To prove that $H$ is a normal subgroup of $G$, we need to show that for all $g \in G$, $gH = Hg$, which is the definition of a normal subgroup. This can also be written as $gHg^{-1} = H$ for all $g \in G$.

Given Condition

The problem states that for all $a, b \in G$, we have: $aHbH = abH.$ This means that the product of any two cosets of $H$ in $G$ is itself a coset of $H$.

Step-by-Step Proof

1. Show that $H$ absorbs conjugation: We want to show that $gHg^{-1} \subseteq H$ for any $g \in G$, and ultimately $gHg^{-1} = H$, which implies that $H$ is normal.

2. Use the given property: Set $a = g$ and $b = g^{-1}$ for any $g \in G$. Then, according to the given property: $gH g^{-1} H = gg^{-1} H = eH = H,$ where $e$ is the identity element of $G$.

3. Interpret the result: Since $gH g^{-1} H = H$, this means that every element of the set $gHg^{-1}$ multiplied by any element of $H$ is still in $H$. Thus, $gHg^{-1}$ must be a subset of $H$ (since $H$ is closed under multiplication and contains the identity).

4. Conclude normality: Since $gHg^{-1} \subseteq H$ for any $g \in G$, we also have that $g^{-1}Hg \subseteq H$ (taking inverses, which are automorphisms in groups). By the subgroup properties and closure, we conclude that $gHg^{-1} = H$ for all $g \in G$.

Therefore, $H$ is normal in $G$, as required.

Conclusion

Since $gHg^{-1} = H$ for all $g \in G$, we conclude that $H$ is a normal subgroup of $G$.

Would you like a deeper exploration of any part of this proof, or do you have further questions on group theory?

Here are some follow-up questions that might interest you:

1. What are the implications of a subgroup being normal in a group?
2. How does the coset multiplication property influence the structure of the group?
3. Can we use similar techniques to identify normal subgroups in nonabelian groups?
4. How does normality relate to the concept of group homomorphisms?
5. What are some examples of groups where all subgroups are normal?

Tip: For any subgroup $H$ of a group $G$, the criterion $gHg^{-1} = H$ for all $g \in G$ is always a direct method to check if $H$ is normal.