Translation Action, Cayley's Theorem and Semi Direct Products

Recall: G acts on itself by $g \cdot a = ga$. More generally G acts on P(x) by $g*X = gX = \{g\cdot x, x \in X\}$. In particular: $H \leq G$, G acts on G/H by $g*(xH) = gxH$.

Examples: $(\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,c\}$ is abelian. $a^2 = b^2 = c^2$ all non identity elements have order 2. ab =c, bc = a, ca = b. Let $\sigma_a : G \rightarrow G$ be defined as $x \mapsto a \cdot x. \sigma_a = (1 a)(b c), \sigma_b(1 b)(a c)...$ then $\sigma_a(1) = a, \sigma_a(a) = 1, \sigma_a(b) = c, \sigma_a (c) = b$.

In $D_8, H = <s>, D_8/H = \{H, rH, r^2, r^3H\}$ so $\sigma_s(H) = sH = H$ $\sigma_s(rH) = srH = r^{-1}sH = r^3H$ $\sigma_s(r^2H) = sr^2H = r^{-2}sH = r^2H$ $\sigma_s(r^3H) = sr^3H = r^{-3}sH = rH$ Hence $\sigma_s = (rH r^3H)$ and $\sigma_r(H rH r^2H r^3H)$

Theorem: G is a group, $H \leq G$, G acts on G/H by translation. 1) There is only 1 orbit (we say that the action is transitive) 2) The stabilizer of xH is $xHx^{-1}$ 3) If $\pi_H : G \rightarrow S_G/H$ is the permutation representation associated to the translation action, then the kernel is ker($\pi_H) = \bigcap_{x \in G}xHx^{-1} = K$ where K is the largest subgroup of H normal in G.

Proof: 1) Let $xH, yH \in G/H$, then $(yx^{-1})xH = yH$ so xH and yH are in the same orbit. 2) the stabilizer $\{ g \in G: gxH = xH\} \leftrightarrow x^{-1}gxH = H \leftrightarrow x^{-1}gx \in H \leftrightarrow g \in xHx^{-1}$ 3) ker$(\pi_H) = \bigcap_{x \in G} S_G(xH) = \bigcap_{x \in G} xHx^{-1}$ so $K \unlhd G$ because it is a kernel and $K = \bigcap_{x \in G} xHx^{-1} \leq H$. Let $N \unlhd G, N \leq H, \forall x \in G, N = xNx^{-1} \leq xHx^{-1}, N \leq \bigcap_{x \in G} xHx^{-1} = K$.

Cayley’s Theorem: let G be a group, then G is isomorphic to a subgroup of a symmetric group (potentially infinite). If $\vert G \vert = n < \infty$ it is isomorphic to a subgroup of $S_n$.

Proof: let $H = \{1\}, ker(\pi_H) = \{1\}$ so $\pi_H$ is a monomorphism. By the first isomorphism theorem $G \cong im(\pi_H) \leq S_G$. If $\vert G \vert = n \leq \infty, G \rightarrow^{\pi_H} S_G \cong S_n$ then $G \cong im(\theta) \leq S_n$ where $\theta$ is a monomorphism from $G \rightarrow S_n$.

Corollary: G is a finite group, p the smallest prime dividing n, $H \leq G$ of index p, then $H \unlhd G$ (note, we are not claiming H exists, if G is simple and nt cyclic such H does not exist)

Proof: let $\pi_H : G \rightarrow S_{G/H} \cong S_p$. Let $K = ker(\pi_H)$, by the first isomorphism theorem $G/K \cong im(\pi_H) \leq S_{G/H}$. Let $k = [H:K], [G:K] = [G:H][H:K] = pk, \frac{\vert G \vert}{\vert H \vert} = [G:H] \cdot \frac{\vert H \vert}{\vert K \vert}$ By lagrange’s theorem $[G:K]$ divides $\vert S_{G/H} \vert = p!$. K divides $\vert G \vert$ by Lagrange Assume $K \neq 1$ let q be the prime that divides K, $q \vert (p-1)! \rightarrow q < p. q \vert n \rightarrow q \leq p$ which is a contradiction So K = 1, $[H:K] = 1$ so $H = K \unlhd G$

Semi direct products

Let G be a group, $H, K \leq G$ and $H \unlhd G$, then $HK \leq G$. If moreover $H \cap K = \{1\}$ then $\vert H \cap K \vert = \frac{\vert H \vert \vert K \vert}{\vert H \cap K \vert} = \frac{\vert H \vert \vert K \vert}{\vert H\times K \vert}$. Let $x, y \in H, y, t \in K$ then $xyst = xysy^{-1}yt$. $(x,y)(s,t) = (x\cdot ysy^{-1}, yt)$ where $ysy^{-1} = y*s$.

If we don’t have an ambiant group G: let H and K be groups. Let $\phi : K \rightarrow Aut(H)$ be a homomorphism that sends $y \mapsto \sigma_y$, then we have a group action $y \cdot x =\sigma_y(x)$. This group action has more properties: $y*(x_1 \cdot x_2) = (y*x_1) \cdot (y*x_2)$. We say that K acts on H by group automorphism.

For example, conjugation is a group action by group automorphism, translation is not ($gxy \neq gxgy$).

Let $G = H\times K$ and $(x,y)\Box (s,t) = (x\cdot(y*s),yt)$. Note that the action is trivial.

Proposition: 1) $(G, \Box)$ is a group 2) $\phi_H : H \rightarrow G$ defined by $x \mapsto (x,1)$ is a group monomorphism. $\phi_k : K \rightarrow G$ is defined by $y \mapsto (y,1)$ and is also a group monomorphism. We can identify H with $\tilde{H} = \{(x,1) : x \in H\} \leq G$ and K with $\tilde{K} = \{(1,y) : y \in K\} \leq G$ 3) $\tilde{H} \cap \tilde{K} = \{1\}, H \unlhd G, HK \leq G/ (x,1) = \tilde{x} \in \tilde{H}. (1,y) = \tilde{y} \in \tilde{K}. \tilde{y}\tilde{x}\tilde{y}^{-1} = (y*x,1)$

Proof (the proof of 3 is left as an exercise): 1) $((x,y) \Box (s,t)) \Box (u,v) = (x \cdot (y*s), yt) \Box (u,v) = ((x \cdot (y*s)\cdot (yt)*u), ytv)$ $(x,y)\Box ((s,t)\Box (u,v)) = (x,y) \Box (s \cdot (t*u), tv) = (x \cdot (y*s)\cdot y(\cdot(t*u)), ytv)$ Note that (1,1) is the identity, and $(y^{-1}*x^{-1}, y^{-1})$ is the inverse of (y,x) and that y*1 = 1. 2) $(x,1) \Box (s,1) = (x \cdot (t*s), 1\cdot 1) = (x\cdot s,1)$ $(1,y)\Box(1,t) = (1 \cdot (g*1), yt) = (1,yt). (x,1) = (1,1)$ if and only if x = 1. (1,y) = (1,1) if and only if y = 1.

Notation: G (as defined above) is called the semidirect product of H and K and we write $G = H\times K$ (for the action *).

Example: $D_{2n} \cong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ for the action $0*i = i, 1*i = -i$. If $\phi(y)$ is the identity transformation for all y then $H \times K$ is just the product.