Some properties implied from the definition of semidirect product.

Question

Problem: Let $G$ be an inner semidirect product of $N$ by $H$. For each element of $H$, denote $\theta_h \colon N \rightarrow N$ is a map defined by $\theta_h (x) = h x h^{-1}, \forall x \in N$. Prove that:

1. $\theta_h$ is a isomorphism of $N$ for all $h \in H$.
2. The map $\theta \colon H \rightarrow \text{Aut}(N)$ defined by $\theta(h) = \theta_h$ for all $h \in H$ is a group homomorphism.
3. All element $g \in G$ can be written uniquely in the form $g=xh$ in which $x \in N$ and $h \in H$.
4. Suppose $g_1 = x_1 h_1$ and $g_2 = x_2 h_2$ with $x_1,x_2 \in N$ and $h_1,h_2 \in H$. Then $g_1 g_2 = (x_1 h_1) (x_2 h_2) = x_1 \theta_{h_1} (x_2) h_1 h_2$.

My proof:

1. $\forall x,y \in N, \theta_h (xy) = h xy h^{-1} = h x h^{-1} h y h^{-1} = \theta_h (x) \theta_h (y)$. So $\theta_h$ is a group homomorphism. For each $h \in H$, we have the map $\theta_{h^{-1}}$ with $\theta_{h^{-1}} (x) = (h^{-1})x(h^{-1})^{-1} = h^{-1} x h$, implies $\theta_h \theta_{h^{-1}} (x) = h h^{-1} x h h^{-1} = x$. Hence $\theta_h$ is a group isomorphism.

2. $\forall h, h' \in H$, we have $\theta (h h') (x) = \theta_{h h'} (x) = h h' x (h h')^{-1} = h h' x h'^{-1} h^{-1} = h \theta_{h'} (x) h^{-1} = \theta_h (\theta_{h'} (x)) = (\theta_h \theta_{h'}) (x)$. This follows that $\theta$ is a group homomorphism.

3. $G$ is an inner semidirect product of $N$ by $H$. So we have $N,H$ are subgroup of $G$, $G=NH$, $N \triangleleft G$ and $N \cap H = \{1\}$. Is that enough to conclude that all element of $G$ can be written quiquely in the form $g=xh$, $x \in N$, $h \in H$?

4. We have $g_1 g_2 = (x_1 h_1)(x_2 h_2) = x_1 h_1 x_2 h_2 = x_1 h_1 x_2 h_1^{-1} h_1 h_2$, this is the mupltiplication in $G$, so it associative, $= x_1 (h_1 x_2 h_1^{-1}) h_1 h_2 = x_1 \theta_{h_1} (x_2) h_1 h_2$.

Please check my proof and show me some hint to prove that $3$. This problem could be very easy to many people, it seems like the definition but I have to prove this. Thank all!

Answer 1

To (3), yes, but you should give the proof. If $n_{i} \in N$, $h_{i} \in H$, and $n_{1} h_{1} = n_{2} h_{2}$, then $N \ni n_{2}^{-1} n_{1} = h_{2} h_{1}^{-1} \in H$, and thus $n_{2}^{-1} n_{1} = 1 = h_{2} h_{1}^{-1}$, as $N \cap H = \{ 1 \}$.

As to (4), $$ (x_1 h_1) (x_2 h_2) = x_{1} h_{1} x_{2} (h_{1}^{-1} h_{1}) h_{2} = x_{1} (h_{1} x_{2} h_{1}^{-1}) h_{1} h_{2}= \dots $$

Answer 2

1 and 2 are fine (maybe in (2) specify that $\theta_h(1)=1$)

Hint (3): Since $G=NH$ by definition certainly every $g \in G$ can be written as $g=nh$ for some $n \in N$, $h\in H$. The part that needs to be proved is the "uniquely". So suppose that $g=n_1 h_1$ and $g=n_2 h_2$, can you prove that $n_1 = n_2$ and $h_1=h_2$?

Hint (4): what is the definition of an inner semidirect product? You are right, that is the definition of the operation... of an outer semidirect product. In the end, the two notions coincide, and what the exercise is asking you is to do a step of the proof of this fact.