# Math Help - 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.

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.
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.
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
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.

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
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?

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?

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.