1. ## group action

I have the following question consisting of two parts: a) Let G be a finite group. Show that if H<G then
G is not the union of conjugates of H ?
b) Show that if G acts transitively on a set X of size at least 2
then some g in G acts without fixed points (Hint: Use a)
For a) I tried to use a counting argument.
Since |gHg^-1|=|H| then The total number of elements
are (|G|/|N(H)|).(|H|-1)<= (|G|/|H|).(|H|-1) <= ...
Does that help me in any thing ?
For b) I can say , let a,b in G then there is g s.t. g.a=b
and we can show that G_a=g^-1.G_b.g
from a) we can conclude there is an element g such that
is not included in any conjugate subgroup of H.
Can any one help on this question ? Thanks

2. Originally Posted by hgd7833
I have the following question consisting of two parts: a) Let G be a finite group. Show that if H<G then
G is not the union of conjugates of H ?
b) Show that if G acts transitively on a set X of size at least 2
then some g in G acts without fixed points (Hint: Use a)
For a) I tried to use a counting argument.
Since |gHg^-1|=|H| then The total number of elements
are (|G|/|N(H)|).(|H|-1)<= (|G|/|H|).(|H|-1) <= ...
Does that help me in any thing ?
For b) I can say , let a,b in G then there is g s.t. g.a=b
and we can show that G_a=g^-1.G_b.g
from a) we can conclude there is an element g such that
is not included in any conjugate subgroup of H.
Can any one help on this question ? Thanks
For (a) use the orbit-stabiliser theorem (your action will be conjugation)! You want to use the obrit-stabiliset theorem to find out stuff about the index of your group.

Again, (b) falls out quite quickly if you use the orbit-stabiliser theorem. However, I am unsure why you would need part (a). Have you come across the orbit-stabiliser theorem yet?..

3. For a) I would say, let H act on G by conjugation, then |G|=|Z(G)/\H|+ sum([H:C(a_i)] = |Z(G)/\H|+|O(H)|
So G is union of conjugates of H only if |G|= |O(H)| but this implies that |Z(G)/\H|= 0 which is impossible ( is this right ?)

For b) If G acts transitively on a set X then X has only one orbit. So |X|= | Fix(X)| + |O(x)| , so for every y in X there is g in G such that g.y=x hence
y belongs to O(x) . BUT we need a single g such that g.a is not a for every a in X.

Actually the question came ion this way, and they gave the hint on b to use a.

For a) I would say, let H act on G by conjugation, then |G|=|Z(G)/\H|+ sum([H:C(a_i)] = |Z(G)/\H|+|O(H)|
So G is union of conjugates of H only if |G|= |O(H)| but this implies that |Z(G)/\H|= 0 which is impossible ( is this right ?)

For b) If G acts transitively on a set X then X has only one orbit. So |X|= | Fix(X)| + |O(x)| , so for every y in X there is g in G such that g.y=x hence
y belongs to O(x) . BUT we need a single g such that g.a is not a for every a in X.

Actually the question came ion this way, and they gave the hint on b to use a.

5. Originally Posted by hgd7833
For a) I would say, let H act on G by conjugation, then |G|=|Z(G)/\H|+ sum([H:C(a_i)] = |Z(G)/\H|+|O(H)|
So G is union of conjugates of H only if |G|= |O(H)| but this implies that |Z(G)/\H|= 0 which is impossible ( is this right ?)
I am not entirely sure what you mean here, although I presume it is a perversion of the class equation. What do you mean by Z(G)/\H? Is it $\displaystyle Z(G) \setminus (Z(G) \cap H)$? This should be the centraliser of H, the elements which stay fixed under conjugation by H. And is O(H) the orbit of H? However, the way you have constructed this is, I believe, incorrect.

As you have covered the class equation, please go and look up the orbit-stabiliser theorem. It will tell you how the centraliser and the orbit of H are connected.

Originally Posted by hgd7833
For b) If G acts transitively on a set X then X has only one orbit. So |X|= | Fix(X)| + |O(x)| , so for every y in X there is g in G such that g.y=x hence
y belongs to O(x) . BUT we need a single g such that g.a is not a for every a in X.

Actually the question came ion this way, and they gave the hint on b to use a.
No! Look up the orbit-stabiliser theorem! $\displaystyle |X| \neq |Fix(X)|+|O(X)|$, the orbit-stabiliser theorem tells you how these are connected!

6. I worked on this problem and using your argument for a while. I think I got it now. I am acting the group G on H by group conjugation, i.e. g.H = gHg^-1
so, number of conjugates of H is the order of orbit of H, If G is the union of conjugates then for every g in G g belongs to one of the orbits,
but Fix(H) is not empty since e.H=e.H.e=H so the identity belongs to Fix(H) hence the result follows.
But this implies b follows trivially without using orbit - stabilizer theorem. Right ?
Thanks Swlabr