rearrangement theorem (group theory)

If \(e,g_1,g_2,...,g_n\) are the elements of a group (algebra) and \(a\) is an arbitrary element then \(ae, ag_1, ag_2, ..., ag_n\) contains each element of the group exactly once

• \(ae, ag_1, ag_2, ..., ag_n\) contains each element at least once
  • consider an arbitrary element \(g_i\). The element \(a^{-1}g_i\) occurs on the LHS of one of the above products, because all elements of the group are represented on the LHS in the list. So, the list contains \(g_i\)
• \(ae, ag_1, ag_2, ..., ag_n\) contains each element at most once
  • suppose there exist \(ag_i = ag_j\) for unique \(g_i\) and \(g_j\) in the above list. Then, multiplying by \(a^{-1}\) gives \(g_i = g_j\), which contradicts the assumption that they are unique

• ashour
• dresselhaus