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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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    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