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 Mar 4 awarded Enlightened Mar 4 awarded Nice Answer Jan 29 awarded Yearling Jun 30 accepted How to compute the automorphism group of split metacyclic groups? Jun 25 comment How to compute the automorphism group of split metacyclic groups? It is a brilliant idea to apply the Frattini Argument to $Q$. But I am just not sure how you get $\text{C}_A(G)=1$. I think you are using the fact that $G$ is embedded in $A$, and so any element cannot centralize $G$ or it will induce a trivial automorphism. But saying $G$ as the inner automorphism group, then the elements in $\text{C}_A(G)$ is just the elements that commutes with the inner automorphisms. In this sense can we still say $\text{C}_A(G)=1$? I am not sure. Jun 23 comment How to compute the automorphism group of split metacyclic groups? @j.p.:$\text{Aut}(\mathbb Z_p)=\mathbb Z_{p-1}$. So the kernel shouldn't be trivial. However, if we can show the kernel is $\mathbb Z_p$ then we are done. Jun 23 comment How to compute the automorphism group of split metacyclic groups? @anon: $k$ can be any integer that divides $p-1$. But I believe it doesn't affect the result. Jun 23 asked How to compute the automorphism group of split metacyclic groups? Feb 4 revised New proof about normal matrix is diagonalizable. added 4 characters in body Feb 4 answered New proof about normal matrix is diagonalizable. Jan 29 awarded Yearling Sep 30 awarded Explainer Jul 2 awarded Curious May 24 revised Does every function with $f_x,f_y>0,f_{xx},f_{yy}<0$ with particular condition have to satisfy $f_{xy}/f_{xx} = -x/y$? deleted 3 characters in body May 24 comment Does every function with $f_x,f_y>0,f_{xx},f_{yy}<0$ with particular condition have to satisfy $f_{xy}/f_{xx} = -x/y$? When you say $\infty$, it is $+\infty$ or $-\infty$? May 21 answered Inverse of a sum of positive definite matrices May 20 answered Proving $A+2B+3C+4D < 2.5$ with given conditions May 20 comment If $f'(x) = 0$ for all $x \in \mathbb{Q}$, is $f$ constant? You may want to consider the fundamental theorem of Lebesgue integral calculus, which requires $f'(x)=0$ almost everywhere to get to your result. However, $\mathbb{Q}$ is far away from almost everywhere.. Feb 12 awarded Nice Answer Jan 29 awarded Yearling