Are all finite simple groups essentially permutation groups? Pardon what may be my stupidity, but I genuinely need some direct clarification on whether or not this is true. I know that by the classification of the finite simple groups that every finite simple group is either cyclic, alternating, a type of lie group, or maybe one of the exceptions called sporadics (side note, is this what they called the "ENORMOUS theorem"?)
I think cyclic and alternating groups are permutation groups. I believe lie groups to be the group formed by symmetries on particular figures (which sounds very permutation-y to my simple brain). I have no idea what sporadic groups are.
That being said, simple groups have a very general definition that don't seem to automatically imply it to be equivalent to a permutation group. I believe this definition to be a nontrivial group with only {e} and the whole set as its normal subgroups. Perhaps when you have any particular finite simple groups it is isomorphic to some group of permutations on some set?
Edit:
I forgot about Cayley's thoerem for a second as I was just focusing on finite simple groups in particular. So am I to assume that cyclic, alternating, lie groups, and sporadic groups are all actually permutation groups?
As an additional question, if they all are, why identify finite simple groups as permutation groups in particular. Is there sometime special about identifying groups with their respective permutation group?
 A: Cayley's theorem says that every group is a permutation group, under the correct definition of "permutation group": for any cardinality $\kappa$ we can define $S_\kappa$ as being the group of permutations of a set of cardinality $\kappa$, where "permutation" is simply defined to be a bijective map. Since left-multiplication by a group element $g \in G$ is automatically bijective, being invertible by the map that left-multiplies by $g^{-1}$, this gives us a map $\rho: G \to S_G$, where $S_G$ again is the group of permutations on $G$, considered as a set. To reiterate, we have $g \mapsto \rho_g$ where $\rho_g(h) = gh$. This embeds $G$ as a subgroup of a group of permutations.
A: To address your additional question, perhaps it worths noting that if $G$ is a finite simple group of order $n$ and $H$ is a proper, nontrivial subgroup of $G$, then $G$ is isomorphic to a subgroup of $S_m$, where $m=[G:H]<n$, namely there's a "sharper" embedding of $G$ into a symmetric group than the one ensured by Cayley's theorem. In fact, the action of $G$ by left multiplication on the left quotient set $G/H$ has trivial kernel.
A: Every group $G$ is a group consisting in permutations, in the sense that, by Cayley's theorem, it can be imbedded in a symmetric group, the group of symmetries of $G$.
In case $G$ is finite, we have that $G$ is isomorphic to a subgroup of the symmetric group $S_n$, where $n=|G|$.
In the infinite case, we still have that $G\hookrightarrow\rm{Sym}G$, the group of symmetries of $G$.
Because of this anything that is true for subgroups of symmetric groups is true for all groups.

What I know is still fairly limited, even though I appear to be headed for gold badges in group theory and abstract algebra, eventually.
The classification of finite simple groups was very difficult and time consuming, and is considered one of the great intellectual achievements.
Furthermore, since every finite group has a composition series, it can be realized as a series of extensions of simple groups.  Thus a solution to the so-called "extension problem", would lead to a classification of finite groups.
