Is Every Element of a Transitive Abelian Permutation Group Not the Identity?

[Bashyboy]

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?

 

[fresh_42]

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.

 

[Bashyboy]

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?

 

[fresh_42]

$\sigma G_a \sigma^{-1}=\sigma \sigma^{-1}G_a = G_a$ 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.

 

[Bashyboy]

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.

 

[fresh_42]

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.

 

[Bashyboy]

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?

 

[fresh_42]

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 $\bigcap_{\sigma \in G}\sigma G_a \sigma^{-1}=\{e_G\}\;$?

 

[Bashyboy]

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.

 

[fresh_42]

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.

 

[WWGD]

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 ?

 

[fresh_42]

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.

(Sorry @Bashyboy, couldn't resist.)

 

[WWGD]

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.

(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 ;).