Abelian Permutation Group

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  • #1
Bashyboy
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Homework Statement


Assume that ##G## is an abelian transitive subgroup of ##S_A## that acts on the set ##A##. Show that ##\sigma(a) \neq a## for all ##\sigma \in G - \{1\}## and all ##a \in A##. Deduce that ##|G| = |A|##.

Homework Equations



A group is said to act transitively on a set if there is only one orbit.

Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{e\}$$

The Attempt at a Solution



If ##G## acts transitively on ##A##, then by the above fact we see that the kernel of the action is faithful. This means that ##\sigma (a) = a## if and only if ##\sigma = e##, which is to say that ##\sigma(a) \neq a## for all ##\sigma \in G - \{e\}## and ##a \in A##. I am now going to switch from the function notation ##\sigma(a)## to the customary group action notation ##g \cdot a##.

Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g \cdot a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A = \cal{O}_a##. Clearly our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g \cdot a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g) \cdot a = a##. We will show this implies that ##(h^{-1}g)\cdot x = x## for all ##x \in A##. If ##x \in A##, then there exists a ##k \in G## such that ##x = k \cdot a##. Hence,

$$(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x$$

In the above we used commutativity several times. Since the kernel is trivial, ##h^{-1}g = e## or ##g = h##, thereby proving injectivity, and hence bijectivity.

Does this sound right?
 

Answers and Replies

  • #2
fresh_42
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Homework Statement


Assume that ##G## is an abelian transitive subgroup of ##S_A## that acts on the set ##A##. Show that ##\sigma(a) \neq a## for all ##\sigma \in G - \{1\}## and all ##a \in A##. Deduce that ##|G| = |A|##.

Homework Equations



A group is said to act transitively on a set if there is only one orbit.

Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{e\}$$

The Attempt at a Solution



If ##G## acts transitively on ##A##, then by the above fact we see that the kernel of the action is faithful. This means that ##\sigma (a) = a## if and only if ##\sigma = e##, which is to say that ##\sigma(a) \neq a## for all ##\sigma \in G - \{e\}## and ##a \in A##. I am now going to switch from the function notation ##\sigma(a)## to the customary group action notation ##g \cdot a##.

Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g \cdot a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A = \cal{O}_a##. Clearly our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g \cdot a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g) \cdot a = a##. We will show this implies that ##(h^{-1}g)\cdot x = x## for all ##x \in A##. If ##x \in A##, then there exists a ##k \in G## such that ##x = k \cdot a##. Hence,

$$(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x$$

In the above we used commutativity several times. Since the kernel is trivial, ##h^{-1}g = e## or ##g = h##, thereby proving injectivity, and hence bijectivity.

Does this sound right?
If ##G## is Abelian, then ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = G_a## and your assumption, that this equals ##\{e\}## is what you want to prove.
 
  • #3
Bashyboy
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If GGG is Abelian, then ⋂σ∈GσGaσ−1=Ga⋂σ∈GσGaσ−1=Ga\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = G_a and your assumption, that this equals {e}{e}\{e\} is what you want to prove.

I am not sure I understand this remark. Have I not proven what I have been asked to prove?
 
  • #4
fresh_42
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##\sigma G_a \sigma^{-1}=\sigma \sigma^{-1}G_a = G_a## for Abelian groups and therefore ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = G_a##. Now you claimed, that transitivity of the action implies ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1}=\{e_G\}##. Combining it, it reads ##G_a=\{g\in G\,\vert \,g.a=a\}=\{e_G\}## for all ##a \in A##, that is ##e_G## is the only operation with fix points.

Nothing to prove anymore.
 
  • #5
Bashyboy
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Nothing to prove anymore.

Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G≤SAG \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

⋂σ∈GσGaσ−1={e}​

On second thought, I think I may have made a mistake when proving this fact. Let ##g \in \sigma G_a \sigma^{-1}##. This occurs if and only if there exists a ##\tau \in G_a## such that ##g = \sigma \tau \sigma^{-1}## iff and only if ##\sigma^{-1} g = \tau \sigma^{-1}##...No matter how much fiddling around with the equation, I can seem to get ##g(\sigma(a)) = \sigma(a)##, which would complete the proof since this implies ##g## stablizes the element ##\sigma(a)##. Here is what makes me even more nervous: ##g = \sigma \tau \sigma^{-1}## seems to imply that ##\sigma^{-1} g \sigma = \tau \in G_a##, which would suggest that the stabilizer is always a normal subgroup...But this isn't always the case...what am I misunderstanding??!

Similarly, ##G_{\sigma(a)} = \{g \in G ~|~ g(\sigma(a)) = \sigma(a) \} = ##\{g \in G , but I am having trouble justifying why the last set in this string of set equalities is the set ##\sigma G_a \sigma^{-1}##.

No matter which way I attack it, it seems that I cannot finish the proof.
 
  • #6
fresh_42
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Or might it be the case, that what you stated under section 2 is actually what you want to show instead of what you want to use? I stopped reading it, as it was already the solution there.
 
  • #7
Bashyboy
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Or might it be the case, that what you stated under section 2 is actually what you want to show instead of what you want to use? I stopped reading it, as it was already the solution there.

No, what I wrote in section 3 was not intended to be used to proof what I mentioned in part 2. However, I just now realized that I incorrectly proved the fact I alluded in section 2 (see my post above).

WAIT! I may have a proof. Give me a few seconds!

##g \in G_{\sigma(a)}## if and only ##g(\sigma(a)) = \sigma(a)## if and only if ## (\sigma^{-1}g \sigma)(a) = a## if and only if ## \sigma^{-1} g \sigma \in G_a## if and only if there exists a ##\tau \in G_a## such that ##\sigma^{-1} g \sigma = \tau## if and only if ##g = \sigma \tau \sigma^{-1} \in \sigma G_a \sigma^{-1}##.

Does this seem right?
 
Last edited:
  • #8
fresh_42
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I don't see the point with the conjugation. In general ##\sigma G_a \sigma^{-1}.\sigma(a)=\sigma(a)##, i.e. ##\sigma G_a \sigma^{-1} =G_{\sigma(a)}##.
On the other hand, your very first sentence was
Assume that ##G## is an abelian transitive subgroup of ##S_A##
so all conjugations are already the identity: ##\sigma g \sigma^{-1}=g\,##.

Therefore my question is: Why is ##\bigcap_{\sigma \in G}\sigma G_a \sigma^{-1}=\{e_G\}\;##?
 
  • #9
Bashyboy
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I don't see the point with the conjugation. In general ##\sigma G_a \sigma^{-1}.\sigma(a)=\sigma(a)##, i.e. ##\sigma G_a \sigma^{-1} =G_{\sigma(a)}##.
On the other hand, your very first sentence was

so all conjugations are already the identity: ##\sigma g \sigma^{-1}=g\,##.

Yes, but the "fact" I alluded to in section 2 of my OP holds whether or not ##G## is abelian. So the proof I gave in post #7 proceeded without this assumption.
 
  • #10
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Yes, but the "fact" I alluded to in section 2 of my OP holds whether or not ##G## is abelian. So the proof I gave in post #7 proceeded without this assumption.
I'm completely confused now, sorry. Let me summarize. We have by commutativity and transitivity
$$G_a= \bigcap_{\sigma \in G}\sigma G_a \sigma^{-1} = \bigcap_{\sigma \in G} G_{\sigma(a)}=\bigcap_{b\in A}G_b$$
This means, every ##\sigma \in G_a## acts as identity on ##A##. Since all elements of ##G## are defined as elements of ##S_A##, it means ##\sigma = e_G## which proves the first part.

