# Prove that if a normal subgroup $H$ of $G$ has index $n$, then $g^n \in H$ for all $g \in G$

I need some help in relation to this exercise

"Prove that if a normal subgroup $H$ of $G$ has index $n$, then $g^n \in H$ for all $g \in G$."

I'm not allowed to use quotient groups in the proof, because the exercise is in the chapter before.

I tried by induction on $n$. The case $n=1,n=2$ are obvious, but even the case $n=3$ is giving me trouble so I give up studying the general case of the inductive step.

My other approach was studying left or right coset of $G$. But I only proved that if $g \in aH$ then $g^2 \notin aH$ if $a \notin H$, and I can't find a way to demonstrate that $g^n \in H$. (my starting idea was to prove that every power of $g$ is in a different coset but then I realize that in this way I don't handle several case, for example $g$ has period strictly lesser than $n$ and in conclusion it doesn't prove the exercise) Maybe I'm missing something about indexes, and this is why I asked here for some help,

(I can't use quotient groups because they are introduced later than this exercise, forgot to add this info in the beginning) Thanks in advance :)

-
Which book are you studying where this exercise appears before quotient groups? – lhf Aug 21 '13 at 10:33
Rotman "An introduction to the theory of groups" – Riccardo Aug 21 '13 at 11:20
Ok, it's exercise 2.39 on page 32 of the 4th edition. – lhf Aug 21 '13 at 11:34
Precisely. So I don't know how to prove the result without the quotient group :( – Riccardo Aug 21 '13 at 11:39
@RicPed, you should write clearly and big in your question that you're asking for a solution without the aid of quotient groups ! – DonAntonio Aug 21 '13 at 12:19

Here is a solution which works in the case where $G$ is finite. (Of course, this assumption is not needed for the theorem to hold)

It was mentioned in the comments that the problem in question is exercise 2.39 from An Introduction to the Theory of Groups by J. Rotman. I am using the fourth edition, so you might have different numbers for exercises and lemmas.

Earlier in exercise 2.28, Rotman asks you to prove the following fact about double cosets:

Let $S, H \leq G$, where $G$ is a finite group, and suppose $G$ is the disjoint union $$G = \bigcup_{i=1}^n S g_i H$$ Prove that $[G : H] = \sum_{i = 1}^n [S : S \cap g_i H g_i^{-1}]$.

To prove this, apply theorem 2.20 to $|Sg_iH| = |Sg_i H g_i^{-1}|$. As an immediate corollary, we get

Let $S, H \leq G$ and suppose that $H$ is a normal subgroup. Then $[S : S \cap H]$ divides $[G : H]$.

To prove exercise 2.39, consider the corollary with $S = \langle g \rangle$. By the corollary, it suffices to prove that $g^{[S : S \cap H]} \in H$. By exercise 2.11, $g^{[S : S \cap H]}$ has order $|S \cap H|$. Since $S$ contains exactly one subgroup of order $|S \cap H|$ (this is lemma 2.15), it follows that $g^{[S : S \cap H]}$ generates $S \cap H$, and in particular $g^{[S : S \cap H]} \in H$.

-
Wow, need some,time to,understand fully,but thanks this seems to be the answer I was looking for. Only one precisation, the corollary you mention is based on a result about finite group let's say. Is it still valid for this exercise where G need not to be finite? – Riccardo Aug 21 '13 at 15:42
It is always true, since when $H$ is normal, $S / S \cap H \cong SH/H$ so $S / S \cap H$ is isomorphic to a subgroup of $G/H$. But you're right, 2.39 is not assuming that $G$ should be finite. I'm not sure right now if there's a way to make this answer work for infinite groups without using quotients. – Mikko Korhonen Aug 21 '13 at 15:51
So,if I understand correctly, the sentence is correct but we have to use the third theorem of isomorphism right? – Riccardo Aug 21 '13 at 16:00
Yes, the corollary is true for infinite groups too (but $[G:H]$ needs to be finite for it to make sense). I'll think about this, but right now the only way I can see to prove the corollary in general (not just for finite groups) is with $SH/H \cong S / S \cap H$. – Mikko Korhonen Aug 21 '13 at 16:03
Agh, and later on we need the fact that $g$ has finite order, so this solution probably can't be fixed for the general case. – Mikko Korhonen Aug 21 '13 at 16:22

Hint: If $H$ is a normal subgroup of index $n$, then $G/H$ is a group of order $n$.

-
Yeah and then it comes easy :) but it is possible to prove without using quotient groups? – Riccardo Aug 21 '13 at 9:48
Thanks for,the answer anyway (I forgot to specify that I need some kind of proof without using quotient groups, sorry ) anyway thanks again :) – Riccardo Aug 21 '13 at 9:49
Probably not, I think. If this theorem is true, then it is true in particular for $H = \{1\}$, which implies that $g^n = 1$ in a group of order $n$. I think any sensible way of proving that special case requires Lagrange's theorem. – spin Aug 21 '13 at 9:52
@spin, that is my pet question: Is Lagrange's theorem the most basic result in finite group theory? – lhf Aug 21 '13 at 10:45

Hint:

• $H$ is a normal subgroup of $G$, then $G/H$ is defined.

• $G/H$, as you pointed, is of order $n$, so $\forall gH\in G/H,~~ (gH)^n=H$.

• $(gH)^n=gHgHgH\cdots gH$ ($n-\text{copy}$)

So ...

-
It is necessary to introduce the notion of quotient group to prove this exercise? Because (I admit to have added later than your quickly asnwer) this is an exercise from the chapter before quotient groups. I apologize for not being precise from the beginning. Thanks for the answer anyway! – Riccardo Aug 21 '13 at 10:09
@RicPed: No. As I think, if we know just this point that for every $g\in G$ and positive integer $n$, $(gH)^n=H$ then we can have $g^n\in H$. But we need the normality :) – Babak S. Aug 21 '13 at 10:39
I don't get the point, are you assuming that $(gH)^n=H$? But this is the thesis not? – Riccardo Aug 21 '13 at 11:16
I think these are great hints! +1 – amWhy Aug 22 '13 at 12:31