Transitive Group Action on a Finite Set

**EntryLevel** · September 5th 2011, 07:57 AM

I am working on problem 4.1.9 in Dummit and Foote's Abstract Algebra. Thought I am in an algebra course, we don't use this book, and this isn't a homework assignment.

G acts transitively on a finite set A and H is a normal subgroup of G. $O_1,...,O_r$ are the distinct orbits of H on A.

I have already done half of the problem, which is to show all orbits of H have the same cardinality and that G is transitive on the set ${O_1,...,O_r}$ where G acts on this set via representative.

My problem is this: Prove that if $a\in O_1$ then $|O_1|=|H:H\cap G_a|$ and prove that $r=|G:HG_a|$ .

I cannot do the second part (the first isn't hard). A hint is to use the 2nd isomorphism theorem, but I don't see how it helps.

Thanks

**Drexel28** · September 5th 2011, 11:13 AM

Originally Posted by EntryLevel

I am working on problem 4.1.9 in Dummit and Foote's Abstract Algebra. Thought I am in an algebra course, we don't use this book, and this isn't a homework assignment.

G acts transitively on a finite set A and H is a normal subgroup of G. $O_1,...,O_r$ are the distinct orbits of H on A.

I have already done half of the problem, which is to show all orbits of H have the same cardinality and that G is transitive on the set ${O_1,...,O_r}$ where G acts on this set via representative.

My problem is this: Prove that if $a\in O_1$ then $|O_1|=|H:H\cap G_a|$ and prove that $r=|G:HG_a|$ .

I cannot do the second part (the first isn't hard). A hint is to use the 2nd isomorphism theorem, but I don't see how it helps.

Thanks

What have you tried? Are the $G_a$ the stabiliers in $G$ or $H$ ?

**EntryLevel** · September 5th 2011, 11:51 AM

Originally Posted by Drexel28

What have you tried? Are the $G_a$ the stabiliers in $G$ or $H$ ?

They are the stabilizers in G.

Actually, I decided to ignore the advice about the 2nd isomorphism theorem and I thought of the following map that I think is a bijection.

$\phi :G / HG_a \rightarrow \{O_1,...,O_r\}$
$\phi (g)=$ the orbit containing g(a), where a is given in the problem.

The map is definitely surjective. It is well defined because if you have two cosets $gHG_a$ and $g'HG_a$ then it must be true that $g=g'hg_a$ for elements $h\in H$ and $g_a\in G_a$

$g(a)=g'hg_a(a)=g'h(a)=g'(a)$

where the 2nd equality is because g_a is a stabilizer and the last equality because in part (a), it is shown that any element of G permutes the different orbits, taking all elements in one orbit to all elements in another...

Finally it is injective because if $g(a)=g'(a)$ then $g^{-1}g'\in G_a\subset HG_a$

This seems to work... perhaps the hint was a red-herring (or perhaps I made a mistake).

**Drexel28** · September 5th 2011, 12:16 PM

Originally Posted by EntryLevel

They are the stabilizers in G.

Actually, I decided to ignore the advice about the 2nd isomorphism theorem and I thought of the following map that I think is a bijection.

$\phi :G / HG_a \rightarrow \{O_1,...,O_r\}$
$\phi (g)=$ the orbit containing g(a), where a is given in the problem.

The map is definitely surjective. It is well defined because if you have two cosets $gHG_a$ and $g'HG_a$ then it must be true that $g=g'hg_a$ for elements $h\in H$ and $g_a\in G_a$

$g(a)=g'hg_a(a)=g'h(a)=g'(a)$

where the 2nd equality is because g_a is a stabilizer and the last equality because in part (a), it is shown that any element of G permutes the different orbits, taking all elements in one orbit to all elements in another...

Finally it is injective because if $g(a)=g'(a)$ then $g^{-1}g'\in G_a\subset HG_a$

This seems to work... perhaps the hint was a red-herring (or perhaps I made a mistake).

Seems to work from what I see. I was going to just note the following. Since $G$ acts transitively on $A$ you have that $\text{card}(A)=[G:G_a]$ for any $a$ , right (this is just the orbit stabilizer theorem and using the fact that the orbit must be the whole set). But, from the orbit decomposition theorem and the fact that all the orbits of $A$ under $H$ are equipotent we have that $\text{card}(A)=r\text{card}(O_1)$ . So, we have that

$\displaystyle \begin{aligned}\frac{|G|}{|HG_a|} &=\frac{|G|}{\displaystyle \frac{|H||G_a|}{|H\cap G_a|}}\\ &=\frac{|G|}{\displaystyle \frac{|H|}{|H\cap G_a|}|G_a|}\\ &=\frac{\displaystyle \frac{|G|}{|G_a|}}{\text{card}(O_1)}\\ &=\frac{\text{card}(A)}{\text{card}(O_1)}\\ &=\frac{r\text{card}(O_1)}{\text{card}(O_1)}\\ &=r\end{aligned}$

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