- #1
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Show that the set [itex]X = \{x : 0 < x < p^m, x \equiv 1 (\mathop{\rm mod}p)\}[/itex] where p is an odd prime, together with multiplication mod [itex]p^m[/itex] forms a cyclic group. It might help to write the x in X in the form:
[tex]x = 1 + a_1p^1 + \dots + a_{m-1}p^{m-1}[/tex]
for [itex](a_1,\, a_2,\, \dots ,\, a_{m-1}) \in (\mathbb{Z}_p)^{m-1}[/itex]. I haven't been able to get anywhere this problem. I've tried to see if [itex]1 + p + p^2 + /dots + p^{m-1}[/itex] generates the group, but haven't had any success in doing so, and I'm not even sure that it does. I've also considered inductive proofs without success. Any hints?
[tex]x = 1 + a_1p^1 + \dots + a_{m-1}p^{m-1}[/tex]
for [itex](a_1,\, a_2,\, \dots ,\, a_{m-1}) \in (\mathbb{Z}_p)^{m-1}[/itex]. I haven't been able to get anywhere this problem. I've tried to see if [itex]1 + p + p^2 + /dots + p^{m-1}[/itex] generates the group, but haven't had any success in doing so, and I'm not even sure that it does. I've also considered inductive proofs without success. Any hints?