# If the index $n$ of a normal subgroup $K$ is finite, then $g^n\in K$ for each $g$ in the group.

Let $K \unlhd G$ be a normal subgroup of some group $G$ and let $|G/K|=n<\infty$. I want to show that $g^n\in K$ for all $g\in G$.

Let $g\in G$, if $g\in K$, then $g^n\in K$ and we are done. If $g^n\notin K$ then consider the set of left cosets $$C=\{K, gK,g^2K,...,g^{n-1}K\}$$ I want to show that these cosets are all disjoint and hence $C=K$, then I want to show that $g^nK=K$, so $g^n\in K$. Suppose $$g^lK=g^mK$$ for some $m,l<n$, then $g^{m-l}\in K$. I am not sure how to proceed from there.

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Note the same questions, but it is answered there: Powers of elements and subgroups –  Martin Sleziak Sep 30 '13 at 15:25
This seems to be basically the same question: Prove that if a normal subgroup $H$ of $G$ has index $n$, then $g^n \in H$ for all $g \in G$ –  Martin Sleziak Sep 30 '13 at 15:27

Consider the canonical projection $\pi : G \to G/K$. We know $G/K$ is a finite group of order $n$ by assumption. Thus $\pi(g)^n = \pi(g^n) = 1$ in $G/K$, i.e. back in $G$ we have $g^n \in K$. Done.

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Although, I think it is duplicate, you can use this fact that: $$[G:K]=n\longrightarrow \forall g\in G, (gK)^n=K\iff g^nK=K\iff g^n\in K$$

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Nice suggestion! +1 –  amWhy Sep 30 '13 at 14:52
Hello, Babak! I've missed you! ;-) –  amWhy Oct 2 '13 at 11:41
Babak, are you okay? I've been worried about you...is it your internet again that's getting in your way? –  amWhy Oct 3 '13 at 23:52
I was just worried, since you hadn't posted. I hope your trip went well! Welcome HOME! –  amWhy Oct 4 '13 at 12:17
@amWhy: Thanks Amy for your charming words. :-) –  B. S. Oct 4 '13 at 13:09

You dont need to go that far. Since you know that $|G/K|=n<\infty$ and $G/K =\{K, gK,g^2K,...,g^{n-1}K\}$. Then for any $gK$ in $G/K$, $(gK)^n= g^nK=K$ which answers your question

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