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Mathematics
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Math Proof Training and Practice
Math Challenge - December 2021
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[QUOTE="Infrared, post: 6571475, member: 467682"] I'll try a few. It's been a while for me. 1. We have the prime factorization ##3129=3\times 7\times 149##. Let ##H## be a subgroup of order ##7.## By one of the Sylow theorems, the number of such subgroups is a divisor of ##3\times 149## and is ##1## modulo ##7.## Checking all possibilities, we see that there can only be one subgroup of order ##7,## and so it must be normal (as otherwise a conjugate subgroup would be another subgroup of order ##7##). Consider the chain of normal subgroups ##\{1\}\leq H\leq G##. The quotient ##H/\{1\}\cong H## is abelian since it has prime order ##7## and is hence cyclic. The quotient ##G/H## has order ##3\times 149## and is therefore abelian as ##149## is not ##1## mod 3 by the following fact: If ##p<q## are primes with ##q## not 1 modulo ##p,## then the only group ##K## of order ##pq## is the cyclic one. This is because by similar counting as above, there is a normal subgroup of order ##q##, so ##K## is a semidirect product of subgroups of order ##p## and ##q.## Since ##q## is not ##1## modulo ##p,## there is only the trivial homomorphism from a ##p##-subgroup to the automorphism group of the ##q## subgroup, so the semidirect product is trivial. 3. We have to check that this subspace of ##\mathfrak{gl}(\mathfrak{g})## is closed under bracket. If ##\alpha,\beta\in\mathfrak{A}(\mathfrak{g})##, then ##[[\alpha,\beta](X),Y]+[X,[\alpha,\beta](Y)]=[\alpha(\beta(X)),Y]-[\beta(\alpha(X)),Y]+[X,\alpha(\beta(Y))]-[X,\beta(\alpha(Y))].## Note that ##[\alpha(\beta(X)),Y]=-[\beta(X),\alpha(Y)]## since ##\alpha\in\mathfrak{A}(\mathfrak{g})## and similarly for the other terms, so they cancel. The non-abelian 2-dimensional Lie algebra has generators ##x,y## with the relation ##[x,y]=x.## If a matrix ##\begin{pmatrix}a & b\\c & d\end{pmatrix}## represents an element ##\sigma\in\mathfrak{gl}(\mathfrak{B})## with respect to this basis, in order for ##\sigma## to be in ##\mathfrak{A}(\mathfrak{B})## it is necessary and sufficient for ##[\sigma(x),y]+[\sigma(y),x]## to be zero. We calculate this to be ##a+d##, so ##\mathfrak{A}(\mathfrak{B})## is isomorphic to ##\mathfrak{sl}(2).## 4. Suppose that ##X## is path-connected. Suppose for contradiction that ##X=U\cup V## where ##U## and ##V## are disjoint, open, non-empty sets. Let ##u## and ##v## be points in these sets, respectively. Let ##\gamma:[0,1]\to X## be a path from ##u## to ##v.## Then ##\gamma^{-1}(U)\cup\gamma^{-1}(V)## is a disconnection of ##[0,1]##. Contradiction. The space ##X=\{(x,\sin(1/x)):x\in\mathbb{R}^{>0}\}\cup\{0\}\times [-1,1]\subset\mathbb{R}^2## is connected but not path-connected. It is connected because if ##X=U\cup V## is as above, then since ##\{0\}\times [-1,1]## is connected, it must be entirely contained within one of these sets, say ##U.## The same must be true for the piece ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}.## Since every neighborhood of ##\{0\}\times [-1,1]## contains points in this second set, this part must also be contained in ##U.## It is not path-connected because there is not path from any point in ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}## to any point in ##\{0\}\times [-1,1].## This is because if ##\gamma:[0,1]\to X## is such a path, then the second component of ##\gamma(t)## attains every value in ##[-1,1]## as ##t\to 1## and hence there is no point that can be assigned to ##\gamma(1)## in order for ##\gamma## to be continuous. The circle ##S^1## is obviously path-connected but it is not simply connected. The (based) loop ##\gamma:[0,1]\to S^1, \gamma(t)=e^{2\pi it}## is not nulhomotopic since it lifts (with respect to the usual covering ##\mathbb{R}\to S^1##) to a map ##\tilde{\gamma}:[0,1]\to\mathbb{R}## in the obvious way, and now the endpoints ##0## and ##1## map to distinct elements of ##\mathbb{R}.## A based homotopy of ##\gamma## lifts to one of ##\tilde{\gamma}## and since the endpoints of the lifted map to ##\mathbb{R}## stay distinct throughout the lifted homotopy, the homotopy cannot end in a trivial map. [/QUOTE]
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Math Proof Training and Practice
Math Challenge - December 2021
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