# Thread: transitive action; conjugation

1. ## transitive action; conjugation

I have been trying these two problems for a long time but cannot seem to get anywhere.

problem 1) Let G be a group acting transitively on a finite set A with |A|>1. Show that there exists $\sigma \in G \text{ such that } \sigma \cdot a \neq a \text{ for all } a \in A$

problem 2) If G be a group of odd order, prove for any non-identity element $x \in G \text{ that } x \text{ and } x^{-1}$ are not conjugate in G.

2. Originally Posted by abhishekkgp
I have been trying these two problems for a long time but cannot seem to get anywhere.

problem 1) Let G be a group acting transitively on a finite set A with |A|>1. Show that there exists $\sigma \in G \text{ such that } \sigma \cdot a \neq a \text{ for all } a \in A$

problem 2) If G be a group of odd order, prove for any non-identity element $x \in G \text{ that } x \text{ and } x^{-1}$ are not conjugate in G.
For 1) what is your definition of transitive? I aks, because there is more than one definition, and if you were to take the definition for all $a,b \in A$ there exists a $g \in G$ such that $x\cdot a = b$' then your result follows quite easily. In fact, I would suggest proving that your definition and the one I just gave are equivalent, and proving the result that way.

For 2) prove that the orbit containing x has even order (i.e. show that if $g$ is in the orbit then so is $g^{-1}$, and that $g\neq g^{-1}$), and then apply the orbit-stabiliser theorem.

3. Originally Posted by abhishekkgp
I have been trying these two problems for a long time but cannot seem to get anywhere.

problem 1) Let G be a group acting transitively on a finite set A with |A|>1. Show that there exists $\sigma \in G \text{ such that } \sigma \cdot a \neq a \text{ for all } a \in A$
If G acts transitively on a finite set A, then A has a single orbit.
Now suppose to the contrary that
For some $a \in A$, $\sigma \cdot a = a$ for all $\sigma \in G$. Then, it has a fixed point. Thus, there exists at least two orbits in A (|A|>2). Contradiction! The result follows.

4. Originally Posted by Swlabr
For 2) prove that the orbit containing x has even order (i.e. show that if $g$ is in the orbit then so is $g^{-1}$, and that $g\neq g^{-1}$), and then apply the orbit-stabiliser theorem.
If i am able to prove that "if $g$ is in the orbit of $x$ then so is $g^{-1}$" then i will be proving the opposite of what the question asks me to prove. Because then i will say "since $x$ is in the orbit of $x$ then so is $x^{-1}$".... which will mean that $x$ and $x^{-1}$ are in the same orbit and this is the opposite of what i have to prove.

It can however be said with certainty that $g \neq g^{-1}$ for each $g \neq 1$ since the order of the group is odd.
Please consider this. Thanks.

5. Originally Posted by abhishekkgp
If i am able to prove that "if $g$ is in the orbit of $x$ then so is $g^{-1}$" then i will be proving the opposite of what the question asks me to prove. Because then i will say "since $x$ is in the orbit of $x$ then so is $x^{-1}$".... which will mean that $x$ and $x^{-1}$ are in the same orbit and this is the opposite of what i have to prove.

It can however be said with certainty that $g \neq g^{-1}$ for each $g \neq 1$ since the order of the group is odd.
Please consider this. Thanks.
I was meaning use proof by contradiction. IF $x$ and $x^{-1}$ are in the same orbit, THEN this orbit has an even number of elements, THUS... and you want to find a contradiction.

6. Originally Posted by TheArtofSymmetry
If G acts transitively on a finite set A, then A has a single orbit.
Now suppose to the contrary that
For some $a \in A$, $\sigma \cdot a = a$ for all $\sigma \in G$. Then, it has a fixed point. Thus, there exists at least two orbits in A (|A|>2). Contradiction! The result follows.
According to me what you have proved is "for each $a \in A$there exists a $\sigma \in G$ which does not fix $a$".

DEFINITION: $g \in G$ fixes $a \in A$ iff $g \cdot a = a$

This was not what was to be proved. I will re-frame the question here :

Given G is a group acting on a finite set A with |A|>1. I t is also given that G has only one orbit. Prove that there exists $g \in G$ which fixes no element of A

7. Originally Posted by Swlabr
For 1) what is your definition of transitive? I aks, because there is more than one definition, and if you were to take the definition for all $a,b \in A$ there exists a $g \in G$ such that $x\cdot a = b$' then your result follows quite easily. In fact, I would suggest proving that your definition and the one I just gave are equivalent, and proving the result that way.
to avoid all the confusion about this i have re-framed the question below:
Given G is a group acting on a finite set A with |A|>1. It is also given that G has only one orbit. Prove that there exists $\sigma \in G$ which fixes no element of A.

