Showing that the union of conjugates is less than |G|

[Mr Davis 97]

Here is a problem statement: Let $H$ be a proper subgroup of a finite group $G$. Prove that the union of the conjugates of $H$ is not all of $G$.

I have proven this statement by considering the action of $G$ on $\mathcal{P}(G)$. But this leads me to wonder: In the problem statement is there any reason why $H$ must be a subgroup? Am I right in saying that the statement seems true for any $S$ such that $S\subset G$?

 

[fresh_42]

Here is a problem statement: Let $H$ be a proper subgroup of a finite group $G$. Prove that the union of the conjugates of $H$ is not all of $G$.

I have proven this statement by considering the action of $G$ on $\mathcal{P}(G)$. But this leads me to wonder: In the problem statement is there any reason why $H$ must be a subgroup? Am I right in saying that the statement seems true for any $S$ such that $S\subset G$?

Define $S := \{\,g \in G=\operatorname{Sym}(3)\,|\,g\neq (13)\,\}.$ Now what is $(12)(23)(12) \in gSg^{-1}\,?$

 

[Mr Davis 97]

Define $S := \{\,g \in G=\operatorname{Sym}(3)\,|\,g\neq (13)\,\}.$ Now what is $(12)(23)(12) \in gSg^{-1}\,?$

$(12)(23)(12) = (13) \not \in S$... For some reason I'm not seeing the significance of this. Is conjugation supposed to be closed?

 

[fresh_42]

The statement was:

Let $H$ be a proper subgroup of a finite group $G$. Prove that the union of the conjugates of $H$ is not all of $G$.

and you asked:

Am I right in saying that the statement seems true for any $S$ such that $S\subset G?$

Now I defined a subset $S=G-\{\,(13)\,\}$ of $G=\operatorname{Sym}(3)$ with $G=eSe^{-1} \cup (12)S(12)^{-1}$, which is a counterexample.

 

[Mr Davis 97]

The statement was:

and you asked:

Now I defined a subset $S=G-\{\,(13)\,\}$ of $G=\operatorname{Sym}(3)$ with $G=eSe^{-1} \cup (12)S(12)^{-1}$, which is a counterexample.

So what property of subgroups am I using in my proof that is not satisfied with just subsets? Would I need to write out my proof for you to know?

 

[fresh_42]

So what property of subgroups am I using in my proof that is not satisfied with just subsets? Would I need to write out my proof for you to know?

Well, I don't know a proof and thought it would be wrong. But then I recognized that conjugation keeps properties like signature, determinant, order etc. invariant, and that this might be the cause why conjugates can't reach every element, same as you cannot get determinant $-1$ matrices from determinant $1$ matrices by conjugation. But that's only a heuristic, along which I would try to prove the statement.

In the counterexample, I made use of the fact that I can switch inside a subgroup $H$ from one element to another, so I took a subset minus an element $(13)$ such that I could still reach my element by conjugation and destroyed the subgroup structure by taking it away. In a subgroup, you can't do this. I suppose you used the orbit-stabilizer formula to divide $G$ in cosets $gHg^{-1}$. They all have the same number of elements, which I think is not true anymore for subsets.

 

[Mr Davis 97]

https://math.stackexchange.com/a/121534/207516 gives a solution to the problem of showing that if $H < G$ and $|G| < \infty$ then $\bigcup_{g\in G}gHg^{-1} \not = G$. I can understand the solution pretty well, but I am a bit confused by what the action actually is. He says "Let $G$ act by conjugation," but what does this mean exactly? Is $G$ acting on the set of all subsets, or is it acting on something else?

 

[fresh_42]

https://math.stackexchange.com/a/121534/207516 gives a solution to the problem of showing that if $H < G$ and $|G| < \infty$ then $\bigcup_{g\in G}gHg^{-1} \not = G$. I can understand the solution pretty well, but I am a bit confused by what the action actually is. He says "Let $G$ act by conjugation," but what does this mean exactly? Is $G$ acting on the set of all subsets, or is it acting on something else?

$G$ acts via conjugation on itself: $(g,h) \longmapsto g.h := ghg^{-1}$. The orbit of an element $h$ are all elements $ghg^{-1}$ and the orbit of a set $H$ is $\{\,ghg^{-1}\,|\,g\in G,h\in H\,\}$. So the corresponding homomorphism (or if you like representation) is $\varphi\, : \,G \longrightarrow \operatorname{Sym}(G)$ defined by $\varphi(g)(h)=ghg^{-1}.$

 

[Mr Davis 97]

$G$ acts via conjugation on itself: $(g,h) \longmapsto g.h := ghg^{-1}$. The orbit of an element $h$ are all elements $ghg^{-1}$ and the orbit of a set $H$ is $\{\,ghg^{-1}\,|\,g\in G,h\in H\,\}$. So the corresponding homomorphism (or if you like representation) is $\varphi\, : \,G \longrightarrow \operatorname{Sym}(G)$ defined by $\varphi(g)(h)=ghg^{-1}.$

Why is it clear that this action is $G$ acting on itself rather than $G$ acting on its set of subsets?

 

[fresh_42]

Why is it clear that this action is $G$ acting on itself rather than $G$ acting on its set of subsets?

I haven't seen any. It is considered what conjugation does to $H$, but there was no set of sets. E.g. one could consider $G/H$ or $G\backslash H$, but then we had to say something about $gg'Hg^{-1}$ or $gHg'g^{-1}$ which didn't occur. It's usually the entire group, because you can still consider orbits of entire subsets, as e.g. $H$. To restrict the conjugation on something else, this something else has to be mentioned in the first place. Additionally we are interested in the entity of all group elements, not just a few in a subset. And the orbit-stabilizer theorem or ... formula is applied to $G$ acting on $X=H$. It makes simply no difference: the conjugation is the same and $X=H$ is a set, not a set of sets, and the operation is elementwise.

Correction: $X=G$ for otherwise we would leave the set $G$ acts on.

 

[Infrared]

This question was posed in the Intermediate August Math Challenge thread https://www.physicsforums.com/threads/intermediate-math-challenge-august-2018.952511/ (problem 6). At the the risk of self-advertisement, I'll mention that I give a solution in post 8.