Now we proceed by
Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g.a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A=\cal{O}_a##
the orbit of ##a##
Clearly
It would be a good habit to learn to avoid such words: clearly, obviously and similar. What might be obvious to you doesn't have to be obvious to others. I remember an occasion an "obviously" took me two days of search and three substitutions of complex variables and written twenty lines of calculations ...
our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g.a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g).a = a##.
I think you can stop here. The first part already showed that a fix point for one element is automatically a fix point for all elements and ##h^{-1}g=e_G##. And I think this has been meant by (my flag):
Deduce that ##|G| = |A|##.

We will show this implies that (h−1g)⋅x=x(h−1g)⋅x=x(h^{-1}g)\cdot x = x for all x∈Ax∈Ax \in A. If x∈Ax∈Ax \in A, then there exists a k∈Gk∈Gk \in G such that x=k⋅ax=k⋅ax = k \cdot a. Hence,

(h−1g)⋅x=(h−1g)⋅(k⋅a)=(h−1k)⋅(g⋅a)=(kh−1h)⋅a=k⋅a=x(h−1g)⋅x=(h−1g)⋅(k⋅a)=(h−1k)⋅(g⋅a)=(kh−1h)⋅a=k⋅a=x​
(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x

In the above we used commutativity several times. Since the kernel is trivial, h−1g=eh−1g=eh^{-1}g = e or g=hg=hg = h, thereby proving injectivity, and hence bijectivity.
 
Last edited:
  • #11
WWGD
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Homework Statement


Assume that ##G## is an abelian transitive subgroup of ##S_A## that acts on the set ##A##. Show that ##\sigma(a) \neq a## for all ##\sigma \in G - \{1\}## and all ##a \in A##. Deduce that ##|G| = |A|##.

Homework Equations



A group is said to act transitively on a set if there is only one orbit.

Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

$$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{e\}$$

The Attempt at a Solution



If ##G## acts transitively on ##A##, then by the above fact we see that the kernel of the action is faithful. This means that ##\sigma (a) = a## if and only if ##\sigma = e##, which is to say that ##\sigma(a) \neq a## for all ##\sigma \in G - \{e\}## and ##a \in A##. I am now going to switch from the function notation ##\sigma(a)## to the customary group action notation ##g \cdot a##.

Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g \cdot a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A = \cal{O}_a##. Clearly our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g \cdot a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g) \cdot a = a##. We will show this implies that ##(h^{-1}g)\cdot x = x## for all ##x \in A##. If ##x \in A##, then there exists a ##k \in G## such that ##x = k \cdot a##. Hence,

$$(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x$$

In the above we used commutativity several times. Since the kernel is trivial, ##h^{-1}g = e## or ##g = h##, thereby proving injectivity, and hence bijectivity.

Does this sound right?
I find your notation unusual: when you mean that the kernel is faithful. I have heard of faithful actions, but not faithful kernels, do you mean that the kernel ( when the action is seen as an injection of ## G ## into the permutation group ## S_{|X|}## , where ## X## is the set you are acting on ) is trivial?.EDIT: And you want to show that if the associated map to the permutation group is injective, then the action is faithful ?
 
  • #12
fresh_42
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I find your notation unusual: when you mean that the kernel is faithful. I have heard of faithful actions, but not faithful kernels, do you mean that the kernel ( when the action is seen as an injection of ## G ## into the permutation group ## S_{|X|}## , where ## X## is the set you are acting on ) is trivial?.
Those kernels ##\{e\}## are especially faithful in group theory. You can factor them out, separate them in direct factors, but you do not get rid of them, never ever. :biggrin:

(Sorry @Bashyboy, couldn't resist.)
 
  • #13
WWGD
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Those kernels ##\{e\}## are especially faithful in group theory. You can factor them out, separate them in direct factors, but you do not get rid of them, never ever. :biggrin:

(Sorry @Bashyboy, couldn't resist.)
And she has your phone number and she will call you even after you break up with her..., er, I mean, I agree, kernels, yes, kernels ;).
 

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