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

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##?

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fresh_42
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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}\,?##

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
Mentor
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##.
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:

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
Mentor
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
Mentor
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
Mentor
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.

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Mr Davis 97