Let |G| = 100, $H\leq G$ with |H| = 25. If $g\in G$ with $|g| = 5^{k}$ for some non-negative integer k, prove that $g \in H$ I'm not exactly too sure how to prove this. I know by Lagrange's theorem that G can have subgroups with orders, 1, 2, 5, 10, 20, 25, 50, 100. We're to assume that the group operation is multiplication.
Any help is appreciated
 A: Here is an elementary approach. Suppose $g\in G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$. 
In this case, $g=e$ (the identity element), and so $g\in H$.
Case 2. $k=1$. Let $K=\langle g \rangle = \{g, g^2, g^3, g^4, e\}$ be the subgroup generated by $g$. Consider the subset $H\cdot K = \{h \cdot k: h\in H \text{ and } k\in K\}$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|H\cdot K| = \frac{|H|\cdot |K|}{|H\cap K|}$. If $g\notin H$, then $H\cap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|H\cap K|=1$. But then 
$|H\cdot K| = \frac{25\cdot 5}{1} = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=\langle g \rangle$ which has size $25$. If $g\notin H$, then $H\cap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $H\cap K\subset K$), we get $|H\cap K|\leq 5$. Again, we get $|H\cdot K|=  \frac{|H|\cdot |K|}{|H\cap K|} \geq \frac{25\cdot 25}{5} = 125 > 100 = |G|$, a contradiction. 
A: Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and 
$$n_5 \equiv 1 \pmod{5} \\
n_5|4$$
