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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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    $\begingroup$ “Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism. $\endgroup$ – Arturo Magidin Dec 3 '18 at 21:57
  • $\begingroup$ See this question. $\endgroup$ – Dietrich Burde Dec 3 '18 at 22:01
  • $\begingroup$ @ArturoMagidin Yes, corrected that. $\endgroup$ – manifolded Dec 3 '18 at 22:02
  • $\begingroup$ ... not everywhere... but now it’s fixed. $\endgroup$ – Arturo Magidin Dec 3 '18 at 22:12
  • $\begingroup$ @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. $\endgroup$ – manifolded Dec 3 '18 at 22:20

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