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When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic? $$U_n=\\ a \\in\\mathbb Z_n \\mid \\gcd(a,n)=1 \\ $$ I searched the internet but. Limit sequence (Un) and (Vn) Ask Question Asked 9 years, 2 months ago Modified 9 years, 2 months ago Estoy haciendo mi reciente evaluación interna del IB Matemáticas HL y mi tema es cómo calcular el área de superficie de un huevo. Quiero aplicar el cálculo conocimiento en esta pregunta, pero mi conocimiento sobre esta área es limitada. J. P. Aubin, Un théorème de compacité, C.R. Acad. Sc. Paris, 256 (1963), pp. 5042–5044. It seems this paper is the origin of the famous Aubin–Lions lemma. This lemma is proved, for example, here and here, but I'd like to read the original work of Aubin. However, all I got is only a brief review (from MathSciNet). Q&A for people studying math at any level and professionals in related fields A remark: regardless of whether it is true that an infinite union or intersection of open sets is open, when you have a property that holds for every finite collection of sets (in this case, the union or intersection of any finite collection of open sets is open) the validity of the property for an infinite collection doesn't follow from that. In other words, induction helps you prove a. The larger was because there are multiple obvious copies of $U (n)$ in $SU (n) \times S^1$. I haven't been able to get anywhere with that intuition though, so it. @Lhf The question is certainly not a duplicate of the linked question, since the author is asking additionally a more general question, namely What are those number theoretic situations? (where the unit group is cyclic). This is an interesting question that is not addressed at all in the proposed duplicate. The Integration by Parts formula may be stated as: $$\\int uv' = uv - \\int u'v.$$ I wonder if anyone has a clever mnemonic for the above formula. What I often do is to derive it from the Product R. Mathematics Stack Exchange is a platform for asking and answering questions on mathematics at all levels.

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