Dihedral group as a semidirect product?

Question

It is known that the dihedral group $D_{2n}$ is isomorphic to the semidirect product $Z_n\rtimes Z_2$, where both $Z_n,Z_2$ are cyclic.

My question is, for a semidirect prouct, the two subgroups should have trivial intersection. But, if $n$ be even, then how can we get trivial intersection? Namely, an order $2$ element would be common to both $Z_n,Z_2$ right? And the order two elements in a dihedral group would be either a $180$ degree rotation about an axis or a reflection about an axis, which are both identical right? Any hints? Thanks beforehand.

Answer 1

I had the same confusion. I used to mistake inner for external semidirect product. When we write $C_{n}\rtimes C_{2}$, we just construct a new binary operation on the cartesian product $C_{n}\times C_{2}$. So $C_{2}\le C_{n}$ (if exists), has nothing to do with $C_{2}$ that acts on $C_{n}$. On the other hand, when we say $G$ is inner semidirect of its two subgroups, we mean that there exists a normal subgroup $N\le G$, and a subgroup $K$ (not necessarily normal), where $G=NK$ and $N\cap K=1$. In this case you can prove that $G=NK\cong N\rtimes K$.

For more detailed explanations, please check: https://en.wikipedia.org/wiki/Semidirect_product

Answer 2

Consider the map \begin{align*} \phi\colon\mathbb Z_2&\to\text{Aut}(\mathbb Z_n)\\ t&\mapsto(c\mapsto(-2t+1)c). \end{align*} It is easy to verify that $\phi(t_1+t_2)=\phi(t_1)\circ\phi(t_2)$, i.e., that $\phi$ is an homomorphism. Then we can define the associated action \begin{align*} \mathbb Z_2\times\mathbb Z_n&\to\mathbb Z_n\\ (t,c)&\mapsto t\cdot c=(-2t+1)c. \end{align*} The multiplication in $\mathbb Z_n\rtimes\mathbb Z_2$ is then $$ (c_1,t_1)(c_2,t_2) = (c_1+(-2t_1+1)c_2, t_1+t_2), $$ where the factor $-2t_1+1$ expresses the difference with the sum operation in $\mathbb Z_n$ (source of your confusion). Note however, that $(0,0)$ is still the identity. In particular, $$ (0,t)^2 = (0,0) $$ and so $(0,1)$ is an involution and \begin{align*} (c,0)^{(0,1)} &= (0,1)(c,0)(0,1)\\ &= ((-2+1)c,1)(0,1)\\ &= (-c,0)\\ &= (c,0)^{-1} \end{align*} because $(c,0)(-c,0)=(c-c,0)=(0,0)$.

The other involutions are $$ (c,0)(0,1) = (c,1), $$ which exactly correspond to the elements of $\mathbb Z_n\rtimes\mathbb Z_2\setminus D$, where $D$ is the image of $\mathbb Z_n$ in $\mathbb Z_n\rtimes\mathbb Z_2$.

Finally, note that the intersection of $\mathbb Z_n$ with $\mathbb Z_2$ inside $\mathbb Z_n\rtimes\mathbb Z_2$ is trivial because $\mathbb Z_n$ is embedded as $\{(c,0)\mid c\in\mathbb Z_n\}$ and $\mathbb Z_2$ as $\{(0,t)\mid t\in\mathbb Z_2\}$.