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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).

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Why isn't Sylow's theorem elementary? – Mariano Suárez-Alvarez Sep 19 '11 at 20:00
Wild's paper can be found at… – lhf Sep 19 '11 at 20:18
@MarianoSuárez-Alvarez "elementary" is relative, my text "classification of the groups of order 16 that is based on elementary facts" was a direct quote from Wild's paper. – Nathan Portland Sep 19 '11 at 20:24
Groups of order 16 are 2-groups, so how would Sylow's theorems even be used at all? – user641 Sep 19 '11 at 21:15
@Steve: Sylow theorems will give existence of subgroups of order 2,4,8; from this we can show that $G$ has normal subgroups of order 2,4,8; from this we can show $G$ has abelian normal subgroup of order 8; center of $G$ intersects each normal subgroup non-trivially...for each statement we need Sylow theorem; all these facts can be used when we face "isomorphism problem" for groups of order 16 constructed. – Marshal Kurosh Sep 20 '11 at 11:41
up vote 2 down vote accepted

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".)

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Great citation, one for the library. – Mark Bennet Sep 19 '11 at 20:30
@Arturo, this is very useful, it also solves the case for order 18, but not 24 ;( – Nathan Portland Sep 19 '11 at 20:31
@Nathan: If you look at Burnside's work for groups of order $p^2q$, you'll find that the difficulties compared to the classification of groups of order $pq$ are much greater. Trying to go to $p^3q$ would likely multiply them even further. You can get a feel for that just from looking at the difference in ease of classifying groups of order $p^2$ vs. group of order $p^3$. The groups of order $p^2$ take half a page; the groups of order $pq$ take half a page. The groups of order $p^2q$ take 4 pages; of order $p^3$ take half a page, but only with many results on $p$-gruops already done. – Arturo Magidin Sep 19 '11 at 20:41
@Mark: It's pretty hard going reading Burnside today; the terminology is just very different (for example, elements of a group are called "operations", and groups are defined to be (what we now call) groups that act faithfully on some set). – Arturo Magidin Sep 19 '11 at 20:44
@Mark: Hall's "Theory of groups" is entirely modern in its language; a very far cry from Burnside's book, which it most definitely not. The effort one needs to read, say, Hall's book or Scott's book, is nothing compared to the one required for Burnside's, because of the entirely different way in which Burnside talks about event the most basic objects (the elements of the group). But if I didn't believe the "old masters" have something to teach us, I wouldn't own the book... (-; – Arturo Magidin Sep 19 '11 at 21:11

(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?).


$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

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

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when $|$Ker$ (\eta )|=6$ , it is not clear to me how it is abelian . All elements of it commute with all element of $H$ , how does that make it abelian ? – user123733 Dec 2 '14 at 13:43

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