Why is $\mathbb{Z_6}=\mathbb{Z_3} \rtimes \mathbb{Z_2}$?

Question

Semidirect product must satisfy 3 axioms:

(1) $N \unlhd G$ (2) $N\cap H = 1$ (3) $NH = G$

But clearly $\mathbb{Z_3} \cap \mathbb{Z_2} = \{0,1\} \neq \{0\}$, and it seems to be the case that $\mathbb{Z_3}\mathbb{Z_2} = \{0,1,2,3\} = \mathbb{Z_4}$ (by adding each element of $\mathbb{Z_3}$ with each element of $\mathbb{Z_2}$). So both (2) and (3) are not satisfied.

What is wrong with my argument?

Answer 1

The misunderstanding in your question is: how is $\mathbb{Z}/3\mathbb{Z}$ a subgroup of $\mathbb{Z}/6\mathbb{Z}$? It's not simply the subset $\{0, 1, 2\} \subseteq \mathbb{Z}/6\mathbb{Z}$, since this is not closed under addition: $1 + 2 = 3 \notin \{0, 1, 2\}$.

Instead, when you are considering $\mathbb{Z}/3\mathbb{Z}$ as a subgroup of $\mathbb{Z}/6\mathbb{Z}$, you are really identifying it with the isomorphic subgroup $2\mathbb{Z}/6\mathbb{Z}$. Similarly, $\mathbb{Z}/2\mathbb{Z} \cong 3\mathbb{Z}/6\mathbb{Z}$.

Answer 2

$\Bbb Z_6$ is isomorphic to the direct product of $\Bbb Z_2$ and $\Bbb Z_3$. Every direct product is a semidirect product. But this direct product does not work the way you think.

If we write $\Bbb Z_6=\{0,1,2,3,4,5\}$ the subgroups $N$ and $H$ can be taken to be $\{0,2,4\}$ and $\{0,3\}$. These are cyclic subgroups and isomorphic to $\Bbb Z_3$ and $\Bbb Z_2$ respectively.