Question

Let A and B be normal subgroups of a group G such that $A\cap B=⟨e⟩$ and $AB=G$ (see Exercise 20). Prove that$A\times B\cong G$ . [Hint: Define$f:A\times B\to G$ by$f\left(a,b\right)=ab$ and use Exercise 21.]

Short answer

It has been proved that $A\times B\cong G$

Step-by-step solution

Define a function

It is given that A and B be normal subgroups of a group G such that $A\cap B=⟨e⟩$ and $AB=G$.

This implies $ab=ba$for every $a\in A$and $b\in B$.

Define $f:A\times B\to G$ by $f\left(a,b\right)=ab$

Prove Injectivity

Let $f\left(a,b\right)=f\left(c,d\right)$, then$ab=cd$.

Multiplying by $a^{-1}$ on the left and $d^{-1}$ on the right,

We get $b{d}^{-1}=a^{-1}c\in A\cap B=\left\{e\right\}$

Therefore, a=c and b=d so that $\left(a,b\right)=\left(c,d\right)$.

Thus, f is injective.

.

Prove Surjectivity

$f\left(A\times B\right)=AB=G$

Thus, f is surjective.

Prove homomorphism

Let $\left(a,b\right),\left(c,d\right)\in A\times B$

Then

$\begin{array}{c}f\left(\left(a,b\right)·\left(c,d\right)\right)=f\left(ac,bd\right)\\ =acbd\\ =abcd\\ =f\left(a,b\right)·f\left(c,d\right)\end{array}$

Hence f is homomorphism.

Conclusion

Being surjective, injective, and homomorphic, f is isomorphism.

Thus, $A\times B\cong G$