$g \in G$ is said to "fix" $a \in A$ iff $g \cdot a = a$

8. Originally Posted by Swlabr
I was meaning use proof by contradiction. IF $x$ and $x^{-1}$ are in the same orbit, THEN this orbit has an even number of elements, THUS... and you want to find a contradiction.
That does the second one! Thanks a ton!

9. Originally Posted by abhishekkgp
According to me what you have proved is "for each $a \in A$there exists a $\sigma \in G$ which does not fix $a$".

DEFINITION: $g \in G$ fixes $a \in A$ iff $g \cdot a = a$

This was not what was to be proved. I will re-frame the question here :

Given G is a group acting on a finite set A with |A|>1. I t is also given that G has only one orbit. Prove that there exists $g \in G$ which fixes no element of A
Proof by contradiction.
Both of your problems can be proven by "proof by contradiction". Read some examples in the link. It is one of the fundamental proof techniques.

Similar to the second one, I used the proof by contradiction, so I started off by supposing to the contrary towards a contradiction. Then derived a contradiction.

10. For your rephrased problem,
If you suppose to the contrary, i.e., for all g in G, it fixes an element of A (|A|>2), then it contradicts that G has only one orbit.
Thus, conclude that there exists g in G which fixes no element of A.

11. Originally Posted by TheArtofSymmetry
For your rephrased problem,
If you suppose to the contrary, i.e., for all g in G, it fixes an element of A (|A|>2), then it contradicts that G has only one orbit.
Thus, conclude that there exists g in G which fixes no element of A.
are you assuming that each $g \in G$ fix the same $a \in A$??
for instance, say $g_1 \in G$ fixes $a_1 \in A$ and,
$g_2 \in G$ fixes $a_2 \in A$ where $a_1 \neq a_2$.
I cannot see how your proof covers this case. can you please elaborate.

12. Originally Posted by abhishekkgp
are you assuming that each $g \in G$ fix the same $a \in A$??
for instance, say $g_1 \in G$ fixes $a_1 \in A$ and,
$g_2 \in G$ fixes $a_2 \in A$ where $a_1 \neq a_2$.
I cannot see how your proof covers this case. can you please elaborate.
Can you describe the orbit of a single element? If there exists an orbit of a single element in the G-set A (|A| >2 ), what does it tell you?

If your confusion comes from the negation of a quantified statement (for the sake of a contradiction), this might help you.

13. Originally Posted by TheArtofSymmetry
Can you describe the orbit of a single element? If there exists an orbit of a single element in the G-set A (|A| >2 ), what does it tell you?

If your confusion comes from the negation of a quantified statement (for the sake of a contradiction), this might help you.
Orbit of G containing an element of A can be described.
let $a \in A$. Then orbit of G containing $a$ is $\{g \cdot a|g \in G\}$.

Originally Posted by TheArtofSymmetry
If you suppose to the contrary, i.e., for all g in G, it fixes an element of A....
to me this means that suppose there exists an $a \in A$ such that $g \cdot a=a$ for each $g \in G$
This will, of course, mean that there are more than one orbits of G.

But my point is that this does not complete the proof because an element of G may fix some $b \in A$ and other element of G may fix $c \in A, c \neq b$

14. Originally Posted by abhishekkgp
Orbit of G containing an element of A can be described.
let $a \in A$. Then orbit of G containing $a$ is $\{g \cdot a|g \in G\}$.

to me this means that suppose there exists an $a \in A$ such that $g \cdot a=a$ for each $g \in G$

This will, of course, mean that there are more than one orbits of G.
But my point is that this does not complete the proof because an element of G may fix some $b \in A$ and other element of G may fix $c \in A, c \neq b$
This completes the proof, because it suffices to draw a contradiction. Of course, your cases may occur, but I don't understand why that makes the proof incomplete. Remember that G acts transitively on A, A has only a single orbit. If we derive that there are at least two orbits in A (by supposing to the contrary), a contradiction occurs and the result follows.

15. Originally Posted by TheArtofSymmetry
This completes the proof, because it suffices to draw a contradiction. Of course, your cases may occur, but I don't understand why that makes the proof incomplete. Remember that G acts transitively on A, A has only a single orbit. If we derive that there are at least two orbits in A (by supposing to the contrary), a contradiction occurs and the result follows.
I think i am no able to make myself clear.i still dont agree but thanks for considering this problem. I got a different solution on the net Every nontrivial transitive permutation group contains elements which are fixed point free « Project Crazy Project
you may check this one out too.