Normal Subset of a Group

We attempt to generalize the notion of “normal subgroup,” used in studying kernels and homomorphisms, to a “normal subset” that does not have the requirement of being a subgroup, and we see whether it is possible for a normal subset to exist that is not a normal subgroup.

Definition. Let $G$ be a group. A normal subset $S \subseteq G$ is such that, for all $g \in G,s \in S,$ we have $g^{- 1}sg \in S$ .

It turns out that with this $S$ could be empty, making it much harder to develop the theory of $S$ . So we also require $S$ to be nonempty.

Conjecture. $S$ is in fact a subgroup.

Attempt. There exists $s \in S$ . Can we show the identity must be in $S$ ?

Let’s say that for some $g,h$ we had $g^{- 1}sg = h^{- 1}sh$ . Then,

$\displaystyle hg^{- 1}sg = sh$

$\displaystyle hg^{- 1}s = shg^{- 1}$

so $s$ commutes with $hg^{- 1}$ . By the same logic, it commutes with $gh^{- 1} = \left( hg^{- 1} \right)^{- 1}$ .

We then have $s = \left( hg^{- 1} \right)^{- 1}s\left( hg^{- 1} \right)$ .

If we reverse this, then if $G$ is abelian, then there may be only one element in the conjugacy class of $s$ , so that for all $g,h$ we have $g^{- 1}sg = h^{- 1}sh$ . In fact, for abelian groups, any subset is a normal subset. Thus, this conjecture is false; a normal subset is not necessarily a normal subgroup. $\blacksquare$

A normal subgroup can be motivated as having the properties for a kernel of a homomorphism. But a normal subset is an entirely different beast.

In fact, using the example of an abelian group, a normal subset need not be closed under multiplication; it need not have identity; it need not be closed under inverses; it may actually just be any old subset. More needs to be known about $G$ to deduce properties of its normal subsets.

Let’s reverse the question, then: if every normal subset of $G$ is a subgroup, what does that say about $G$ ?

We conjecture that this means that $G$ is “least abelian”, in the sense that no two elements commute. Wait, that wouldn’t work with powers …

Can we actually use this to develop an “abelian index” of a group? (Like a numerical measure of “how close the group is to being abelian.”) Something like:

$\displaystyle \frac{\#(normal\ subsets\ that\ are\ not\ subgroups)}{\#(subsets\ that\ are\ not\ subgroups)}$

Is it true that this number (or its analogue for infinite groups) is 1 if and only if the group is abelian? Can we show that the group is “least abelian” if the number of normal subsets that are not subgroups is 0?

How would we define “least abelian” here?

Let $G$ be a group; assume for now that $G$ is finite. For all $a,b \in G$ , we can write $(ab)(ba)^{- 1}$ . If $a$ and $b$ commute, then this element is $e$ . Thus, the number of distinct elements formed from $(ab)(ba)^{- 1}$ out of all choices of $a,b$ is a measure of “how much $G$ departs from being abelian”.

But we know that $a,b$ automatically commute if they are both powers of something. We have to be careful.

So consider all the possible $(a,b)$ that aren’t guaranteed to commute. For example, this excludes $a$ and $b$ being powers of something. Maybe that is the only case of guaranteed commutativity, although I’d want to do more work to check that. More precisely, we can define “guaranteed commutativity” as follows:

Definition. Let $s$ be a first-order statement about two elements of $G$ . (For example, $s$ could be “there exist $x$ and $m,n \in \mathbb{Z}_0$ such that $a=x^m,b=x^n$ . Actually, quantifying over positive integers is not something I think first-order logic supports, so we may have to modify this definition a bit.) We say that $s$ guarantees commutativity if the group axioms imply that any two elements satisfying it commute.

Actually, rather than formalizing this via mathematical logic, we can just try to show that there exists a group where the only elements that commute are powers. Is this true? In other words, does there exist $G$ such that $a,b\in G$ commute if and only if there exist $x$ and $m,n \in \mathbb{Z}_0$ such that $a=x^m,b=x^n$ ?

Regardless of how these questions turn out, let’s assume that we have some notion of identifying pairs of elements that have guaranteed commutativity. Then, we could define an “antiabelian index” like

$\displaystyle antiabelian\ index = \frac{\#(number\ of\ (a,b)\ that\ don^{'}t\ commute)}{\#(number\ of\ (a,b)\ without\ guaranteed\ commutativity)}$

Alternatively, we could define an antiabelian index like

$\displaystyle antiabelian\ index = \frac{\#(number\ of\ distinct\ (ab)(ba)^{- 1})}{\#(number\ of\ possible\ distinct\ such\ elements)}$

The denominator refers to number of pairs $(a,b)$ such that, among those pairs, $(ab)(ba)^{- 1}$ for one $(a,b)$ isn’t guaranteed to be equal to $(ab)(ba)^{- 1}$ for another.

Are these indices related or even equivalent?

Are the $(ab)(ba)^{- 1}$ for $(a,b)$ and $(b,a)$ equal? (Does order matter?) Let $x = (ab)(ba)^{- 1} \rightarrow xba = ab$ ; then $(ba)(ab)^{- 1}$ is what? If $y = (ba)(ab)^{- 1}$ , then $xy = xba(ab)^{- 1} = (ab)(ab)^{- 1} = e$ , so these are inverses of each other. What if they are the same? Then $(ab)(ba)^{- 1} = (ba)(ab)^{- 1} \rightarrow aba^{- 1}b^{- 1} = bab^{- 1}a^{- 1}$ .

How are our antiabelian indices related to the normal-subsets abelian index above?

I suspect that generators will play a role in figuring this out. Let $G$ have generators $g_{1},\ldots,g_{n}$ with orders $k_{1},\ldots,k_{n}$ . Then the order of $G$ is $k_{1}\ldots k_{n}$ . (EDIT, 02/2023: I’m not sure if I made a mistake of forgetting about generating relations between generators when first writing this. Also, I may have neglected order?)

It remains to investigate this further.

Written August 28, 2020. Edited and clarified some ideas on February 1, 2023.

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Edit (Oct 2025): The term “normal subset” is in fact standard and already present in the literature, as seen in the ProofWiki page and the GroupProps page (both edited with the same definition before this article was written, so not circular), as well as in earlier research (see this paper.) So far, I’m not sure whether the abelian or anti-abelian indices are already present in the literature. It remains to investigate those further.

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