I came across this question and I can't figure it out.
Full problem is:
Suppose the order of each non-trivial member of non-trivial group G is infinite. Prove that if H is a subgroup of G with finite index then intersection of each non-trivial subgroup of G like K with H is non-trivial.
All I could figure it out by myself is that in a group the intersection of an infinite subgroup with a finite index subgroup is infinite. but I don't know what to look for in next steps.
Welcome to MSE!
Here's a hint: Since $H$ is finite index in $G$, there are only finitely many cosets $H, g_1 H, g_2 H, \ldots, g_n H$, and each $g \in G$ is a member of (exactly one) coset.
To say that $K \cap H$ is nontrivial, we want to find some $k \neq e \in K$ so that $k \in H$.
So fix some $k \neq e \in K$. We know $k, k^2, k^3, k^4, \ldots$ are all distinct elements (since $k$ has infinite order). Can you use the pigeonhole principle (where the "holes" are the cosets of $H$) to get what you want?
I hope this helps ^_^