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How possible methods for finding groups of given order we have?

Sylow theorems - and next? We know something about center of group, about possible orders of these elements and subgroups. But what is most effective way, to find all non-isomorphic groups of given order? Is there any other strong "weapon" for this (like Sylow theorems)?

For example, I took groups order 15.

Commutative group is only $\mathbb{Z}_{15}$.

About non-commutative we know, they have 1 Syllow 3-subgroup and 1 Syllow 5-subgroup. And next, that $M_3 \cup M_5 = G$ (if $M_3$, resp. $M_5$ is Syllow 3-subgroup, resp. Syllow 5-subgroup).

$M_3$ and $M_5$ are single generated, than we can choose elements $f$ and $g$ satysfaing $<f>=M_3$ and $<g>=M_5$. Because there are characterictis, we get these equations: $f^{-1}gf=g^m$, where $m \in \{1,2,3,4\}$ and $g^{-1}fg=f^n$, where $n \in \{1,2\}$. $m$ and $n$ must supplying $|g|=|g^m|$ and $|f|=|f^n|$.

But there is end of my way, i don't know how continue... Can anyone help? (Sorry for bad English)

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There's the construction of semi-direct products of groups. That plus the sylow theorems would help you to some extent. There are more sophisticated methods from representation theory for the complete classification. – user1119 Nov 29 '10 at 10:56
I don't like semi-direct products of groups, becouse is bounded to some automorphism, and i don't know how to make with him in abstract view... How can help me in this way? – tomas.lang Nov 29 '10 at 11:06
If your group has only one $p$-Sylow group, then this $p$-Sylow subgroup is normal. $M_3 = \langle f \rangle$ and $M_5 = \langle g \rangle$ are cyclic, and $G$ is generated by the two elements $f, g$. Now $[f, g] = f^{-1}g^{-1}fg = f^{-1}f^g \in M_3$ as $M_3$ is normal and $[f, g] = (g^{-1})^{f^{-1}}g \in M_5$ as $M_5$ is normal. Hence $[f, g] \in M_3 \cap M_5 = 1$ is trivial, i.e., both elements commute, and as $G = \langle a, b\rangle$ your group $G$ is abelian. – j.p. Nov 29 '10 at 11:17
up vote 5 down vote accepted

If you don't like semi-direct products, then you will not get very far in your endeavour. I recommend that you play around with them until you start liking them. The reason is that the single most powerful result in trying to classify all finite groups of given order is the so-called Schur-Zassenhaus theorem:

Schur-Zassenhaus Theorem: Let $N$ be a normal subgroup of a finite group $G$ and suppose that the order of $N$ is co-prime to its index. Then $G$ is a semidirect product $G=N\rtimes H$ for a subgroup $H$ of $G$.

Why is this useful? I will show you, how to finish your project for $n=15$ essentially in one line: as you have noted, there is one Sylow 5-subgroup in $G$, so it's normal. Since its index is 3, Schur-Zassenhaus applies and tells us that $G=C_5\rtimes C_3$. But the automorphism group of $C_5$ is cyclic of order 4, so there are no non-trivial maps from $C_3$ to the automorphism group of $C_5$ and $G$ must in fact be a direct product.

Exercise: classify all finite groups of order 20 using this approach.

P.S. One more thing: don't ever take unions of subgroups! What you wanted to say is that $G=\langle M_3,M_5\rangle$, i.e. is generated by (all possible products of elements of) the subgroups.

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Here's how to finish without assuming the Sylow 5-subgroup is normal.

(Assume by way of contradiction:) If $n=2$, so $g^{-1} f g = ff$, then: $$g^{-2} f g^2 = g^{-1}( g^{-1} f g ) g = g^{-1}( ff )g = g^{-1}( fg g^{-1}f)g = (g^{-1}fg) (g^{-1}fg) = (ff) (ff) = f^4.$$ Every time you apply $g^{-1}( * )g$, you double the number of $f$, so $g^{-i} f g^i = f^{(2^i)}$.

Now we use $g^5 = 1$. If you apply $g^{-1}( * )g$ five times, you get $g^{-5} f g^5 = 1 f 1 = f$ on the one hand, but $f^{(2^5)} = f^{32} = f^2$ on the other, using the fact that 2 ≡ 32 mod 3.

How can $f = f^2$? It does not, since $f$ has order 3. Assuming $n=2$ gives a contradiction, so $n=1$, and $g^{-1} f g = f$. Multiply on the left by $g$ to get $f g = g f$. The group must be abelian, since all of its generators commute.

Now try this with "5" replaced by "2" and you will still get the Sylow 3-subgroup is normal, but the Sylow 2-subgroup need not be normal. The same argument will get you that both n=1 (cyclic group of order six), and n=2 (dihedral group of order six) are possible.

More precisely, you'll check that $g^{-2} f g^2 = f^{(2^2)} = f^4 = f$ just as it should be.

Congratulations, you have just analyzed some semi-direct products using automorphisms! With cyclic Sylow subgroups it makes a lot of sense. We didn't even need to call "n" an automorphism, nor did we need to call M3∪M5 a semi-direct product.

If the subgroups are not Sylow subgroups, then it is harder to check (1) that they are normal or (2) that M∩N = 1, but there are more general techniques for this.

If the subgroups are not cyclic, then you cannot just use one "n" for the Sylow 3-subgroup. You get a matrix of "m"s and "n"s. Not just any matrix will do, but the ones that do work are called automorphisms.

In other words, the general case of semi-direct products is similar to what you are now doing. Don't give up.

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