If $G$ is finite, and $H$ a proper subgroup (which existence is often easy to show by using Sylow Theory), it would be sufficient to prove $core_G(H)=\bigcap _{g\in G}gHg^{-1}\neq \{ 1\}$ to conclude that $G$ is not simple since we know $core_G(H)\lhd G$.
Also, some proofs regarding groups of particular order (say, groups of order $p^nq$ for primes $p$ & $q$) not being simple are somewhat long and I'm just in hopes of not having to learn them for my exam.
In the infinite case there is some merit to $core_G(H)$ in this regard, since by Poincare's Theorem (namely, if $G$ has a subgroup $H$ with finite index, then $core_G(H)$ also has a finite index in $G$) if an infinite group has a subgroup of finite index, then it is not simple.
For a finite group $G$ and $H< G$, I've tried without success to find useful conditions for which $core_G(H)\neq 1$, so I'm wondering if there are non-trivial sufficient conditions for which $$core_G(H)\neq 1$$
For example, let $G$ be a group and $H$ a subgroup of $G$. Let us say that $G$ is finite, and $[G:H]=n$. The action of $G$ on $G/H$ induces a group homomorphism from $G$ to $S_{n}$ whose kernel is exactly $\cap_{g \in G}gHg^{-1}$. An example where this is used : Let G be a finite group , and $H$ a subgroup of index $p$, where $p$ is the smallest prime dividing $|G|$. Show that $H$ is normal in $G$. You will need to use basic group theory and basic number theory!. (In fact, if $H$ is normal in $G$, $\cap_{g \in G}gHg^{-1}=H$, so one obvious condition is that $H$ is normal in $G$).
Edited: If I am not mistaken, one can use this theorem, along with the Sylow theorem to prove that a group of order $105$ is not simple.