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

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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 $g^n$. –  Ted May 26 '12 at 20:43
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if you mean $ng$ for $g^n$ then it is in $G$ since it is closed under its own operation. –  sxd May 26 '12 at 20:45
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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?" –  Ted May 26 '12 at 20:45
    
I've improved the formulation –  klm May 26 '12 at 21:15

2 Answers 2

Take $G$ any nontrivial group, $n=1$ and therefore $N$ the trivial subgroup.

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Certainly not necessarily in $N$ (of course it has to be in $G$: $G$ is closed under all finite products, and in particular all finite powers). For a counterexample, take $G$ to be a product of a cyclic group of order $n$ and one of order $m$ with $\gcd(n,m)=1$, $C_n\times C_m$, and let $N=C_n\times\{1\}$. Any element $g$ with nontrivial second coordinate will not satisfy the condition that its $n$th power lies in $N$. Likewise the result does not hold for a not-necessarily-normal subgroup of order $n$.

What is true is that if the index of $N$ is $n$, then $g^n\in N$ for every $g\in G$. This is a consequence of Lagrange's Theorem applied to $G/N$.

If $H$ is a not-necessarily-normal subgroup of index $n$, it is not the case that $g^n$ must lie in $H$ for all $g$. Take $G=S_3$, $H=\langle (12)\rangle$, and $g=(13)$. The index of $H$ is $3$, but $g^3\notin H$.

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Thank you for your help :D –  klm May 26 '12 at 21:20

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