# Theorem If $e, g_1, g_2, \ldots, g_n$ are are the [[Group Element|elements]] of a [[group]], and $a_k$ is an arbitrary element, then the assembly of elements $ a_ke, a_ka_1, a_k a_2, \ldots a_ka_n $ contains each element of the group once and only once. ### Proof First, we prove that each element in the group exists in the assembly. Let $g$ be an arbitrary element of the group. Since the elements form a group, there's an element $a_r = a_k^{-1} g$ in the group (by closure). Then, the element of the assembly $a_k a_r$ is given by $a_k a_r = a_k a_k^{-1} g = g$. Thus, we can always find an arbitrary element of the group $g$ in the assembly shown above after multiplication of the appropriate group elements. Second, we prove that each element appears in the assembly once and only once. We prove this by contradiction. First, assume that $g$ appears twice in the assembly, i.e., $g = a_k a_r = a_k a_s$. Multiplying on the left by $a_k^{-1}$, we get $a_r = a_s$, which contradicts the uniqueness of the original group's elements the way we listed them. # Consequences Due to this theorem, every row and column of a [[Group Multiplication Table]] contains each element once and only once. Say row $k+1$ of the table corresponds to element $a_k$. Then, the entries of this row are $a_k e, a_k a_1, \ldots a_k a_n$, which are just the assembly shown above. We know from the rearrangement theorem that the assembly contains each element of the group once and only once, hence the statement above.