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If we define a function $\phi : D_{2n} \rightarrow D_n \times Z_2$ for odd $n$ and we want to show that it is an isomorphic function, I am not very sure how to do it. We know that $D_{2n} = \{e, r, \ldots, r^{2n-1}, s, rs,\ldots, r^{2n-1}s\}$ and $D_n = \{e, t, \ldots, t^{n-1}, u, tu, \ldots, t^{n-1}u\}$.

I think that if we define $\phi(r) = (t, 1)$, then we can show that it is a homomorphism by showing that $\phi(r)\phi(r) = (t, 1)(t, 1) = (t^2, 0) = \phi(r^2)$, but I do not know if this is enough already or if I have to do it for all possibilities?!

And then the bijection. I know that it is injective if $\phi(a) = \phi(b) \iff a = b$, but how do I prove it in this particular example? And how do I show that every element of $D_n \times Z_2$ has an element in $D_{2n}$ to prove surjectivity?

E W H Lee
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FlyingDutchman
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Hint: You will also need to define $\phi(s)$; since $D_{2n}$ is generated by $r$ and $s$, this will fully define $\phi$. It then suffices to show that the homomorphism property holds for $r^2$, $rs$, $sr$ (and $s^2=e$). For bijection, note that you need only show either injectivity or surjectivity since the two sets have the same size.

rogerl
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  • Thank you very much! I showed that the homomorphism holds for $r^2, rs, sr$ and $s^2$. I said that it has to be injective, because every element $D_{2n}$ has a different outcome under $\phi(x)$ and the order is the same. Is this the correct way to prove the injection? – FlyingDutchman Oct 07 '14 at 18:45
  • Almost. Just showing that every element of $D_2n$ maps to a distinct element in $D_n\times\Z/2$ is enough to show injectivity since that is the definition of being injective. The fact that the two sets have the same size and are finite implies that any injective map is surjective. – rogerl Oct 07 '14 at 21:22