I know that in the semidirect product of $A$ and $B$, the homomorphism $\phi:A\rightarrow Aut(B)$ should be $\phi_y(x) = yxy^{-1}$ but have no idea how to construct one for $\phi:Z_3\rightarrow Aut(Z_{13})$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.
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Arturo Magidin
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2“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism. – Arturo Magidin Dec 03 '18 at 21:57
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See this question. – Dietrich Burde Dec 03 '18 at 22:01
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@ArturoMagidin Yes, corrected that. – Dec 03 '18 at 22:02
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... not everywhere... but now it’s fixed. – Arturo Magidin Dec 03 '18 at 22:12
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@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms. – Dec 03 '18 at 22:20
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@ArturoMagidin Thanks. – Dec 03 '18 at 22:22
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It's $y \mapsto (x\mapsto x^3)$. – jgon Dec 04 '18 at 01:24