I'm having trouble understanding the following method for determining the number of semidirect products between two groups in simple cases that arise when trying to classify certain groups of small order. Here is an example, but I'm really interested in understanding the "general" (i.e. applies in lots of special cases for classifying small order groups) argument.

For example, I was just trying to classify groups of order $140=2^2\cdot 5\cdot 7$. Sylow's theorems immediately give $n_5=n_7 = 1$, so we have that the group is either $(\mathbb{Z}_2\oplus \mathbb{Z}_2) \ltimes (\mathbb{Z}_5 \oplus \mathbb{Z}_7)$ or $\mathbb{Z}_4 \ltimes (\mathbb{Z}_5 \oplus \mathbb{Z}_7)$ depending on the 2-Sylow subgroup. Now, so clearly for each map $$\mathbb{Z}_4 \to \operatorname{Aut}(\mathbb{Z}_5 \oplus \mathbb{Z}_7) \simeq \mathbb{Z}_4 \oplus \mathbb{Z}_6$$ and $$\mathbb{Z}_2\oplus \mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_5 \oplus \mathbb{Z}_7) \simeq \mathbb{Z}_4 \oplus \mathbb{Z}_6$$

I can count these, but I'm not exactly sure how to prove that they all lead to unique semidirect products which I think happens to be the case. I don't really understand the argument given here, for example. I particularly don't understand how to deal with the fact that in this case, both factors in the semidirect product have order divisible by 2, so it seems like an isomorphism could mix things up in ways that you could probably prove can't happen when the orders are relatively prime.


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    $\begingroup$ Note: it is not true that "both factors in the semidirect product have order divisible by $2$". One factor has order $4$, the other has order $35$. I think you are confusing the automorphism group of that cyclic subgroup of order $35$ with the factor in the semidirect product. $\endgroup$ Commented Sep 7, 2011 at 20:20
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    $\begingroup$ I don't think the link you give provides a complete argument: while it is true that non-isomorphic semidirect products of $N$ by $H$ must correspond to distinct homomorphisms $H\to \mathrm{Aut}(N)$, it is indeed possible to have distinct homorphisms that yield isomorphic semidirect products. The link you give simply asserts that the groups you get are non-isomorphic (and omits one possibility in each grouping, possibly because it yielded a group isomorphic to another one already obtained), and does not prove it (at least not explicitly). $\endgroup$ Commented Sep 7, 2011 at 20:23
  • $\begingroup$ Oh, you are completely right! Can you explain why the link's argument works? Thanks! $\endgroup$
    – Otis
    Commented Sep 7, 2011 at 21:35
  • $\begingroup$ I'll have to think about this, but I wanted to make sure to mention Keith Conrad's handout on groups of order 12 which, if I remember correctly, goes through the reasoning for why the groups he obtains are non-isomorphic. $\endgroup$ Commented Sep 7, 2011 at 21:38
  • $\begingroup$ @Otis: Have you heard of "modules" or "character theory"? I can explain it using some fancier algebra, but especially with the 2×2 case, I think I'd more or less be teaching you those in the answer anyways. This argument would apply anytime you are using the "normal Sylow" trick, at least if the Sylow is abelian. $\endgroup$ Commented Sep 7, 2011 at 21:51

1 Answer 1


Here is a general answer that mostly just sums up the consequences of Schur–Zassenhaus.

Hypothesis and goal: Suppose G is a group with a normal Hall subgroup N, that is the order and index of N are coprime. We want to describe G as extension of a group of the same order as N, and determine all descriptions up to isomorphisms of G.

Observation 1: N is the unique subgroup of G of the same order as N. In particular, the isomorphism type of N is determined, and the specific copy of N in G is determined as a subset of G, and up to an element of Aut(N) for the embedding.

Observation 2: G is a semi-direct product of QG / N, and Q is determined up to conjugacy in G (Schur–Zassenhaus, or Sylow's theorem in the 140 case). In particular, the isomorphism type of Q is determined, the specific copy of Q in G is determined up to conjugacy by an element of N, and and the embedding into a specific copy is determined up to an element of Aut(Q).

Observation 3: The action of Q on N is determined up to a pair of elements of Aut(Q)×Aut(N). In particular if N is abelian, then this is the same as an Aut(Q)-conjugacy class of Q-module structures on N.

Observation 4: The reverse is true: given an Aut(Q)-conjugacy class of Q-module structures on N we get an isomorphism class of G, the semi-direct product.

Result: Hence we have a one-to-one correspondence between isomorphism classes of groups G with normal abelian Hall subgroup N and quotient QG / N and the Aut(Q)-conjugacy classes of Q-module structures on N.

This uses that N is a normal Hall subgroup on several occasions to radically simplify the arguments. At the very least, I always work with characteristic N, but in this case we have vanishing first and second (and third) cohomology which simplifies several steps.

More down to earth comments:

Hence we just count homomorphisms from Q to Aut(N) up to Aut(Q)×Aut(N) conjugacy. For G of order 140, N of order 35, that means N is cyclic of order 35, and Q is either cyclic or elementary abelian of order 4, and there are 6 and 5 such classes of homomorphisms, respectively. They are listed in the pdf you mentioned.

For Q cyclic, you can distinguish each of the actions by what it does on the Sylow 5-subgroup and the Sylow 7-subgroup as the effect can be described intrinsically as: centralize, invert, or (in the case of the Sylow 5-subgroup) "do an order 4 thing". The three possibilities on the (normal) Sylow 5-subgroup combine in all possible ways (due to CRT) with the two possibilities on the (normal) Sylow 7-subgroup, giving both at most and at least six distinct groups.

For Q elementary abelian, things are a little more complex, but basically on the Sylow 5-subgroup it must either centralize, or have two of its elements invert. If it centralizes, then we have the same two possibilities on the Sylow 7-subgroup ( so (1)(2) + (1)(?) groups so far). If it inverts, then the question is how many of those inverters of the 5 remain inverters of the 7. Answers are 0, 1, 2 giving a total of (1)(2) + (1)(3) = 5 groups.

One should get similar results for groups of order 4pq where p, q are primes with p not dividing q−1 and q not dividing p−1, with exactly one of p, q equivalent to 1 mod 4.

For instance, there are 11 groups of order (4)(5)(19) = 380 and of order (4)(7)(13) = 364. If both p and q are equivalent to 1 mod 4, then you get an extra possibility on the cyclic Q side too: (3)(3) actions individually, but now it matters if the order 4 on the 5 and the order 4 on the 7 match, giving (3)(3)+1 = 10 groups with cyclic Q and 5 with elementary abelian Q.


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