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Nilpotent Elements of ##\mathbb{Z}/n\mathbb{Z}##
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[QUOTE="fresh_42, post: 5633280, member: 572553"] Yes, that's right. So you have ##n=p_1^{c_1}\cdot \ldots p_k^{c_k}##. Now if you defined ##c:= max\{c_1, \ldots ,c_k\}##. What is ##(x(p_1\cdot \ldots \cdot p_k))^c##? And what does it mean for the elements of your subgroup. I don't really understand the rest under 3.) Since we are talking about nilpotent elements, I assume we are talking about rings. ##\mathbb{Z}_n## is a ring, so far so good. The only guaranteed group in a ring is the additive group. (Because e.g. ##3\cdot 24=0## in ##\mathbb{Z}_{72}##, multiplication is in general no group.) So if you talk about subgroups, you have to mean according to addition. These groups however don't automatically inherit the multiplication structure of the ring they are from. Structures which do that are either subrings or ideals. This is why I don't understand, what you actually mean, esp. by ##x\cdot p##. Do you mean ##x## is a natural number and ##xp## is an abbreviation for ##p+p+\ldots +p## (##x## times)? So the short answer to your question is probably, yes. (I just don't follow your argument. Try the one above (first 3 lines)). [/QUOTE]
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Nilpotent Elements of ##\mathbb{Z}/n\mathbb{Z}##
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