Assumption throughout the post:
Suppose that $G = \bigoplus_{i=0}^{\infty} \mathbb{Z}$ and $N = \bigoplus_{i=1}^{\infty} \mathbb{Z}$.
Question #1:
Is the following attempt correct?
My attempt:
Since $\mathbb{Z}$ is an abelian group, then $G$ is an abelian group. Therefore, every subgroup of $G$ is a normal subgroup of $G$. Hence, $N \le G$, implies $N \trianglelefteq G$.
Question #2:
Is $N = \bigoplus_{i=1}^{\infty} \mathbb{Z} := \{(0, a_1, a_2, \dots , a_k, 0, 0, \dots) \, \mid \, (\forall i \in \mathbb{N})([1 \le k \le i] \Rightarrow [a_k \ne 0])\, \wedge \, (\forall k > i)(a_k = 0)\}$, where $a_j = f(j)$, for some $f \in \prod_{i=1}^{\infty} \mathbb{Z}$?
Here, $\prod_{i=1}^{\infty} \mathbb{Z}$ is the infinite direct product of groups.