Let $H$ be a normal subgroup of index $n$ in a group $G$. Show that for all $g \in G, g^n \in H$

Question

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.

Answer 1

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 $a=e, b=g^{n}$).

Answer 2

The multiplication in the quotient group is defined this way: $$(aH)(bH) = (ab)H$$ no need to multiply $H$ by it self.

Answer 3

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 $H^n \subset H$ ).

Answer 4

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 $g^n\in\ker\pi=H$.

Answer 5

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, $g^n \in H$