The links in the comments already give very good answers, but here is something similar:
There are very few coincidences amongst the orders of the various finite simple groups in the various families. Many of these coincidences are explained by exceptional isomorphisms for groups that have more than one characteristic:$
\newcommand{\PSL}{\operatorname{PSL}}
\newcommand{\PSU}{\operatorname{PSU}}
\newcommand{\PSp}{\operatorname{PSp}}
\newcommand{\PSO}{\operatorname{P\Omega}}
$
- Order 60, $A_5 \cong \PSL(2,4) \cong \PSL(2,5)$
- Order 168, $\PSL(2,7) \cong \PSL(3,2)$
- Order 360, $A_6 \cong \PSL(2,9)$
- Order 20160, $A_8 \cong \PSL(4,2)$
- Order 25920, $\PSO(5,3) \cong \PSU(4,2)$
Also one has various low dimension isomorphisms in general:
- $\PSU(2,q) \cong \PSL(2,q)$
- $\PSO(5,q) \cong \PSp(4,q)$
But also a slightly weird one that is only an isomorphism in characteristic 2:
- $\PSO(2n+1,2^f) \cong \PSp(2n,2^f)$
This slightly weird one forces the orders of $\PSO(2n+1,q)$ and $\PSp(2n,q)$ to be the same for all $q$, however. So we get the order coincidence:
- $|\PSO(2n+1,q)| = |\PSp(2n,q)|$, but $\PSO(2n+1,q) \not\cong\PSp(2n,q)$ for odd $q$
The only other order coincidence is the very special $A_8 \cong \PSL(4,2)$ versus $\PSL(3,4)$ of order 20160.
These results are proved for the classical and exceptional groups in Artin (1955). The original observation of the two distinct simple groups of order 20160 is from Schottenfels (1899). While Artin's technique work for all groups of Lie type (the twisted types are no hard), it might be nice to see a post CFSG version in Garge (2005), which also handles direct products of simple groups and knows the orders of all sporadic simple groups.
Schottenfels, Ida May.
“Two non-isomorphic simple groups of the same order 20,160.”
Ann. of Math. (2) 1 (1899/00), no. 1-4, 147–152.
MR1502265
DOI:10.2307/1967281
Artin, Emil.
“The orders of the classical simple groups.”
Comm. Pure Appl. Math. 8 (1955), 455–472.
MR73601
DOI:10.1002/cpa.3160080403
Garge, Shripad M.
“On the orders of finite semisimple groups.”
Proc. Indian Acad. Sci. Math. Sci. 115 (2005), no. 4, 411–427.
MR2184201
DOI:10.1007/BF02829803
In fact, every solvable group can be obtained by a sequence of semidirect product and subgroup operators applied by starting with abelian groups. The trick is to start at top, rather than at the bottom, to avoid the issue I mentioned in the comments.
This relies on the theorem of Kaloujnine and Krasner that every extension of $G$ by $K$ can be embedded into the regular wreath product $G\wr K$. (Note: in my nomenclature, "extension of $G$ by $K$" means a group with a normal subgroup isomorphic to $G$ and corresponding quotient isomorphic to $K$). The regular wreath product is constructed by taking a direct product of $|K|$ copies of $G$, and then letting $K$ act on this product by acting on the coordinates via the regular action. Then you take the semidirect product $G^{|K|}\rtimes K$.
So proceed by induction on the solvability length. A solvable group of length $1$ is abelian. And a solvable group of solvability length $n+1$ can be realized as an extension of an abelian group $A$ by a group of solvability length $n$ $K$ (e.g., take the $n$th derivated group as the normal subgroup). This extension can be embedded into the wreath product $A\wr K$. Now, this is a semidirect product of an abelian group, $A^{|K|}$, by $K$; and $K$ can be obtained, inductively, as a sequence of subgroup and semidirect product operators with abelian groups. This proves that so can $G$.
Explicitly, if $G$ is solvable, let $n$ be the largest integer such that the $n$th derived subgroup $G^{(n)}$ is nontrivial. Start with $G^{\rm ab}=G/G'$; then construct the extension of $G'/G''$ by $G/G'$ by taking the corresponding subgroup of the wreath product $(G'/G'')\wr (G/G')$. This can be done under your restrictions because the base, $(G'/G'')^{|G/G'|}$, is actually abelian. Continue this way until you get to the extension of $G^{(n-1)}/G^{(n)}$ by $G/G^{(n-1)}$, where again the base is abelian.
You can certainly do this to obtain all finite groups if you allow simple groups, since a finite direct product that occurs in a base can be realized as a finite sequence of (trivial) semidirect products. But the trick of just looking at the base as an abelian group won't work if we are considering simple groups when the extension has an infinite quotient.
Best Answer
The easiest examples of non-simple groups, that are not a semidirect product of two non-trivial subgroups, are cyclic groups of order $p^n$, where $p$ is prime and $n \geq 2$, and generalized quaternion groups