370 reputation
19
bio website math.joedub.net
location Phoenix, AZ
age 25
visits member for 2 years, 8 months
seen Dec 12 at 19:56

I'm a graduate student in Mathematics program at Arizona State University. My research interests lie in algebraic topology and hyperbolic geometry.


Dec
11
awarded  Caucus
Nov
29
asked Counting tamely ramified Galois extensions of $\mathbb{Q}_p$ with a given Galois group.
Aug
18
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 <i>a priori</i> 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 $G = C_{35}$.
Jul
2
awarded  Curious
Apr
30
accepted Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology
Apr
30
answered Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology
Apr
15
asked Help Understanding/Completing Proof of Prop 3.18/3.19 in Hatcher's Algebraic Topology
Feb
5
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.
Dec
13
comment Anti-Derivative where f(n) = m
According to the original post, $f'(1)=1$, so $C=3$.
Dec
13
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
Dec
13
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.
Dec
13
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 $\varphi^{-1} (3).$
Dec
8
accepted Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$.
Dec
8
asked Compact hypersurface of $\mathbb{R}^{n+1}$ with positive curvature is diffeomorphic to $S^n$.
Dec
7
accepted Volume of a complete, simply connected Riemannian manifold of constant negative curvature
Dec
7
asked Volume of a complete, simply connected Riemannian manifold of constant negative curvature
Dec
5
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?
Dec
5
revised Commutator subgroup of a dihedral group.
Tidied up TeX and, in general, attempted to make it more readable.
Dec
5
comment Groups elements
Hmm, since $\langle (123) \rangle$ is cyclic, that notation really should be $(\tau \bmod (123), n \bmod 2)$. I've edited my response.
Dec
5
revised Groups elements
edited body