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

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

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 ...

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