I am trying to prove the following:
If G is a group of order 35 and operates nontrivially on a set of five elements, then there exists a normal subgroup of order 7.
I saw that it is possible to solve this problem using Sylow's theory, for example Group of order 35, and normal subgroup of order 7. But with the hypothesis of my problem is it necessary to use this theory? How would this be done?
group actions are simply homomorphisms into a symmetric group ( https://gowers.wordpress.com/2011/11/06/group-actions-i/ ). In the case of this problem, we have $\phi:G\to S_5$ where $|G|=35$. However, any homomorphic images of $G$ necessarily have orders 35, 5, 7, or 1 (by the first isomorphism theorem and logic on factor groups). If it isn't 1 (that would be the trivial action), and 35 and 7 are out of the question (as homomorphic images are subgroups, and $7\not\mid5!$ and $35\not\mid5!$ - lagrange's theorem), this means that $|\text{Im}(\phi)|=5$. Hence the kernel of this homomorphism has order $35/5=7$ and every kernel of a homomorphism is a normal subgroup.
Maybe this should give you an idea of where to start, assuming you are talking about group actions.