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let $G$ be a group. and $A$, $B$ be two subgroups of $G$. suppose we have an action of $B$ on $A$ : $\phi:B\rightarrow Aut(A)$ then we can turn the set $AB$ into a group by defining the multiplication law: $(a_1b_1)(a_2b_2)=a_1\phi_{b_1}(a_2)b_1b_2$. the group we get we call the semidirect product of $A$ by $B$ corresponding to $\phi$.

This does not require $A$ to be normal in $G$ unless we choose $\phi $ to be conjugation and i don't understand the need for the conditions $G=AB$ and $A\cap B=\{1\}$?

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Isn't your definition of the semi direct product wrong? If i remember correctly, it is a part of A $\times$ B instead of AB. Thus your multiplication law is wrong. –  sxd Jul 28 '11 at 14:25
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up vote 3 down vote accepted

If you want the multiplication to be well-defined, then you must have AB = 1. If g in AB, how do you know if g = g⋅1 or g = 1⋅g. In particular, you need some very serious compatibility conditions for AB and φ.

If you want to say that G, rather than just AB, is a semi-direct product, then obviously you need G = AB.

If you want AB to be a subgroup of G, then you must have A is normalized by B, since clearly A is normal in the group structure you define on AB. Assuming you want G to be a semi-direct product, then that means A is normal in G.


As a specific example, consider G to be the symmetric group on 3 points, and let A be generated by (1,2) and B be generated by (2,3). Let φ be the unique function (which happens to be a homomorphism) from B to Aut(A). Then AB = { (), (1,2), (2,3), (1,3,2) } has order 4, and the group law you define is well-defined since AB = 1, but it gives the group a very weird multiplication: (1,2)⋅(2,3) = (2,3)⋅(1,2) in the new AB. The subgroup generated by A and B is actually G itself, of order 6. So we've somehow defined a group of order 4 as a subset of a group of order 6, and Lagrange tells us we do not have a subgroup.

To see the problems with overlap, consider two Sylow 2-subgroups of the symmetric group of order 4. Letting φ be an isomorphism (spooky choice, eh), I think you'll find the multiplication is not well-defined on the set AB.

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so the normal subgroup condition is a result and not part of the definition is that correct? –  palio Jul 28 '11 at 14:42
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There are two definitions of semi-direct product. The internal definition concerns a normal subgroup A, a subgroup B, such that B normalizes A and AB = 1. It is a result that AB is a group in this case, and the definition is just that AB is called a semi-direct product. The second definition is the external definition, and it concerns two groups A and B and a homomorphism φ from B to Aut(A), and defines a group structure on the set A × B, called the semi-direct product. It is a result that A × 1 is a normal subgroup of that semi-direct product. –  Jack Schmidt Jul 28 '11 at 14:46
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It is also a result that A × 1 ≅ A, and the set 1 × B is a subgroup of the external semi-direct product, that 1 × BB, and that (A × 1) ∩ (1 × B) = 1×1 = 1, so that the external semi-direct product is also an internal semi-direct product. The actual subgroups involved in the internal vs. external are different sets, but isomorphic as groups. In the external, you write them as tuples (a, b). In the internal you can think of them as just products ab, but to be able to find the a and b from ab, you need to have AB = 1. –  Jack Schmidt Jul 28 '11 at 14:49
    
@ Jack Schmidt: in the internal definition don't we have to specify the action $\phi:B\rightarrow Aut(A)$ as i did or we suppose always that it is conjugation and in this case is the internal semidirect product always UNIQUE since we don't require to specify an action? –  palio Jul 28 '11 at 14:53
    
In the internal we always specify that φ is conjugation, so yes the internal semidirect product is always unique: it is the subgroup AB of G. In other words, the multiplication in AB is the same as the multiplication in G, and that forces φ to be conjugation. $$ $$Be careful that the set AB can be a subgroup of G without being a semi-direct product. Fo instance, if A = G, then any BG works since GB = B. There is a group G of order 16 where every subgroup A and every subgroup B forms a subgroup AB, but neither A nor B need be normal in AB). –  Jack Schmidt Jul 28 '11 at 15:00
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