A basic question on Relative Homology And so this week, our algebraic topology class starts with relative homology groups. But there are some (REALLY) basic parts of the definition of the relative homology group that I don't understand why...
Our class is currently using Hatcher's Algebraic Topology.
Given that $A$ is a subspace of a topological space $X$. 
Hatcher defines the chain group $C_n(X,A)=C_n(X)/C_n(A)$.
I understand that $C_n(A) \subset C_n(X)$.
1) Why is $C_n(A)$ even a subgroup of $C_n(X)$ in the first place?
2) I understand why the boundary map $\partial$ takes $C_n(A)$ to $C_{n-1}(A)$, but why does it induce a quotient boundary map from $C_n(X,A)$ to $C_{n-1}(X,A)$?
I apologize if it is really simple and probably obvious to most but I really can't see why.

Ad 1), since $A\subset X$, every chain in $A$ is also a chain in $X$. Then $C_n(A)$ is the subgroup of chains such that the image of every singular simplex happens to lie in $A$. The empty chain $0$ is a chain in $A$, since it contains no (singular) simplex whose image is not contained in $A$. If $c_1,c_2 \in C_n(A)$, then every simplex in $c_1+c_2$ lies in $A$, hence $c_1+c_2\in A$, and finally, for $c\in C_n(A)$, the chain $-c$ also consists only of simplices in $A$, hence $-c \in C_n(A)$. So $C_n(A)$ is a nonempty subset of $C_n(X)$ that is closed under the group operations, hence a subgroup.
Ad 2), we have the composition
$$\varphi = \pi \circ \partial \colon C_n(X) \to C_{n-1}(X) \to C_{n-1}(X,A)$$
of the boundary map and the canonical projection. That is a homomorphism, and
$$C_n(A) \subset \ker \varphi$$
since $\partial C_n(A) \subset C_{n-1}(A) = \ker \pi$. Thus we have an induced homomorphism
$$\overline{\varphi} \colon C_n(X)/C_n(A) \to C_{n-1}(X,A).$$
This induced homomorphism is the quotient boundary map.
