“…Then, by Propositions 4.1-4.3, the twisted Alexander ideals of G i,j are I 1 i,j = I 2 i,j = L and I 3 i,j = (1 + y d + y 2d ). As 3|(i + j ), 3 | j and l is odd, by Proposition 4.3 we have I 3 i ,j = ((1 + y d + y 2d )/(1 + y + y 2 )). Note that as d > 1, I 3 i ,j = L. Since G i,j ∼ = G i ,j , two of the twisted Alexander ideals of G i ,j must coincide with L. For this to happen, we must have I 1 i ,j = I 2 i ,j = L. Since the automorphism ϕ must map a non-trivial ideal to a Since the automorphism ϕ must map a non-trivial ideal to a non-trivial one, we get ϕ(I 3 i,j ) = I 3 i ,j .…”