Is $D_n$ isomorphic to $C_n \times C_2$?

Question

I am doing a proof for an exercise where if the dihedral group $D_{n}$ $\cong$ $C_{n} \cdot C_{2}$ where $\cong$ means isomorphic and $\cdot$ is the direct product, I would be able to finish it, but I don't know it this is true. $C_{n}$ is the cyclic group of order $n$.

Answer 1

The exercise is wrong. It's the semidirect product $$D_{2n}\cong C_n \rtimes C_2.$$

Answer 2

Perhaps the notion $\cdot$ is not what you claim\think it is.

I mean, usually given any $H,K\leq G$ we can define: $$HK=\{h\cdot k : h\in H,k\in K\}$$ this is not necessarily a direct product!

With respect to this definition, the answer is yes. One can show that $D_n = C_n\cdot C_2$ where $D_n = \left<\tau,\sigma\right>$ with a rotation $\sigma$ and translation $\tau$ with $C_n$ being identified with $\left<\sigma\right>$ and $C_2$ being identified with $\left<\tau\right>$.

Indeed, every element in $D_n$ can be written as $\sigma^k\cdot \tau^l$ for $0\leq k \leq n-1$ and $0\leq l \leq 1$.

The last line is an easy, yet not completely trivial exercise and I bet that this is what the question in your exercise sheet is about.