Let G be a group of order 25. Prove that G is cyclic or g^5=e for all g in G.
The group is not necessarity cyclic, how about $\displaystyle \mathbb{Z}_5\times \mathbb{Z}_5$.
If $\displaystyle G$ is cyclic then it is clearly true because $\displaystyle G\simeq \mathbb{Z}_{25}$ and $\displaystyle \left< 5 \right> = \{ 0,5,10,15,20\}$.
If $\displaystyle G$ is not cyclic then choose $\displaystyle x\in G$ so that $\displaystyle x\not = e$. Then $\displaystyle \left< x\right>$ is a subgroup of $\displaystyle G$. It cannot (subgroup) have order 1 because it is not identity and it cannot be 25 (the full group) because we assumed it is not cyclic. Thus it must be 5 by Lagrange's theorem.