Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Group Actions (Pt. III G-Space Isomorphisms and the Fundamental Theorem of G-Spaces)

Point of post: In this post we describe what it means for two G-spaces to be isomorphic and describe the Fundamental Theorem of G-Spaces.



Just as with all other structures it’s fruitful to define the maps between G-spaces that preserve the structure we’re interested in. It’s intuitively clear that for G-spaces this kind of map should preserve the action, in the sense that if we act on an element and then map it over we should get the same result if we map it over and then act on it.

After we have rigorously defined the notion of G-space isomorphism we describe the Fundamental Theorem of G-spaces.


G-space Isomorphisms


Let S and S' be two G-spaces with G-actions \cdot_1 and \cdot_2 respectively. We say that S and S' are G-space isomorphic if there exists a bijective map f:S\to S' such that

\displaystyle f\left(g\cdot_1 s\right)=g\cdot_2 f(s)

for all s\in S and g\in G. We call such a f a G-space isomorphism.


Some theorems which become immediately obvious are


Theorem: Let S and S' be G-spaces with G-actions \cdot_1 and \cdot_2 respectively. Then, if f:S\to S' is a G-space isomorphism then


for every s\in S. In particular \#\left(\mathcal{O}_{f(s)}\right)=\#\left(\mathcal{O}_s\right)

Proof: We merely note that

\displaystyle \begin{aligned}\mathcal{O}_{f(s)} &= \left\{g\cdot_2 f(s):g\in G\right\}\\ &= \left\{f\left(g\cdot_1 s\right):g\in G\right\}\\ &= f\left(\mathcal{O}_s\right)\end{aligned}

The fact that \#\left(\mathcal{O}_{f(s)}\right)=\#\left(\mathcal{O}_s\right) then follows immediately from the fact that f is a bijection. \blacksquare


Corollary: Let S and S' be G-spaces and f:S\to S' a G-space isomorphism. Then, for every s\in S it’s true that


Transitive G-actions and the Fundamental Theorem of G-spaces


Let S be a G-space with G-action \cdot.  We call S a transitive G-space if S=\mathcal{O}_s for every s\in S. The first theorem which results from considering transitive G-spaces is


Theorem: Let G be a group and H\leqslant G. Then, the G-action g\cdot hH=(gh)H turns G/H into a transitive G-space.

Proof: We merely note that by definition that for any hH\in G/H

\displaystyle \begin{aligned}\mathcal{O}_{hH} &= \left\{g\cdot hH:g\in G\right\}\\ &= \left\{(gh)H:g\in G\right\}\\ &= G/H\end{aligned}

since \theta:G\to G:g\mapsto gh is a bijection (this is Cayley’s Theorem). The conclusion follows. \blacksquare


and with this we are able to state the Fundamental Theorem of G-spaces


Theorem: Let S be a transitive G-space with G-action \cdot_1. Then, for any s\in S G/G_s acted upon by the G-action g\cdot_2 hG_s=(gh)G_s is G-space isomorphic to S.

Proof: For a fixed s'\in S defined Q_{s'}\subseteq G to be

Q_{s'}=\left\{g\in G:g\cdot s=s'\right\}

Then, consider that if s\in Q_{s'} then y\in Q_{s'} if and only if x^{-1}y\in G_s in other words y\in xG_s. Thus, it follows that Q_{s'}=xG_s and thus the set of Q_{s'},\text{ }s'\in S are precisely the elements of G/G_s. Thus, consider the map

f:S\to G/G_s:s'\mapsto Q_{s'}

we have then that

f\left(g\cdot_1 s'\right)=gQ_{s'}=g\cdot_2 f(s')

since it’s evident that f is a bijection the conclusion follows. \blacksquare


Corollary: Let S be a transitive finite G-space with |G|<\infty, and s\in S. Then, \#(S)|G_s|=|G|. In particular \#(S)\mid |G|.



1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

3. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.



January 5, 2011 - Posted by | Algebra, Group Theory | , , ,

1 Comment »

  1. […] elements of with -sets, but what about the arrows of ? Can we interpret these as nothing more than -set homomorphisms? Indeed, suppose we had two elements of , in other words to functors . From the above we can […]

    Pingback by Functor Categories (Pt. I) « Abstract Nonsense | January 7, 2012 | Reply

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