Prove this isomorphism about semidirect product $((\oplus_1^k \mathbb{Z}/2) \oplus \mathbb{Z}/n) \rtimes_{\varphi} \mathbb{Z}/2$.

Edited: Considering the comments it turns out that my guess was true, and we have this isomorphism:

$$((\oplus_1^k \mathbb{Z}/2) \oplus \mathbb{Z}/n) \rtimes_{\varphi} \mathbb{Z}/2 \equiv (\oplus_1^k \mathbb{Z}/2) \oplus (\mathbb{Z}/n \rtimes_{\varphi} \mathbb{Z}/2), $$

but yet I have no idea how to prove this isomorphism. How can I prove this isomorphism?

Let $e, c$ be the elements of the cyclic group of order $2$, and let $\varphi_e(x)=I_N(x)=x$, and let $\varphi_c:N\rightarrow N$ be equal to $\varphi_c(x)=x^{-1}$.

I can show that the nontrivial semidirect product $\mathbb{Z}/n \rtimes_{\varphi} \mathbb{Z}/2$ is the Dihedral group $D_{2n}$. Also, I can show that $(\oplus_1^k \mathbb{Z}/2) \rtimes_{\varphi} \mathbb{Z}/2 \equiv (\oplus_1^{k+1} \mathbb{Z}/2)$, for any positive integer $k$.

What can we say about

$$((\oplus_1^k \mathbb{Z}/2) \oplus \mathbb{Z}/n) \rtimes_{\varphi} \mathbb{Z}/2?$$

What is this group? Is it isomorphic to

$$(\oplus_1^k \mathbb{Z}/2) \oplus (\mathbb{Z}/n \rtimes_{\varphi} \mathbb{Z}/2)\equiv (\oplus_1^k \mathbb{Z}/2) \oplus D_{2n}?$$

(If it is true then we should have $D_{4n} \equiv \mathbb{Z}/2 \oplus D_{2n}$ (for odd $n$), and some othe similar isomorphisms.) If the answer is no, then are there some conditios to have $$((\oplus_1^k \mathbb{Z}/2) \oplus \mathbb{Z}/n) \rtimes_{\varphi} \mathbb{Z}/2 \equiv (\oplus_1^k \mathbb{Z}/2) \oplus (\mathbb{Z}/n \rtimes_{\varphi} \mathbb{Z}/2)?$$

Solution 1:

Note in the below I shall use additive notation so $\phi_e(x)=x$ and $\phi_c(x)=-x$ as these are usually written as additive groups. I think the key observation is that if we let $(x,y)\in \oplus_1^k\mathbb{Z}/2\oplus\mathbb{Z}/n$ where $x\in \oplus_1^k\mathbb{Z}/2$ and $y\in\mathbb{Z}/n$, then $-x=x$, so $\phi_c((x,y))=(-x,-y)=(x,-y)$.

How about we just write down an isomorphism $$\psi:(\oplus_1^k\mathbb{Z}/2\oplus\mathbb{Z}/n)\rtimes_{\phi}\mathbb{Z}/2\rightarrow\oplus_1^k\mathbb{Z}/2\oplus(\mathbb{Z}/n\rtimes_{\phi}\mathbb{Z}/2)$$ we note that an element of $(\oplus_1^k\mathbb{Z}/2\oplus\mathbb{Z}/n)\rtimes_{\phi}\mathbb{Z}/2$ can be written as a three tuple say $(x,y,z)$ where $x\in\oplus_1^k\mathbb{Z}/2$, $y\in\mathbb{Z}/n$, and $z\in\mathbb{Z}/2$. Then we shall let $\psi(x,y,z)=(x,y,z)$ as we have an element of $\oplus_1^k\mathbb{Z}/2\oplus(\mathbb{Z}/n\rtimes_{\phi}\mathbb{Z}/2)$ can also be represented by such a three tuple. We note that by definition, it is clear that this function is a bijection, and maps the identity to the identity. Thus, we just need to check that multiplication works out correctly. Thus, say we have $(x,y,z),(x',y',z')\in(\oplus_1^k\mathbb{Z}/2\oplus\mathbb{Z}/n)\rtimes_{\phi}\mathbb{Z}/2$, then $$ \psi((x,y,z)+(x',y',z'))=\psi((x+\phi_z(x'),y+\phi_z(y),z+z'))=\psi((x+x',y+\phi_z(y),z+z'))=(x+x',y+\phi_z(y'),z+z') $$ Similarly, we have that $$ \psi((x,y,z))+\psi((x',y',z'))=(x,y,z)+(x',y',z')=(x+x',y+\phi_{z}(y'),z+z') $$ Thus, we have that $\psi$ respects the group operation and we are done.