Discover Top Posts Tagged with #group actions

Is there a name for when you have a group action G x X -> X alongside an morphism G -> X such that G x X -> X extends the self-action G x G -> G in the obvious way?

#group theory #groups #group actions

Reconstructing groups and rings from faithful actions

Groups and group actions

Let $G \curvearrowright X$ be a faithful group action. It is not necessarily transitive, and the actions of $G$ on its orbits do not have to be faithful either. For each orbit $O$ of $G \curvearrowright X$, let $K_O$ be the kernel of the associated group homomorphism $\phi_O : G \to \operatorname{Sym}(O)$. Faithfulness is equivalent to $\bigcap_O K_O = 1$.

The motivating question is:

Question. How do we recognize the image $G' \subseteq \operatorname{Sym}(X)$ of $\phi_X : G \hookrightarrow \operatorname{Sym}(X)$?

Clearly, a necessary condition for a permutation $\sigma \in \operatorname{Sym}(X)$ to be in $G'$ is for all the orbits of $G \curvearrowright X$ to be invariant under $\sigma$.

Generalizing slightly, note that $G$ (and $\operatorname{Sym}(X)$) have a natural product action on any power $X^{I}$ of $X$, by $g.(x_i)_{i \in I} \overset{\operatorname{df}}{=} (g.x_i)_{i \in I}$. We call an orbit of the product action on $X^I$ an $I$-orbit of $X$. For $\sigma$ to belong to $G'$, it is also necessary that for every $I$, every $I$-orbit of $X$ is $\sigma$-invariant.

Suppose that we restrict to the permutations with respect to which all the orbits of $G$ are invariant. Unless if $G$ is a product of the permutation groups of those orbits, the subgroup of all such permutations is larger than $G$.

This leads to the next natural question:

Question 2. Can we specify a collection of index sets $(I_{\alpha})_{\alpha}$ such that \[\left\{\sigma \in \operatorname{Sym}(X) \operatorname{\big{|}} \text{for all $\alpha$, for every $I_{\alpha}$-orbit $O$, $O$ is $\sigma$-invariant}\right\} = G'?\]

So in the previous paragraph, we saw that the answer to this question is “no” if $(I_{\alpha})_{\alpha}$ is just $\{1\}$.

Proposition. We can answer “yes” to Question 2 by putting $(I_{\alpha})_{\alpha} = \{|X|\}$.

Proof. Fix an enumeration $(x_i)_{i \in |X|}$. Its orbit $O\left((x_i)_{i \in |X|}\right)$ is an $|X|$-orbit of $X$, and looks like \[\left\{(g . x_i)_{i \in |X|} \operatorname{\big{|}} g \in G\right\}.\] If $\sigma$ is a permutation such that $O\left((x_i)_{i \in |X|}\right)$ is $\sigma$-invariant, then there exists some $g \in G$ such that \[\sigma . (x_i)_{i \in |X|} = (\sigma . x_i)_{i \in |X|} = (g. x_i)_{i \in |X|}.\]

Therefore, for every $x \in X$, $\sigma(x) = g . x$, and so $\sigma = \phi_X(g)$, where $\phi_X$ is the action map $G \to \operatorname{Sym}(X)$. $\square$

Actually, this previous argument works to reconstruct $G'$ even when the action is not faithful, and in fact only requires just that one $|X|$-orbit, not any of the others.

Here is an alternate way of reconstructing $G$. Recall that if $G \curvearrowright X$ is a faithful and transitive action, then it is isomorphic as a $G$-set to the regular action $G \curvearrowright G$, and $\operatorname{Aut}_{G\text{-}\mathbf{Set}}(G \curvearrowright G) \simeq G$ (by looking at what the action does on the identity $e_G$).

Here’s the trick: we can define what it means for a permutation $\sigma$ to be an automorphism of a $G$-set $G \curvearrowright X$ by requiring $\sigma$-invariance for certain subsets of $X \times X$.

Lemma. A permutation $\sigma : X \to X$ is compatible with a $G$-action $G \curvearrowright X$ (i.e. $\sigma$ is $G$-equivariant) if and only if for every $g \in G$, the graph $\Gamma(g) \subseteq X \times X$, given by $\Gamma(g) \overset{\operatorname{df}}{=} \left\{(x, g.x) \operatorname{\big{|}} x \in X \right\}$, is $\sigma$-invariant.

Proof. Suppose $\sigma$ is already $G$-equivariant, so that for all $x$ and for all $g$, $g. \sigma(x) = \sigma(g.x)$. Then for every $g$, $\sigma.(x, g.x)$ = $(\sigma(x), \sigma(g.x)) = (\sigma(x), g.\sigma(x)) \in \Gamma(g)$.

Conversely, suppose that for every $g$, $\Gamma(g)$ is $\sigma$-invariant. Then for every $x \in X$, $\sigma(x, g.x) = (\sigma x, \sigma(g.x)) \in \Gamma(g)$, which means that $\sigma(g.x) = g.\sigma(x)$. $\square$

So, suppose that $G$ acts faithfully on $X$. As we mentioned in the beginning, this action doesn’t need to be faithful on any of the orbits. However, let’s look at the product action on $X^X$. Fix an enumeration $(x_i)$ of $X$, which will be an element of $X^X$. Then $G$’s action on the orbit of $(x_i)$ is faithful: if $g_1 \neq g_2$, then since the original action was faithful, there exists some $x_j$ such that $g_1. x_j \neq g_2.x_j$. Then $g_1.(x_i)$ and $g_2.(x_i)$ do not agree at the $j$th component, and so are not the same. Since an orbit is by definition transitive, $G \curvearrowright O\left((x_i)\right)$ is faithful and transitive (and therefore we have found an isomorphic of $G \curvearrowright G$ inside $G \curvearrowright X^X$!) Now, name the graphs of the action of each $g \in G$ on $O(\left((x_i)\right))$. These are subsets of $X^{X \times X}$, and if they are simultaneously invariant under some $\sigma \in \operatorname{Sym}(X)$, then $\sigma$ acts as a $G$-set automorphism on $O((x_i)_{i \in I})$ and as a $G$-set automorphism on $X$.

