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As for $\varphi : G \to H$ the order of $\varphi(g)$ divides the order of $g$, we have for $N \unlhd G$ that if $g \notin N$ and $g^p \in N$ for some prime $p$, then $p$ divides the order of $g$. This result I know.

Does a similar result hold for non-normal $U \le G$, i.e. if $g^p \in U$ for $g \notin U$, then $p$ divides the order of $g$? Specifically I ask what happens if we drop the assumption of normality, is the result still valid?

StefanH
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2 Answers2

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Assume that $g$ has some finite order $n$ and $p\not\mid n$. Then $p$ has an inverse mod $n$, so there is some $r\in\mathbb{Z}$ such that $(g^p)^r=g$. So if $g^p$ is in some subgroup $U$, then so is $g$.

Another way to think about this is that in your question, you might as well restrict everything to the subgroup $H\subseteq G$ generated by $g$, replacing $U$ by $U\cap H$. But $H$ is cyclic, so every subgroup of $H$ is normal in $H$, so the argument from the normal case applies.

Eric Wofsey
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Let $g^p=h\in U$. Consider the subgroup $\langle h\rangle$. Then $g^p\in \langle h\rangle$, and $g$ can not be in $\langle h\rangle$.

Thus, simply consider subgroup $\langle g\rangle$, which contains subgroup $\langle h\rangle$ and has index $p$.

Thus, $p$ must divide order of $\langle g\rangle$, which is equal to order of $g$.

p Groups
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    Other way: by contradiction; suppose $p$ does not divide $o(g)$. Then $o(g)=o(g^p)$, hence the map $x\mapsto x^p$ from $\langle g\rangle\rightarrow \langle g\rangle$ is isomorphism. Thus, $\langle g^p\rangle \subseteq U$ implies $\langle g\rangle\subseteq U$, contradiction. (This is exactly what Eric said in first two lines of his answer) – p Groups Jan 11 '16 at 03:19
  • Thank you. Just a minor addition. By looking at your comment I noticed that if $g = x^p$, then either $|x| = |g|$ with to $p$ coprime order, or $|x| = p|g|$. This could be seen by considering the map $x \mapsto x^p$, if $p$ is coprime to $|x|$ this is an automorphism, hence $|x| = |x^p|$, otherwise the cyclic subgroup $\langle x^p \rangle$ has order $|x|/p$ in $\langle x \rangle$. – StefanH Jan 11 '16 at 12:02