Both a semi and a direct product

We all know what a direct product of two groups A and B is. It is isomorphic to the group defined as follows : As a set it is  A \times B and has the group operation  (a,b)*(c,d)=(ac,bd) where ac and bd make sense inside A and B respectively. And clearly you identify A\times\{1\} with A and \{1\}\times B with B and both these are normal subgroups inside A\times B.

To make a semidirect product of two groups A and B, we need more information. Namely, we need a homomorphism of  f: B\to Aut(A) which just means we need an action of B on A. This is what links the groups A and B together.

Then we define A\rtimes_f B to be the semidirect product of A and B via f as follows :

  • As a set, it is still A\times B
  • (a,b)*(c,d) = (a f(b)(c), bd)

Notice that in the first coordinate, the map f is involved. A direct product A\times B is just A\rtimes_f B where f:B\to Aut(A) sends every b\to id_A. Thus f(b)=id_A and hence (a,b)*(c,d)=(ac,bd).

It is generally taught in any group theory course that you can find an isomorphic copy of A as a normal subgroup in A\rtimes_f B and an isomorphic copy of B as a subgroup (not neccesarily normal) in A\rtimes B.

Now here is a question :

Can a group be isomorphic to a nontrivial semidirect product of A and B (that is f:B\to Aut(A) is not the trivial map) and also simultaneously be isomorphic to the direct product of A and B. In symbols, can you think of an example of A, B and f where

A\rtimes_f B\equiv A\times B

Note that we don’t demand that the isomorphism preserves A or B or anything like that.

OK, here’s an answer. Take A=B to be some nonabelian finite group and let f: B\to Aut(A), b\to [a\to bab^{-1}]. f is not the trivial map because there is some a, b such that bab^{-1}\neq a because A=B is nonabelian.

\alpha : T\rtimes_f A\to A\times A, (a,b)\leadsto (ab,b).

  1. T is a group homomorphism because
    1. T((a,b)*(c,d))=T((abcb^{-1}, bd))= (abcb^{-1}bd, bd) = (abcd,bd)
    2. T((a,b))T((c,d))=(ab,b)(cd,d)=(abcd,bd)

     

  2. T is 1-1 because (ab,b)=(1,1) implies b=1 and ab=1 and hence (a,b)=(1,1).
  3. Since the domain and codomain are finite and have the same size, T is onto and hence is a group isomorphism.

Here’s the “picture” behind the proof [I am not sure whether there really is a mathematical analogy, but this picture helped me formalize the example]. The real plane is the span of the x and the y-axis. It is also the span of the line y=x and the x-axis… With this as an intuition, can we try to symbolize everything into groups ?

Well, look at A \times A. It has A\times\{1\}=C as a normal subgroup. It has also the “diagonal” D=\{(a,a)|a\in A\} as a subgroup which is not normal. Clearly C and D are isomorphic to A and even more clearly C\cap D=\{(1,1)\}. Also CD=A\times A for any (a,b)=(ab^{-1},1)(b,b). Thus A\times A is the semidirect product (internal, if you will) of subgroups C and D ..

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