# 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.

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.

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

• Nice suggestion! +1 Sep 30, 2013 at 14:52
• Hello, Babak! I've missed you! ;-) Oct 2, 2013 at 11:41
• Babak, are you okay? I've been worried about you...is it your internet again that's getting in your way? Oct 3, 2013 at 23:52
• I was just worried, since you hadn't posted. I hope your trip went well! Welcome HOME! Oct 4, 2013 at 12:17
• @amWhy: Thanks Amy for your charming words. :-) Oct 4, 2013 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