Let $H$ be a normal subgroup of index $n$ in a group $G$. Show that for all $g \in G, g^n \in H$ I am having a lot of trouble understanding the solution to this problem. 
$(gH)^n = g^nH \implies g^n \in H$
Why does $H^n$ just turn into the identity? I am very confused, any help is appreciated. 
 A: The index of $H$ in $G$ is the order of the quotient group $G/H$. Since $[G:H] = |G/H| = n$, by Lagrange, every element of $G/H$ has order dividing $n$. Hence, every coset $gH \in G/H$ has order dividing $n$, i.e. $(gH)^{n} = g^{n}H = eH =H$, the identity coset. Thus, $g^{n} \in H$, since if $aH=bH$, then $a^{-1}b \in H$ (here $a=e, b=g^{n}$). 
A: The multiplication in the quotient group is defined this way: $$(aH)(bH) = (ab)H$$
no need to multiply $H$ by it self.
A: I'm not sure what you mean by the identity, but if you're asking why $H^n=H$, then the reason is because $H^n = \{ h_1\cdot\ldots\cdot h_n \,|\, h_i\in H \, \forall i\} $, and since $H$ is a subgroup, we get that $e\in H$ ( this gives us $H \subset H^n$ ) and that $H$ is closed under the group operation ( this gives us $H^n \subset H$ ).   
A: A basic result on finite groups is that if $|G|=n$, then $g^n=1_G$ (the identity element in $G$), for all $g\in G$. Consider the projection homomorphism
$$
\pi\colon G\to G/H.
$$
Since $|G/H|=n$ by assumption, we have, for all $g\in G$,
$$
\pi(g^n)=(\pi(g))^n=1_{G/H}
$$
which means that $g^n\in\ker\pi=H$.
A: We have $gH = \{ gh \mid h \in H\}$, so in particular, if $gH = H$, then $e = gh$ for some $h \in H$ - i.e. $g^{-1}\in H$ and hence $g \in H$.
The converse is also true, so $$gH = H \iff g \in H$$
The proof is saying that since $H$ has index $n$, if $gH \in G/H$, then by Lagrange, $$H = (gH)^n = g^nH$$ so by the above, $g^n \in H$
