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Let $(G,\circ)$ be a group and $N\subseteq G$ a normal subgroup of order $n<\infty$ and let $g\in G$. Is the element $g^n$ in $N$?

Given a subgroup $H\subseteq G$ of order $n$, is element $g^n$ in $H$ for all $g\in G$?

Edit: $g^n=\underbrace{g\circ g\circ\ldots\circ g}_{n\text{ times}}$

  • 3
    Your notation is a bit inconsistent. You denote the group operation as "+" but then use power notation $g^n$. Usually, if the group operation is written "+" then we use "multiple" notation $ng$ instead. When the group operation is written as multiplication, then we use power notation $g^n$.2012-05-26
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    if you mean $ng$ for $g^n$ then it is in $G$ since it is closed under its own operation.2012-05-26
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    Also: (1) did you mean to ask $g^n \in N$? (2) You already stated that $N$ is normal, so what do you mean by "is it necessary for $N$ to be normal?"2012-05-26
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    I've improved the formulation2012-05-26

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