How is $\lbrace a_1, a_2, ..., a_n : a_i \in \Bbb Z_2\rbrace$ a group? I was asked to prove that if we define 
\begin{equation*}
\Bbb Z_2^n = \lbrace a_1, a_2, ..., a_n : a_i \in \Bbb Z_2\rbrace 
\end{equation*}
then it's a group under the operation of addition like $$(a_1, a_2, \ldots, a_n) + (b_1, b_2, \ldots, b_n) = (a_1+b_1, a_2+b_2, \ldots , a_n+b_n)$$
It doesn't look like a group, because if we let $a_i = 1 = b_i$, then the sum is $(2,2,\ldots, 2)$ because it's not modular addition, but $2\notin \Bbb Z_2$. 
What am I missing?
 A: It is modular in each component. So 
$$
(1,\ldots, 1)+(1,\ldots,1)=(0,\ldots,0)
$$
and the set is closed under component-wise addition mod 2.
A: As the comments have pointed it out, it is indeed modular addition.
$\mathbb{Z}^n_2$ is $\underbrace{\mathbb{Z}_2 \times \dots \times \mathbb{Z}_2}_{n \text{ times}}$, so think about it as just dealing with $\mathbb{Z}_2$ (where the sum is modular) with $n$ separate components.
So we have $(1, \dots, 1) + (1, \dots, 1) = (0, \dots, 0)$ since $1 + 1 \equiv 0 \pmod 2$, and $0 \in \mathbb{Z}_2$.
A: If $(G_1,\phi_1)$,...,$(G_n,\phi_n)$ is a collection of $n$ groups with $n$ respective binary operations, we define their product as $(G_1\times...\times G_n, \phi_1\times...\times\phi_n)$ where $\phi_1\times...\times\phi_n:(G_1\times...\times G_n)\times(G_1\times...\times G_n)\rightarrow G_1\times...\times G_n$ by computing the group operation componentwise. In $\mathbb{Z}_2$ (which I would really recommend writing as $\mathbb{Z}/2\mathbb{Z}$), we have the operation $\phi$ as being addition modulo $2$.
A group is not just the set $\{0,1\}$. It is also the binary operation of addition modulo $2$. Whenever a group is used, do not forget about the implicit notion of the operation which gives it a group structure.
