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Isomorphism of dihedral with a semi-direct product
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[QUOTE="fresh_42, post: 6098165, member: 572553"] I suggest to write it in terms of ##r## and ##s## first, and then see how it translates to ##\mathbb{Z}_m##, resp. ##\mathbb{Z}_2##. And I will write the representatives, aka elements of ##\mathbb{Z}_m## as ##[k] = k+m\mathbb{Z}## which is more convenient and easier to read. In general a semidirect product ##G = N \rtimes_\varphi H## goes by ##(n_1\, , \,h_1)\cdot (n_2\, , \,h_2) \stackrel{(*)}{=} (n_1 \cdot \varphi(h_1)(n_2)\, , \,h_1\cdot h_2)##, see e.g. [URL]https://en.wikipedia.org/wiki/Semidirect_product[/URL]. We have ##\varphi\, : \,H \longrightarrow \operatorname{Aut}(N)##. Here we have products ## P =(r^k\; , \;s^\varepsilon)\cdot (r^n\; , \;s^\eta) = r^k\cdot s^\varepsilon \cdot r^n \cdot s^\eta## with ##\varepsilon \in \{\,0,1\,\}##. Now write ##P## in the form ##P=r^l \cdot s^\mu=(r^l,s^\mu)## and compare that with ##(*)## to see how ##\varphi(s^\varepsilon)(r^n)## has to be defined. If you're done, you can write this as ##\varphi([\varepsilon])([n])##. You have only defined ##\varphi([1])([1])##. What are ##\varphi([0])## and ##\varphi([1])([n])## if ##n>1## or short: What is ##\varphi([\varepsilon])## with ##\varepsilon \in \{\,0,1\,\} = \{\,[0],[1]\,\}\,?## [/QUOTE]
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Isomorphism of dihedral with a semi-direct product
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