In my lecture notes it says that the group generated by the automorphisms $\sigma(t)=it, \tau(t)=\tfrac{1}{t}, G=<\sigma,\tau>$ is the Dihedral 8 Group .
Now $D_8=<\sigma,\tau|\sigma^4=\tau^2=1, \tau \sigma \tau^{-1}=\sigma^{-1}>$
And we see that:
$\sigma^4=\sigma(\sigma(\sigma(\sigma(t))))=\sigma(\sigma(\sigma(it)))=\sigma(\sigma(-t))=\sigma(-it)=t$
$\tau(\tau(t))=\tau(\tfrac{1}{t})=t$
But for the final condition that $\tau \sigma \tau^{-1}=\sigma^{-1}$ I seem to be making a mistake . First we note that $\tau^{-1}(t)=\tau(t)=\tfrac{1}{t}$, while $\sigma^{-1}(t)=-it$
I get that
$\tau \sigma \tau^{-1}(t)=\tau \sigma (\tfrac{1}{t})=\tau(\tfrac{1}{it})=it$
What mistake am I making ?
Your mistake is it should be $\sigma (\tfrac{1}{t})=(i\tfrac{1}{t}),$ not $\tfrac{1}{it}$