$N⋊G$对$G$的右陪集的忠实作用

hbghlyj

我想知道一般情况下是否存在 $N⋊G$ 对 $N$ 的忠实作用，如果存在，如何构造它。
The geometry of the action of the semidirect product
Suppose $K\circlearrowright X$ is an action of $K$ on some set. We have the structure homomorphism $\varphi:K\rightarrow \text{Sym}(X).$

Let $H$ act on $K$ by automorphisms, and suppose these automorphisms can be realized as inner automorphisms within $\text{Sym}(X)$. That is, the action is given by some $\theta:H\rightarrow\text{Sym}(X)$ such that conjugations by $\theta(H)$ leave $K$ invariant. Equivalently, the $H$-action takes stabilizer subgroups of $K\circlearrowright X$ to stabilizers (it may permute them nontrivially).

Then we can construct an action of $K\rtimes H$ on $X$ by $(k,h).x = \varphi(k)\cdot\theta(h).x$, where the product $\varphi(k)\theta(h)$ is just taken in $\text{Sym}(X)$. Here, the multiplication rule for $K\rtimes H$ is
$$(k_1, h_1)\cdot (k_2,h_2) = (k_1 \cdot (h_1.k_2), h_1 h_2). $$

Any action arises in this way, since $H$ acts on $K$ by conjugation in the semidirect product $K\rtimes H$ and therefore it also acts by conjugation in the image under the structure morphism $K\rtimes H \rightarrow \text{Sym}(X)$ of an action of the semidirect product on $X$. This explains when an action of $K$ can be extended to an action of $K \rtimes H$.

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It is not clear how to visualize the above. So let's pass to a nice special case.

Given another action of $H$ on a set $Y$, we can extend $H\circlearrowright Y$ to an action $K\rtimes H \circlearrowright Y$ (let the latter act via the quotient $K\rtimes H\twoheadrightarrow H$). Then we can produce an action of $K\rtimes H \circlearrowright Y \times X$. This action descends to the quotient $H\circlearrowright Y$ so that $K$ fixes each fiber $\{y\}\times X$, and $H$ acts by permuting fibers and "twisting" $X$.

If the $H$-action is faithful, the action of any element $(k,h)$ can be nicely separated into an $h$-part and a $k$-part, the $h$-part being uniquely identified by the action on $Y$. Thus given a permutation $\sigma$ of $Y\times X$ we can write that it is of the form $(k,h)$ for a known $h$, and then compute the permutation $(y,x)\mapsto \sigma\left((e,h^{-1}).(y,x)\right)$, which acts the same as $(k,h)\cdot(e,h^{-1}) = (k,e)$.

One gets some examples which appear different, but are isomorphic to these, by choosing different identifications between fibers than $\text{id}_X : \{y_1\}\times X \rightarrow \{y_2\}\times X$. These identifications may be analogous to the connection described in the question.

hbghlyj

Faithful Action of N ⋊ G on N

To see this action is faithful let $(n,g)∈N⋊G$. $(n,g)=(1,g)(n⋅g,1)$

为什么$N⋊G$乘法满足$(n,g)=(1,g)(n⋅g,1)$呢

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明白了:
定义$G$在$N$上的作用为$n^g=ng^{-1}$
则$(n,g)=\bigl(1⋅(n⋅g)g^{-1},g⋅1\bigr)=\bigl(1⋅(n⋅g)^g,g⋅1\bigr)=(1,g)(n⋅g,1)$

Czhang271828

hbghlyj 发表于 2023-5-11 08:34
Faithful Action of N ⋊ G on N

为什么$N⋊G$乘法满足$(n,g)=(1,g)(n⋅g,1)$呢

$N\rtimes G$ 的乘法定义作
\[
((n_1,g_1),(n_2,g_2))\mapsto (n_1 n_2^{g_1} ,g_1g_2)
\]
其中 $n_2^{g_1}$ 就是 $g_1$ 在 $n_2$ 上的作用. 原回答里规定了是在右陪集上的作用, 所以不妨令 $n_2^{g_1}=n_2g_1^{-1}$.