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 Mar26 comment Help with old exam question relating a spanning set of covectors of $T_{p}^{*} M$ to a coordinate chart containing $p$. Doh! That makes perfect sense and is so much cleaner than what I was trying to do. Thank you very much. Mar26 accepted Help with old exam question relating a spanning set of covectors of $T_{p}^{*} M$ to a coordinate chart containing $p$. Mar26 asked Help with old exam question relating a spanning set of covectors of $T_{p}^{*} M$ to a coordinate chart containing $p$. Dec11 awarded Caucus Nov29 asked Counting tamely ramified Galois extensions of $\mathbb{Q}_p$ with a given Galois group. Aug18 comment If G is a group of order n=35, then it is cyclic As others have said, this proof does not work because you don't know a priori that it is abelian. It's entirely possible that $gh$ has infinite order (see: above matrix example). Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. We know that $n_p \equiv 1 \pmod p$ and $n_p \mid 35$. We deduce from this that $n_5 = n_7 = 1$, so both the $5$-subgroup, $C_5$, and $7$-subgroup, $C_7$, are normal in $G$. By normality, we have that $C_5 C_7 < G$. Moreover, since $C_5 \cap C_7 = \{1\}$, $C_5 C_7 = C_5 \times C_7 \cong C_{35}$, and finiteness of $G$ implies that $G = C_{35}$. Jul2 awarded Curious Apr30 accepted Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology Apr30 answered Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology Apr15 asked Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology Feb5 comment Open sets are $\mu*$-measurable in a metric space with a given condition Notice that it's from your second approach, where you strive for contradiction. You already have that $\mu^{*}(E) < \mu^{*}(E \cap U) + \mu^{*}(E \cap U^{C})$, so the suggestion was to find a $U' \subseteq U$ that is just big enough to maintain the inequality $\mu^{*}(E) < \mu^{*}(E \cap U') + \mu^{*}(E \cap U^{C}) \leq \mu^{*}(E \cap U) + \mu^{*}(E \cap U^{C})$, and just small enough to also yield $d(E \cap U', E \cap U^{C}) > 0$. From there, your calculation shows that the inequality cannot be strict, and therein lies your contradiction. Dec13 comment Anti-Derivative where f(n) = m According to the original post, $f'(1)=1$, so $C=3$. Dec13 revised For the mapping $\varphi: \Bbb{Z}+\Bbb{Z} \to \Bbb{Z}$ given by $(a,b) \to a-b$, describe the set $\varphi^{-1} (3).$ TeXified it Dec13 comment Anti-Derivative where f(n) = m Almost had it. When you take the antiderivative of a constant $C$ (with respect to $x$), you end up with $Cx+D$, where $D$ is another constant. You should plug in for your constants along the way once you've solved for them. So the first derivative becomes $f'(x)=4x^3 - 6x^2 + 3$. Then take the antiderivative of that and find the new constant based on the initial values. Dec13 suggested approved edit on For the mapping $\varphi: \Bbb{Z}+\Bbb{Z} \to \Bbb{Z}$ given by $(a,b) \to a-b$, describe the set $\varphi^{-1} (3).$ Dec8 accepted Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$. Dec8 asked Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$. Dec7 accepted Volume of a complete, simply connected Riemannian manifold of constant negative curvature Dec7 asked Volume of a complete, simply connected Riemannian manifold of constant negative curvature Dec5 comment Groups elements I would prefer to keep contact here on StackExchange so that other users can chime in while I'm away. What exactly is it that you're still stuck on?