Element in subgroup and closure I wasn't certain how I should have worded the title. Anyway, I have this problem:
Let $H$ be a subgroup of a finite group $G$. Suppose that $g$ belongs to $G$ and $n$ is the smallest positive integer such that $g^n$ is in $H$. Prove that $n$ divides $|g|$.
The solution I have ( http://i.imgur.com/o7A0Ezk.png ) says that since $g^n$ is in H, then by closure $(g^n)^2$, $(g^n)^3$, ... are also in H, and those are the only elements in H.
Is this true? Why are they the only elements in H? Can't there be some element $a$, and then $a^2$, $a^3$, ... and $ag^n$, ... and so on are also in H?
 A: Your objection is entirely valid.
The given solution has (1) misworded: $g, g^2, \dots, g^m$ are not the elements of $G$, but only are some elements of $G$. Likewise, $g^n, g^{2n}, \dots$ are not the elements of $H$, but are only some elements of $H$, as you say. For a concrete example, consider $G = (\mathbb{Z}_{12}, +) \times \{ 0, 1 \}$, and $H = \{ (0, 0), (6, 1) \}$, $g = (2, 0)$. Then $\{(2, 0), (4, 0), (6, 0), (8, 0), (10, 0), (0, 0) \}$ are the $g^i$, so the least $g^n \in H$ is $(0, 0)$. This clearly doesn't generate $H$.

However, the proof is nearly correct.
Let $g \in G$ where $|G| = k$, $H \leq G$, and $o(g) = m$. Then $\langle g \rangle = \{ g^1, g^2, \dots, g^m = e \}$ is a subgroup of $G$; we are given that $g^n \in H$ for $n$ least positive.
By the closure property of $H$, have that $\{ g^n, g^{2n}, \dots \} \subseteq H$, and $e = g^0 = g^m \in H$ also.
Suppose $n$ does not divide $m$, so $m = x n + y$ with $1 \leq y < n$. Then $g^m = g^{x n + y} = (g^n)^x g^y$; but $g^n \in H$ so it has an inverse in $H$. Therefore $((g^n)^{-1})^x (g^n)^x g^y \in H$, so $g^y \in H$.
This contradicts the definition of $n$ as minimal, since $y < n$.
Therefore, $n$ must divide $m$ after all.
