Groups of order 12 without Sylow It is clear that Sylow theorems are an essential tool for the classification of finite groups.
I recently read an article by Marcel Wild, The Groups of Order Sixteen Made Easy, where he gives a complete classification of the groups of order $16$ that is based on
elementary facts, in particular, he does not use Sylow theorem.
Did anyone encounter a complete classification of the groups of order $12$ that does not use Sylow theorem?
What about order 24? (I'm less optimistic there, but who knows).
 A: Burnside's Theory of groups of finite order determines all distinct groups of orders $p^2q$ with $p$ and $q$ primes in Section 59; Sylow's Theorems do not appear until Chapter IX, section 120. However, he assumes that a group of order $p^2q$ is solvable without proof, and says "The truth of this statement, which is not difficult to verify directly, follows immediately from Sylow's theorem." So he 'farms out' a proof of solvability to Sylow's, though he asserts it can be done directly.
(My copy is the Dover edition from 1955, described as "an unabridged republication of the second edition, published in 1911".)
A: (I think Cauchy's theorem is proved before Sylow's theorem; I don't know exactly. I will use Cauchy's theorem.)
Let $G$ be a non-abelian group of order 12. By Cauchy's theorem, it has an element, hence subgroup, $H$, of order $3$.


*

*If $H$ is not normal in $G$, then as $[G\colon H]=4$, there is a homomorphism $\phi\colon G\rightarrow S_4$, with $ker(\phi)\subseteq H$ and so $ker(\phi)=\{1\}$ (since $H$ is not normal). $G$ is then isomorphic to a subgroup of $S_4$, it must be $A_4$ (it is only subgroup of order 12 in $S_4$, which can be proved without using Sylow theorem);



If $G$ has non-normal subgroup of order 3 then $G\cong A_4$.



*

*Let $H\triangleleft G$. Then it must be unique subgroup of order 3: if $K$ is another subgroup of order 3, then $HK\leq G$, $H\cap K=\{1\}$, $|HK|=9$, contradiction.


Now there is a homomorphism $\eta \colon G\rightarrow$ Aut($H $ ) $ \cong C_2$, $g\mapsto \{h\mapsto ghg^{-1} \}$. Then $ker(\eta)=C(H)$, the centralizer of $H$ in $G$.
It follows that $G$ has abelian (hence cyclic) subgroup of order 6:  (1) if $\eta(G) \cong C_2$ then $ker(\eta)$ will be a subgroup of order 6, and it should be abelian  since all its elements commute with $H$.  (2) if $\eta(G)=\{1\}$, then $H\leq Z(G)$, the center of $G$. So, if $u$ is an element of order $2$ in $G$ (Cauchy's theorem), then $\langle H,u\rangle \cong C_3\times C_2\cong C_6$.
Let $K=\langle x\rangle$ be an abelian subgroup of order 6. Now $G$ has unique subgroup $H$ of order 3, $H\leq K$, so elements of $G$ outside $K$ can have oprder 2,4 or 6.
If $z\in G\backslash K$ is an element of order 6 then $\langle z \rangle$ has a subgroup of order 3, it must be $H$, so $\langle z\rangle \cap K=H=\{1,z^2,z^4\}$, so $z^3$ can not be in $K$. We found an element $y=z^3$ of order 2 outside $K=\langle x\rangle$. Then $K=\langle x\rangle $ and $y$ will generate $G$.
Since $\langle x\rangle \triangleleft G$, $y^{-1}xy\in \langle x\rangle$, and so $y^{-1}xy\in \{x,x^{-1}\}$, since these are only  elements of order 6 in $\langle x\rangle$. But we can not have $y^{-1}xy=x$ (why?).
Therefore 

$G=\langle x,y\colon x^6, y^2, y^{-1}xy=x^{-1}\rangle $ (well-known group!)

If $z\in G\backslash K$ is an element of order 2, then as in previous paragraph, $G$ will be isomorphic to $......(?)$
If $z\in G\backslash K$ is an element of order 4, then as $H=\langle x^2\rangle \triangleleft G$, $\langle z\rangle \cong C_4$, we see that $H$ and $z$ generate $G$: $H\langle z\rangle \leq G$, and $|H\langle z\rangle| =(3)(4)/1 =12$. Then $z^{-1}(x^2)z\in \langle x^2\rangle$ and $z^{-1}(x^2)z\neq x^2$ ($G$ is non-abelian). Therefore $z^{-1}x^2z=x^4=x^{-2}$, and if we let $x_1=x^2$, then 

$G=\langle x_1,z\colon x_1^3,z^4,z^{-1}x_1z=x_1^{-1}\rangle \cong C_3\rtimes C_4$

