Cosets & Groups. I am a newbie to group theory. I am confused by something that I would like clarified.
Say $G$ is the integers, a group under addition, and $H$ is the even integers, a subgroup of $G$.
It is usually stated that left cosets of $G$ are a partition of $G$. 
Now for all $g \in G$, the left coset of $G$ by $g$ is defined as $gH$ in multiplicative notation, which for this example is really $g+H$.
So, here a left coset of $H$ by $g$ should be the set $\left\{g+h_1, g+h_2, g+h_3,\ldots\right\}$, where $h_1, h_2,h_3,\ldots$ are the elements of $H$.
How can these sets be disjoint?
For example, the two 
cosets $2+H$ and $4+H$ both contain the element 8. 
I don't understand what am I getting it wrong. 
Thanks in advance.
 A: The distinct cosets of $H$ partition $G$. If $g_1+H\cap g_2+H \neq \emptyset$, then $g_1 + 2m_1 = g_2 + 2m_2$ for integers $m_1$ and $m_2$, and therefore $g_2 = g_1 + 2n$ for $n = m_2-m_1$. It follows that $g_1 + H = g_2 + H$.
The more precise statement is that if we choose a complete family of representatives $g_1,\ldots,g_n$ (or more generally $g_i$ for $i \in I$ with $I$ a (possibly infinite) index set), then $G$ is the disjoint union of $g_i + H$ for $i = 1,\ldots,n$.
Here a complete family of coset representatives is a set of distinct elements $g_1, \ldots, g_n$ of $G$ such that the cosets of $H$ in $G$ are exactly $g_1+H, \ldots, g_n+H$ and $g_i + H \neq g_j + H$ for $i \neq j$.

Edit: For your particular example, notice that $2+H = 4+H$. Indeed $g \in 2+H$ if and only if $g = 2+2n$ for some integer $n$, which in turn is true if and only if $g = 4 + 2m$ for some integer $m$. (Here $m = n-1$ since $g = 2+2n = 4 + 2(n-1)$.) Consequently $g \in 2+H$ if and only if $g \in 4+H$, i.e., $2+H = 4+H$.
