If $|H|=4$ and $gH$ has order 3 in $G/H$. Show that $K=H\cup gH\cup g^2H$ forms a subgroup of $G$.

Question

Suppose that H is a normal subgroup of G. If $|H|=4$ and $gH$ has order 3 in $G/H$. Show that $K=H\cup gH\cup g^2H$ forms a subgroup of $G$.

We know $g^3 \in H$, and that $K=\{h_1,...,h_n,...,gh_1,...gh_n,...,g^2h_1,...,g^2h_n\}$. I am stuck here.
I don't know how to choose a general element in $K$ and then use the subgroup tests. I know $g^3 \in H$ will come in handy since $gh*g^2h'=g^3(hh') \in K$ , but how to choose a general element to account for elements that don't have a $g$?

Answer 1

I figured it out. Let $a=g^ih \in K$, for $0\leq i \leq2$