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 ..