Partition of a group such that an operation can be defined I'm struggling with Problem 43 of 3.1 of Dummit's algebra book. The problem is:

Assume $P=\{A_i\}$ is any partition of $G$ with the property that a "quotient operation" is defined as follows: to compute the product of $A_i$ and $A_j$ take any element $a_i$ of $A_i$ and $a_j$ of $A_j$ and define $A_i∗A_j$ to be an element of $P$ that contains $a_i∗a_j$ (assume this operation is well defined).
Prove that an element of $P$ that contains an identity of $G$ is a normal subgroup of $G$ and elements of $P$ are cosets of this normal subgroup.

I tried to first prove that the element of $P$ that contains an identity is a subgroup and normal... but it seems that I have not enough decks in my hands to play the game. Any help will be greatly appreciated.
 A: Denote by $A_g$ the element of $P$ that contains $g \in G$. 
The proof mostly uses the facts that the operation is well-defined and $A_g=A_1 \iff g \in A_1$

Let's check that $A_1$, the element of $P$ containing the identity, is a subgroup:
Identity: It's given that $1 \in A_1$
Closure: Suppose $a, b \in A_1$. Then $A_a = A_1$ and $A_b=A_1$.  
By the definition of the operation, $A_1\ast A_1$ is the element of $P$ containing $1 \ast 1 =1$, so $A_1 \ast A_1=A_1$.  
But $A_1=A_a$ and $A_1=A_b$, so, since the operation is well defined, the equation becomes $A_a \ast A_b = A_1$, which means that the element of $P$ that contains $a \ast b$ is $A_1$.
Inverses: suppose $a \in A_1$.
  The set $A_{a^{-1}} \ast A_a $ is the element of $P$ containing $a^{-1}\ast a =1$, so $A_{a^{-1}} \ast A_a =A_1$.  
But since $A_a=A_1$, the equation becomes $A_a= A_{a^{-1}} \ast A_1$, which shows that the element of $P$ containing $a$ must also contain $a^{-1}\ast 1 = a^{-1}$.

Now let's check it's normal:
Let $g \in G$ and $a \in A_1$. We must check that $gag^{-1} \in A_1$, meaning $A_g \ast A_a \ast A_{g^{-1}}=A_1$.
Since $a \in A_1$, we have $A_a = A_1$.
Hence, $A_g \ast A_a \ast A_{g^{-1}}=A_g \ast A_1 \ast A_{g^{-1}}$, which is the element of $P$ containing $g\ast 1 \ast g^{-1}=g\ast g^{-1}=1$, and this element is $A_1$.
