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4

It only holds for the trivial group. As it was mentioned in the comments, we consider $\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module and in this way the augmentation map $\epsilon \colon \mathbb{Z}G \to \mathbb{Z}$ is a morphism of $\mathbb{Z}G$-modules. Let $ f \colon \mathbb{Z} \to \mathbb{Z}G$ be a morphism of $\mathbb{Z}G$-modules and say $ f(1) = ...


4

A nice instructive example is to take the group $\mathbb{Z}$ and let it act on $S^1$ by an irrational rotation. That is, let $\alpha\in[0,1]$ be some irrational number and define $n\cdot e^{i\theta}=e^{i(\theta+2\pi n\alpha)}$ for $n\in\mathbb{Z}$. The action is free since $\alpha$ is irrational, so $e^{2\pi i n\alpha}\neq 1$ for any nonzero integer $n$. ...


3

It sounds like you're describing the quotient space $\mathbb{R}^n/S_n$. Some people call this the $n^{th}$ "symmetric power" (of $\mathbb{R}$), although be a little careful with that terminology because it can be used to refer to two other related but different constructions. This quotient is not a manifold, but can be thought of as an orbifold. For ...


2

The map is smooth because we can construct it by passing the smooth map $G \to G_p$ via the natural quotient map $G \to G / G_p$. It follows immediately from the definitions of the two group actions that the map is equivariant. Thus, it has constant rank, and any bijective, smooth map of constant rank is a diffeomorphism. (See $\S$9 of Lee's Introduction to ...


2

This is a consequence of the slice theorem of Koszul https://en.wikipedia.org/wiki/Slice_theorem_%28differential_geometry%29


1

Your example works since you can take the non-zero vector $v := e_1 + e_2 + \dots + e_n$ which is fixed by all $g \in G \leq S_n$. It works also (probably a little bit easier) if you simply take $G \hookrightarrow S_n \hookrightarrow S_{n+1}$, where $S_n$ is the subgroup of permutations which fix $n+1$. In the picture of your linear action (but now) onto ...


1

The action of $G$ by multiplication on the the left cosets of $B$ is equivalent to the projective action of $G$ on the $1$-dimensional subspaces of ${\mathbb C}^2$, since $B$ is precisely the stabilizer of $\langle (1\ 0)^{\rm T} \rangle$. Now $w$ maps $\langle (1\ 0)^{\rm T} \rangle$ to $\langle (0\ 1)^{\rm T} \rangle$, which is stabilized by $T$, so ...


1

2) Clearly any $Y$ that has such a $H$ must be equal to $H\cdot Y$, and by transitivity all such $Y$ must be of the form $H\cdot x$ for at least one $x \in Y$. Now clearly for any $x \in X$ and any $H \leq G$, the orbit $H\cdot x$ has the property that $H\cdot (H\cdot x) = (HH)\cdot x = H\cdot x$, so $H$ acts transitively on $H\cdot x$, and freeness is ...


1

Hint: show that the subset of permutations whose signature is 1 and the subset whose signature is $-1$ are preserved by the action.


1

You have the action of $S_n$ on itself by conjugation. The equivalence classes, or orbits, under an action by conjugation are called conjugacy classes. What you need to show is that two elements in $S_n$ are conjugate if and only if they share the same cycle type. You can find a proof of that on page 126 of Dummit and Foote under Proposition 11, for one. ...



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