Given a cyclic group $G$, how do I find the index of subgroup $H = \{x^n : \forall x \in G\}$ for some fixed $n$?
I kind of figured from answers to similar questions that it should be the number of solutions to $x^n = 1$, but I can't quite connect the dots.
Edit: I want to remark I'm assuming finite groups (due to the comments).
Let's first check $H = \{x^n : x \in G\}$ is a subgroup for fixed $n \geq 1$. Notice that $e^n = e$, so $e \in H$. Next, let $a \in H$. Then $a = x^n$ for some $x \in G$, and $a^{-1} = x^{-n} = (x^{-1})^n \in H$. Hence it is closed under inverses. Finally if $a, b \in H$, then $a = x^n$, $b = y^n$ for $x,y \in G$, so $ab = x^n y^n$. Now here is where we need that $G$ is cyclic (so abelian). Recall that for an abelian group we have $(ab)^2 = a^2 b^2$, since $(ab)^2 = abab = aabb = a^2b^2$. By induction, we get this holds for all $n$. Notice that we then have $x^ny^n = (xy)^n \in H$, so it is closed under multiplication. This tells us that it is a subgroup.
Now the index of a subgroup is $[G:H] = |G|/|H|$. Consider the map $f : G \rightarrow H$ defined by $f(x) = x^n$. This is clearly surjective. Let's check it is a homomorphism. We have $$ f(xy) = (xy)^n = x^n y^n = f(x) f(y)$$ by our above remark. Finally we observe $$\ker(f) = \{x \in G : f(x) = x^n = e\}.$$ By the first isomorphism theorem, $$ G/\ker(f) \cong H,$$ so $$ |G/\ker(f)| = |G|/|\ker(f)| = |H|,$$ and $$ [G:H] = |G|/|H| = |G|/(|G|/|\ker(f)|) = |\ker(f)|.$$ Thus the index is the number of solutions to the equation $x^n = e$.