Showing that $\phi$ is a homomorphism

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$G=\mathbb{Z}^2$ is a group with product $(a,b)\cdot(c,d)=(a+c,(-1)^cb+d)$. Show that the image $\phi: G \to D_{10}$ with $(a,b) \mapsto s^ar^b$ is a homomorphism ($D_{10}$ is the dihedral group of order $10$).


I know we have to show that $\phi(xy)=\phi(x)\phi(y)$.

Let's choose some random elements in $G$: $(a,b)$ and $(c,d)$.

$\phi((a,b))\phi((c,d)) = s^{a+c}r^{-b+d}=\phi((a+c,-b+d))$

I can only make this equal $\phi((a,b)(c,d)$ if I put a restriction on $c$: $c$ can only be odd. But that doesn't prove that $\phi$ is a homomorphism.

I'm pretty sure I have to use the fact that we're working with $D_{10}$ but I don't know what I'm doing wrong.

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You have to show that $$s^ar^bs^cr^d=s^{a+c}r^{(-1)^cb+d}$$ or - after cancelling - that $$r^bs^c=s^{c}r^{(-1)^cb}$$ Can you take it from here?