prove that : $D_{4n} $ is isomorphic to $ D_{2n} \times Z_2 $ when $n$ is odd

Question

let $n$ be odd integer , prove that : $D_{4n} $ is isomorphic to $ D_{2n} \times Z_2 $

it's an example which the text proves ! but i can't understand any thing from the argument !

but i tried to prove it by constructing the isomorphism function directly but every time there was a tiny gap !so anyone knows nice proof fot this ?

Answer 1

Note that $Z_n\times Z_2\cong Z_{2n}$ if $n$ is odd and the inveres of $(x,y)\in Z_n\times Z_2$ is $(-x,y)$, that is we have an operation of $Z_2$ on $Z_{2n}$ by inversion that boils down to the trivial operation on the summand $Z_2$ and again inversion on $Z_n$. Therefore $$ D_{4n}\cong Z_{2n}\rtimes Z_2\cong(Z_n\times Z_2)\rtimes Z_2\cong(Z_n\rtimes Z_2)\times Z_2\cong D_{2n}\times Z_2.$$

Answer 2

Hagen's proof is great, and certainly more generalizable (a great advertisement for semidirect products).

In this particular case, however, it may be more intuitive to see what's going on...

If we have an even-sided regular polygon, say with $2n$ sides, then we may form a regular polygon inside it by connecting every other vertex, in turn. Actually, we can form two such, by offsetting by one. Doing this with a hexagon should produce a familiar star-shape, which can also be found, for example, on the Israeli flag ;)

Now, pick one of these $n$-gons. Any symmetry of the $2n$-gon either sends this $n$-gon to itself (in which case it is a symmetry of the $n$-gon), or else sends it to the other $n$-gon (possibly with a symmetry applied). This gives a map $D_{4n} \rightarrow D_{2n} \times C_2$. Now note that once you've said where the $n$-gon goes you've uniquely determined a symmetry of the $2n$-gon, so this map is bijective.