We'll denote $A \lhd G$ for $A$ a normal subgroup of $G$; and $A \leq G$ to mean $A$ is a subgroup of $G$.
I don't know if it's a tricky question. But it seems strange to me. The question asks for an example of $A \le G$ solvable, $B \lhd G$ solvable, but $AB$ is not solvable.
We have on page 122 of the book A Course in the Theory of Groups of Derek J. S. Robinson. It states that:
The product of two normal solvable subgroups of a group is solvable.
And here's it's proof:
Let $M \lhd G$, $N \lhd G$ are two normal solvable subgroups of $G$. Then, since $N \lhd G$, we have $N \lhd MN \leq G$. So: $MN / N \cong M / (M \cap N)$. Since $M$ is solvable, $M/ (M \cap N)$ is too. Which means that $MN / N$ is solvable. Now by the fact that $N$ is also solvable, we have $MN$ is solvable. (Q.E.D)
My question is that: Maybe I am missing something, but I fail to see where in the proof above is the fact that $M, N$ are both normal subgroups used?
Is it a tricky question? Do such subgroups exist?
Thank you guys so much,
And have a good day, 😡
Best Answer
The same proof works if one take $A$ to be non-normal instead of being normal.
One has $AB/B \cong A/A \cap B$ is soluble since it is a homomorphic image of a soluble one. As $B$ is soluble and the class of soluble groups is closed under taking extensions of its members, $AB$ is soluble.
If I remember, $AB$ is also soluble, if one drop the normality condition and assume that both $A$ and $B$ are nilpotent.