So the subgroup of all such $\sigma$ embeds via some map $\psi$ into $\operatorname{Aut}(O((x_i))) \simeq \operatorname{Aut}(G \curvearrowright G) \simeq G$.1

Conversely, let $\rho \in \operatorname{Aut}(G \curvearrowright G)$. Then $\rho$ is given by right-multiplication by some group element $g$; chasing $g$ through the isomorphism $G \curvearrowright O((x_i)) \simeq G \curvearrowright G$ yields another group element $g'$ such that right-multiplication by $g'$ gives an automorphism $\rho'$ of $O((x_i))$, and $\psi(\rho') = \rho$, showing that $\psi$ is bijective, so the subgroup of all such $\sigma$ is isomorphic to $G$.

Rings and modules

The last thing I want to mention is that this all works equally well for rings and modules in place of groups and sets, and linear self-maps in place of permutations. Indeed, the two arguments presented above go through mutatis mutandis.

Proposition. Let $R \curvearrowright M$ be an $R$-module. Fix an enumeration $(m_i)$ of $M$. Then the image $\phi_M(R)$ of the action map $\phi_M : R \to \operatorname{End}(M)$ is equal to the following set: \[\left\{\sigma \in \operatorname{Hom}_{R\text{-}\operatorname{\mathbf{Mod}}}(M, M) \operatorname{\big{|}} O((m_i)) \text{ is $\sigma$-invariant under the product action} \right\}.\]

Proof. It is clear that for each $r \in R$, scalar multiplication by $r$ belongs to the set displayed above. Conversely, let $\sigma$ belong to the set displayed above. Since $O((m_i))$ is $\sigma$-invariant, there exists some $r \in R$ such that \[\sigma.(m_i) = (\sigma . m_i) = (r . m_i),\] and so for every $m \in M$, $\sigma.m = r.m$. $\square$

Similarly, if $R \curvearrowright M$ is faithful, then $R$’s product action on the orbit of $(m_i)$ is faithful and transitive, and thus $O(m_i)$ is isomorphic to $R \curvearrowright R$, which is isomorphic to $R$. And $\operatorname{Hom}_{R\text{-}\operatorname{\mathbf{Mod}}}(R, R)$ is again $R$, so naming the graphs of the scalar multiplications does the trick as in the case for groups.

These isomorphisms are obtained as follows. We can choose a $G$-$\mathbf{Set}$ isomorphism $\iota : O((x_i)) \simeq G$. $\iota$ induces by conjugation an isomorphism $\operatorname{Aut}_{G\text{-}\mathbf{Set}}(O((x_i))) \simeq \operatorname{Aut}_{G\text{-}\mathbf{Set}}(G)$, and then there is a canonical isomorphism $G \to \operatorname{Aut}_{G \text{-}\mathbf{Set}}(G)$ by right-multiplication.↩

#math #groups #group actions #rings

Actions and their equivalence

Group actions are, for me at least, an integral part of studying groups. I may have mentioned that intuitively they are natural (just as an adjective, not naturality in the sense of category theory) ways in which a group interacts with sets.

It turns out that a group can act on any set, if that action satisfies certain properties – no elements end up outside the set, identity does nothing (which should not be surprising) and acting by an element h and then element g, is the same as acting by their composition hg.

There are many interesting types of actions, and the ones I like thinking about the most are transitive. Transitivity means that, for any two points in the set on which the group acts, there is always an element of the group, that moves one to the other (the action allows us to ‘go anywhere’ in our set).

Being transitive is also important, because it allows a certain very useful result to hold – any such action of a group G is equivalent to some coset action. It follows, that although a group could transitively act on quite outlandish sets, that action is ‘the same’ (we can relabel points in our set to be cosets, and the action is compatible with that relabelling) as the group acting on cosets of one of its subgroups – which is often easier to describe and imagine.

As an aside, in the project I consider mostly actions on cosets, but I am becoming increasingly interested in how a group can act on shapes and topological spaces. Hopefully it will be something I explore in the coming academic year.

#group theory #group actions #maths

Apparently "faithful" is a thing in Abstract Algebra

A (group) action is "faithful" if it changes everything about the other (set)

(That is, induces a distinct permutation)

Math... who hurt you

#Math #abstract algebra #group actions

The p-group fixed point theorem

An easy but surprisingly useful result.

Theorem. Let $G \to \mathrm{Sym}(X)$ be the action of a $p$-group $G$ on finite $X$. Then the fixpoints of the action are $ \equiv |X| \mod p$. In particular, if $p \not{|} |X|$ then $|X|$ has a fixed point, if $|X|$ has 1 fixed point, then $|X| \equiv 1 \mod p$, and if $p | |X|$, then it has $0 \mod p$ fixed points.

Proof. Let $S$ be the fixpoints of $|X|$. By orbit-stabilizer,

\[|X| = \sum \text{ orbits } = |S| + \sum [G : \mathrm{Stab}_x]\] where the final final sum is taken over representatives of distinct nontrivial orbits; that sum vanishes mod $p$, giving the result. The other statements follow.

#group theory #p group fixed point theorem #group actions

GAH

I'm trying to prove Burnside's Formula

I did not sign up for this set theoretic summation

HELP

#mathematics #math #group theory #abstract algebra #group actions #l's life