A question about cosets. Say we have a group $G$ under the operation $+$ and a subgroup $H$. We'll consider cosets of $H$. Say we have $g_{1}+H=g_{2}+H$, where $g_{1},g_{2}\in G-H$. 
Does this mean that $g_{1}+h_{i}=g_{2}+h_{\phi(i)}$, where $i=1,2...|H|$ and $\phi(i)$ is a permutation of $i$?
Because if it does, then $g_{1}-g_{2}=h_{\phi(i)}-h_{i}$. Hence, there are |H| elements in $H$ (the elements are in the form $h_{\phi(i)}-h_{i}$), with the same value (the value is $g_{1}-g_{2}$). This is a property I've never read about before. 
 A: It seems to me that what you have shown is that there are $\lvert H\rvert$ ways to write each element of $H$ $(g_i-g_j$ is an element of $H)$ as a difference of elements of $H$.  But this is true, not just of $H$, but of any finite group.
Let $G$ be a finite group, and let $g\in G$.  Then for every $g'\in G$, there exists a unique $g''\in G$ such that $g'-g''=g$.  The element $g''$ is simply $-g+g'$.
Added: You can imagine constructing the "subtraction table" of the group, that is, the table whose $i^\text{th}$ row and $j^\text{th}$ column contains $g_i-g_j.$  The subtraction table, like the addition table, is a Latin square, that is, it contains each group element exactly once in each row and exactly once in each column.  So each group element appears $\lvert G\rvert$ times in the table.
A: $g_1 + H = g_2 + H$ means that $g_1 + h_i = g_2 + h_j$ for some $i,j\in S = \{1,2,\ldots,\left|H\right|\}$, or equivalently, $g_1 = g_2 + (h_j - h_i) = g_2 + h_k$, for some $k\in S$. For each fixed $i$ we can find a unique $\phi(i)$ such that $g_1 + h_i = g_2 + h_{\phi(i)}$, so we do indeed get a permutation $\phi$ of $S$. It is also true that each coset contains the same number of elements, $\left|H\right|$. This follows from the fact that if $g + h = g + h'$ in a group, $h = h'$ by cancellation, so $\left|g + H\right| = \left|\{g + h\mid h\in H\}\right| = \left|H\right|$ (each element $g + h\in g + H$ corresponds to the element $h\in H$, giving the required bijection, and by the above remark, each $g + h_i$ is unique). I'm not sure what you mean by $H\to\left(h_{\phi(i)} - h_i\right)$, but maybe you mean the function
\begin{align*}
f : S&\to G\\
i&\mapsto h_{\phi(i)} - h_i.
\end{align*}
This function is indeed the same as
$$
f : i\mapsto g_1 - g_2,
$$
by definition of $\phi$.
Edit: It's certainly not true that $H = \{h_{\phi(i)} - h_i\mid i\in S\}$; after all, this set has only one element, $g_1 - g_2$ (as shown above), and a general subgroup can have more than one element (remember that a set cannot have multiple instances of the same element - $\{a,a\}$ = $\{a\}$ - so it doesn't make sense to say that $H$ has more than one occurrence of the element $g_1 - g_2$). If you wanted to get all of $H$, you would need to fix $i$ and then look at $\{h_j - h_i\mid j\in S\}$. This runs through all of $H$ for the same reason that $G = \{g - g^*\mid g\in G\}$ ($g^*\in G$ fixed), which is essentially the "suduko" property of the Cayley table of a group. That is, in the Cayley table of any (finite) group $G$, each element of $G$ appears exactly once in each row and column. (And this property holds because cancellation holds in any group.)
