p\times C_q$ where if $p\neq q$ are prime it's zero) as nontrivial automorphisms may preserve all subgroups. In the case of the Klein group, the map does embed $S_3 \to Bij(SG)$. So the naturalness criterion here is that $Image(Aut(G)) \subset Aut(SG)$ inside of $Bij(SG)$. $\endgroup$
    – Ben
    Jan 5, 2019 at 10:31

We can open up our conditions a little bit to get something "natural." Namely, we can take the vector space $\mathcal{S}(G)$ spanned by the subgroups of $G$, over some field $F$, with the product being the linear extension of $$A\cdot B = AB$$ Then if we have a homomorphism $f:G\to H$, we can extend this to a homomorphism of $F$-algebras $f_*:\mathcal{S}(G)\to\mathcal{S}(H)$ given by $$f_*(A) = f(A)$$ In particular, automorphisms $a:G\to G$ induce automorphisms $a_*:\mathcal{S}(G)\to \mathcal{S}(G)$.

Note this only works directly for abelian groups, or the rare nonabelian groups like the quaternion group where every subgroup is normal. If you want to do this for general groups, you have to extend the vector space to consist of all nonempty subsets of the group. Either that, or you can declare the subgroups to just be generators. Then the vector space is actually larger, with basis elements that are not subgroups but rather are products of subgroups. Then you'll end up with something noncommutative, with $HK\neq KH$ if $H$ and $K$ are not normal. Then normal subgroups will be in the center of the algebra.

Note that a subgroup in this case is never invertible. In fact, subgroups are idempotent elements, with $A^2=A$ for all subgroups $A$. Then $$A(A-\{e\}) = 0$$ for all $A$, so every subgroup is a zero divisor. We thus lose invertibility, but we do have an algebraic construction that preserves homomorphisms (in the category theory sense, it is functorial). You also maintain your equation given by the isomorphism theorem. Namely, if $\langle H-\{e\}\rangle$ is the ideal generated by a normal subgroup, then $\mathcal{S}(G)/\langle H-\{e\}\rangle\cong \mathcal{S}(G/H)$, and $HH'\mapsto (HH')/H$, which is the same as the image of $H'$. This is really isomorphic to $H'/(H\cap H')$ however, and this is the subgroup it maps to in the quotient. If $H$ and $H'$ intersect trivially, this is just $H'$.

In general, if we want to do constructions involving groups we may get to a point where we can't do what we want because groups are extremely rigid. However, we have a lot of freedom with algebras over a field, and this can is an example where it allows us to basically do what we want. A similar situation is that of quantum groups, where we deform a group to obtain a $q$-analog.


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