show that there exist an element $g$ of a group $G$ such that $g^q$ is in $H$

Question

$H$ is normal in $G$ and $q$ is prime with $q\mid[G:H]$. Show that there exist $g$ in $G$ such that $g^{q}$ is in $H$.

I am not sure how to use $q\mid[G:H]$. Should I try to show that $f:G→H$, with $f(g)=g^q$ or there is better way to do it?

g can't be the identity

Thanks in advance.

Answer 1

As suggested in the comments, if we allow $g$ to be in $H$ the problem is trivial. Here is how to find an element $g\notin H$ such that $g^q\in H$:

Look at the quotient group $G/H$. This group has order divisible by $q$. By Cauchy's theorem there is an element of $G/H$ of order $q$. Hence it's representative $g$ satisfies $g^q\in H$, while $g\notin H$.

Also notice that the map you suggested defining in your question, $f(g)=g^q$, has image lying in $H$ iff every element $g\in G$ satisfies $g^q\in H$. This will of course not be true in general since the quotient may have elements of order not dividing $q$. For example, if some other prime $p$ divides $[G:H]$, the quotient will have an element of order $p$, again by Cauchy's theorem.

Answer 2

Your question is incomplete, if we take the identy of $G$, it is obvius that for any $n$ $1^n=1$ is in $G$, but since $H$ is subgroup of $G$ then $1\in H$ so for any $n$ in particular for $q$ we have $1\in G$ and $1^q\in H$

Answer 3

With the hyphotesis, I will show that exist an element $g \in G$, and $g\not \in H$ such that $g^q\in H$. So We can take the group $G/H$ and since $q|[G:H]$ then $q$ divide the order of the group $G/H$, so by Couchy exist and elemente $g+H\in G/H$ with order $q$(since q is prime) therefore $g\not\in H$ because if that is true we will have that the order of $g+H$ will be 1, and it is $q$. So now we will se that $g^q\in H$, remenber $(g+H)^q=H$(H is the identy of $G/H)$, then $(g+H)^q=g^q+H=H$ in other words $g^q\in H$. So we conclude.