I'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:
- Given two groups $A$ and $B$, construct a group $G$ which admits a normal subgroup $H$ isomorphic to $A$, such that $G/H$ is isomorphic to $B$.
I think I have a proof that the cartesian product $A \times B$ (with the usual component-wise operation) verifies (1), but since I'm just starting out I'm not totally confident in my construction. Furthermore, if I'm right, is this the only group up to isomorphism satisfying (1)?
Edit: I just noticed Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H, which seems to positively answer my question. In that case I'd like to draw attention to the follow up question above (uniqueness up to isomorphism).
Best Answer
Yes, the direct product $A \times B$ satisfies the property, as you've noticed. But it's not unique up to isomorphism. For example, the dihedral group $D_n$ has a normal subgroup $H \simeq \mathbb Z/n \mathbb Z$, with $G/H \simeq \mathbb Z/2 \mathbb Z$, but $D_n$ is not isomorphic (for $n > 1$) to $\mathbb Z/2\mathbb Z \times \mathbb Z/n\mathbb Z$.
More generally, you're asking whether, given an exact sequence of the form $1 \to N \to G \to H \to 1$, is $G$ isomorphic to $N \times H$? The answer is no, as I've shown. Many counterexamples are provided by semidirect products (something you'll learn soon enough if you're studying elementary group theory).
For abelian groups, the concept of Ext functor allows one to classify all such extensions (given abelian groups $A,B$, "how many" groups $G$ are there with an exact sequence $0 \to B \to G \to A \to 0$ is given by $\mathrm{Ext}(A,B)$), but this is much more advanced.