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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.2011-07-28

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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?2011-07-28
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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 *A* ∩ *B* = 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.2011-07-28
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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 × *B* ≅ *B*, 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 *A* ∩ *B* = 1.2011-07-28
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    @ 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?2011-07-28
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    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 *B* ≤ *G* 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*).2011-07-28