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

Question

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.

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
If normality is not assumed, see https://math.stackexchange.com/questions/108952/ — Watson, Nov 26 '18 at 21:05

score 8 · Answer 1 · answered Sep 30 '13 at 14:30

8

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

answered Sep 30 '13 at 14:30

Mikasa

67,374

Nice suggestion! +1 – amWhy Sep 30 '13 at 14:52
Hello, Babak! I've missed you! ;-) – amWhy Oct 02 '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 03 '13 at 23:52
I was just worried, since you hadn't posted. I hope your trip went well! Welcome HOME! – amWhy Oct 04 '13 at 12:17
@amWhy: Thanks Amy for your charming words. :-) – Mikasa Oct 04 '13 at 13:09
"Meeting" you online each day warms my heart. You're always welcome, dear Babak! – amWhy Oct 04 '13 at 13:12
welcome back dear friend! – Adi Dani Oct 06 '13 at 19:56
@B.S. Nice hint! – Jan 14 '14 at 13:15

score 8 · Accepted Answer · answered Sep 30 '13 at 14:30

8

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.

answered Sep 30 '13 at 14:30

score 2 · Answer 3 · answered Sep 30 '13 at 14:33

2

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

answered Sep 30 '13 at 14:33

tattwamasi amrutam

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

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