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Let $G$ be a group of order $2014$. Let $\theta\in G$ such that $|\theta|=19$ and $\alpha\in G$ such that $|\alpha|=2$. Show that $\alpha\theta\alpha=\theta^{\pm1}$. Since the order of $\alpha$ is 2 I know that $\alpha=\alpha^{-1}$ so $\alpha\theta\alpha=\alpha^{-1}\theta\alpha$ so the element is conjugate to $\theta$ and thas must be $\theta^x$ for some positive integer x such that $0<x<19$ (I already showed that the group generated by $\theta$ is a unique sylow-p group in $G$). I would like a hint on how to continue from here

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Hint: What is the group of automorphims of $\mathbb Z/19\mathbb Z$? (Or at least: What is the order of it?)

There are $18$ automorphisms of $\mathbb Z/19\mathbb Z$: for $1\le k\le 18$ we obtain one from $1\mapsto k$. Among these is $x\mapsto x^{-1}$, which is of order $2$ and (because $18=2\cdot 9$) is the only automorphism of order $2$. Since conjugation with $\alpha$ is an automorphism of order at most $2$, the claim follows.

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  • $\begingroup$ Can you explain why the order of this automorphism is exactly 2? $\endgroup$ – Gyt Feb 14 '14 at 16:12
  • $\begingroup$ @Gyt because the inverse of the inverse is the element itself (and because some elements are note their own inverses) $\endgroup$ – Hagen von Eitzen Feb 14 '14 at 16:25
  • $\begingroup$ von Etzen And why is it the only automorphism? $\endgroup$ – Gyt Feb 14 '14 at 17:12

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