# How to find the semi direct product of group explicitly?

Classification of finite group upto Isomorphism of order $$6$$ :

$$|G|=6=2\cdot 3$$

$$H:=\textrm{Syl}_3(G)$$ and $$K:=\textrm{Syl}_2(G)$$

Then $$n_3\cong{1}\mod(3)$$and $$n_3|2$$

$$\boxed{n_3=1}$$

$$n_2\cong{1}\mod(2)$$and $$n_2|3$$

$$\boxed{n_2=1\text{or}3}$$

We have so far -

$$H \vartriangleleft G$$

$$K \le G$$

$$H\cap K=(0)$$

$$|HK|=6=|G|$$

Hence $$G=H\rtimes_{\phi}K$$

Where $$\phi: K\to \textrm{Aut}{H}$$ is a homomorphism.

Let $$H=\langle x\rangle, K=\langle y\rangle$$

$$\textrm{Aut}(H) =\{\textrm{id}(x\to x) ,\psi(x\to x^2)\}$$

Since $$\phi: K\to \textrm{Aut}(H)$$ is a homomorphism, $$\phi(y) =\textrm{Id}$$ or $$\phi(y) =\psi$$

Case $$1$$ :

$$G=\langle x, y |x^3=1=y^2,y^{-1}xy=x\rangle\cong\mathbb{Z}_3\times \mathbb{Z}_2$$

Case $$2$$ :

$$G=\langle x, y |x^3=1=y^2,y^{-1}xy=x^2\rangle\cong D_3\cong S_3$$

Now I want to compute the semidirect product $$\mathbb{Z}_3,\mathbb{Z}_2$$ explicitly.

$$G=\Bbb{Z_3}\rtimes_{\phi}\Bbb{Z_2}=\{ (h,k):h\in \Bbb{Z_3},k\in \Bbb{Z}_2 \}$$

Where $$(h_1, k_1)•(h_2, k_2) =(h_1+\phi_{k_1}(h_2), k_1+k_2)$$

Let $$\psi:\Bbb{Z}_3\to \Bbb{Z}_3$$ is an automorphism.

Then $$\psi$$ respect the generating set i.e $$\psi(1) = 1 \text{or} 2$$

$$\textrm{Aut}(\Bbb{Z_3}) =\{\textrm{Id}(x\to x),\psi(x\to 2x)\}$$

Case $$1$$ : $$\phi=\textrm{Id}$$

$$G=\Bbb{Z_3}\rtimes_{\textrm{Id}}\Bbb{Z_2}=\{ (h,k):h\in \Bbb{Z_3},k\in \Bbb{Z}_2 \}$$

Where \begin{align}(h_1, k_1)•(h_2, k_2)&=(h_1+h_2,k_1+k_2)\end{align}

$$G=\Bbb{Z_3}\rtimes_{\textrm{Id}}\Bbb{Z_2}\cong \Bbb{Z_3}\times\Bbb{Z_2}\cong \Bbb{Z}_6$$

Case $$2$$ : $$\phi=\psi$$

$$G=\Bbb{Z_3}\rtimes_{\psi}\Bbb{Z_2}=\{ (h,k):h\in \Bbb{Z_3},k\in \Bbb{Z}_2 \}$$

Where

\begin{align}(h_1, k_1)•(h_2, k_2)&=(h_1+\psi_{k_1}(h_2), k_1+k_2)\\&= (h_1+2h_2,k_1+k_2)\end{align}

Computing the order:

$$|(0, 0) |=1$$

$$(1, 0) •(1, 0) =(1+2, 0) =(0, 0)$$

$$(2, 0) •(2, 0) =(2+4, 0) =(0, 0)$$

$$(0, 1) •(0, 1) =(0, 2) =(0, 0)$$

$$(1, 1) •(1, 1) =(1+2, 2) =(0, 0)$$

$$(2, 1) •(2, 1) =(6, 2) =(0, 0)$$

What I have done is completely wrong.

I think I have made a mistake in the group operation.

• Your calculations are wrong. Look at the definition of the product in the semidirect product. You need to do $\phi_{k_1}(h_2)$, but you are just doing $\phi(h_2)$ always. Commented Feb 24, 2023 at 15:22
• Use $x\mid y$ for $x\mid y$. Compare this to $x|y$, which looks like $x|y$. Commented Feb 24, 2023 at 16:17

To do the semidirect product of $$H$$ by $$K$$, you do not select an automorphism of $$H$$. You select a homomorphism $$\psi\colon K\to \mathrm{Aut}(H)$$, and you use $$\psi$$ to define the multiplication. You define the multiplication by $$(h_1,k_1)\bullet (h_2,k_2) = \Bigl( h_1\bigl(\psi(k_1)(h_2)\bigr), k_1k_2\Bigr),$$ where $$\psi(k_1)\in\mathrm{Aut}(H)$$, so we can evaluate it at $$h_2$$ to get an element of $$H$$.
You are instead selecting an automorphims $$\phi\in\mathrm{Aut}(H)$$ and trying to define the product as $$(h_1,k_1)\bullet(h_2,k_2) = (h_1\phi(h_2),k_1k_2),$$ which of course does not work. It's not even associative in general: for example, with $$H=\mathbb{Z}_3$$ and $$\phi(x)=x^{-1}$$, you would have $$\Bigl( (x,e)(x,e)\Bigr)(x,e) = (xx^{-1},e)(x,e) = (x^{-1},e)$$ but $$(x,e)\Bigl( (x,e)(x,e)\Bigr) = (x,e)(xx^{-1},e) = (x,e).$$
So what you need is a morphism from $$\mathbb{Z}_2$$ to $$\mathrm{Aut}(\mathbb{Z}_3)$$. You can select the one that sends every element of $$\mathbb{Z}_2$$ to the identity automorphism; that's your first construction. Or you can select the one that sends the identity element of $$\mathbb{Z}_2$$ to, necessarily, the identity automorphism; and sends the nonidentity element of $$\mathbb{Z}_2$$ to the map sending $$h$$ to $$2h$$. Then the correct formula for the multiplication is $$(h_1,k_1)\bullet (h_2,k_2) = \left\{\begin{array}{ll} (h_1+h_2,k_1+k_2)&\text{if }k_1=0,\\ (h_1+2h_2,k_1+k_2)&\text{if }k_1=1. \end{array}\right.$$
• Dear professor ,thank you. I have understood where I have done mistake. $\phi$ can't be equal to $\psi$ as $\phi: K\to \textrm{Aut}(H)$ homomorphism and $\psi:H\to H$ automorphism. For a fixed $k\in K$ , $\phi_k\in\textrm{Aut}(H)$ . As $K$ cyclic, image of $\phi$ can be completely determined by $\phi(1)$ which can be $\textrm{Id}$ or $\psi$ . In both cases $\phi(0) =\textrm{Id}$ . Two homomorphisms are $\phi(k) =\textrm{Id}$ for all $k\in K$ and other homorphism $\phi(k) =\begin{cases}\textrm{Id}&k=0\\\psi&k=1\end{cases}$ Commented Feb 24, 2023 at 17:41