# Abstract Nonsense

## 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.

Motivation

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

$\mathcal{O}_{f(s)}=f\left(\mathcal{O}_s\right)$

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

$\left(G:G_s\right)=\left(G:G_{f(s)}\right)$

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 $\#(S)\mid |G|$.

References:

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.