3
$\begingroup$

In group theory, we often deal with subsets that are denoted by 'lifting' the group operation to subsets. For instance, the coset $gH = \{gh:h\in H\}$, or the conjugate $gHg^{-1} = \{ ghg^{-1}: h\in H \}$, or even $HK=\{hk:h\in H, k\in K\}$.

We also reason on these sets. For instance, if we have $$gHg^{-1}\subseteq H$$ then we can say that for any $h\in H$, by assumption $ghg^{-1} = h'$ for some $h'\in H$, and therefore that $h=g^{-1}h'g\in g^{-1}Hg$, and so $H\subseteq g^{-1}Hg$. Alternatively, we can do the more slick one line argument \begin{equation}\tag{1} H= g^{-1}gHg^{-1}g \subseteq g^{-1}Hg. \end{equation}

My question is something like -- when and why are we justified in doing the slick algebraic manipulation on subsets like in (1)?

What is the general thing that's going on here? E.g. this also works when $H$ is just a subset, not a subgroup. What if we're not working with groups at all (e.g. this would also work if we were working with monoids)? Or what if we're reasoning about longer strings of elements and subsets, like $abCdEF = GhIjkL$ where lowercase denotes elements and uppercase denotes subsets?

One idea I had was to think of a subset like $gS=\{gs: S\in S\}$ as the result of a group action of $g\in G$ on the powerset $\mathcal{P}(G)$. But even this view seems to require some further coherence conditions between e.g. left and right group actions. And even then we would want to generalize this to non-group settings. It feels like there should be an easier characterisation just in terms of sets and functions.

$\endgroup$
4
  • $\begingroup$ Your final sentence is unclear. You are quite right that the left and right actions of $G$ on ${\cal P}(G)$ give rise to identities like the one that your (1) justifies. What could be easier than that? $\endgroup$
    – Rob Arthan
    Commented Aug 11 at 21:18
  • $\begingroup$ You might need some of what semigroup theory uses initially, at least in Howie's, "Fundamentals of Semigroup Theory". What I'm referring to is congruence relations (as opposed to equivalence relations). I don't have time right now to explain my thoughts however. Sorry. $\endgroup$
    – Shaun ♦
    Commented Aug 11 at 21:19
  • $\begingroup$ @RobArthan yes my thought there isn't very clear. firstly, it's that a subset like $xHy$ could be expressed as first a left action by $x$ on $H$, then a right action by $y$ on $xH$, or a right action by $y$ on $H$ and then a left action by $x$ on $Hy$. But these two should be obviously the same. So at least extra equations seem to be required? Further, there is the point that something like this should also work for e.g. monoids, so at the very least this behaviour shouldn't be something that is specific to group actions. $\endgroup$
    – Merk Zockerborg
    Commented Aug 11 at 21:23
  • 1
    $\begingroup$ You need to show an associativity property: $(xH)y = x(Hy)$ but that's easy. It's still unclear what your question is. $\endgroup$
    – Rob Arthan
    Commented Aug 11 at 21:38

1 Answer 1

Reset to default
4
$\begingroup$

The powerset $\mathcal{P}(G)$ admits commuting left and right actions of $G$. These actions have the property that if $X \subseteq Y$ then $gX \subseteq gY$ and $Xg \subseteq Yg$, neither of which is hard to prove. So from

$$gHg^{-1} \subseteq H$$

we deduce by multiplying by $g^{-1}$ on the left that

$$Hg^{-1} \subseteq g^{-1} H$$

and then by multiplying by $g$ on the right that

$$H \subseteq g^{-1} H g.$$

So this is not so bad. It is moreover true that left and right multiplication by elements of $G$ induces isomorphisms $\mathcal{P}(G) \to \mathcal{P}(G)$ of Boolean algebras, so they not only preserve containment but also union, intersection, and complement. All of this is also true for the conjugation action $X \mapsto gXg^{-1}$ (which is a restriction of the commuting left and right actions, in a suitable sense), which has the additional benefit of preserving subgroups.

Products of subsets are a bit trickier. $\mathcal{P}(G)$ inherits a product $ST = \{ st : s \in S, t \in T \}$ from the product in $G$ but of course it no longer has inverses so one has to be a bit more careful with it. But it is still associative, for example, and the left and right actions of $G$ above correspond to the special cases when either $S$ or $T$ has one element. The product is "monotonic" in both variables, by which I mean that if $S_1 \subseteq S_2$ then $S_1 T \subseteq S_2 T$ and similarly on the right, and this is also not hard to prove. None of this requires $G$ to have inverses or an identity so it's still true with $G$ a semigroup.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .