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 :)

Solution 1:

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

Solution 2:

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 ...

Solution 3:

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 $g^{[S : S \cap H]} \in H$.