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Dummit and Foote - Let $p$ and $q$ be primes with $p < q$. Prove that a non-abelian group $G$ of order $pq$ has a non normal subgroup of index $q$, so that there exists an injective homomorphism into $S_q$. Deduce that $G$ is isomorphic to a subgroup of the normalizer in $S_q$ of the cyclic group generated by the $q$-cycle $(1\ 2\ \dots\ q)$. I've gotten as far as to see that $(1\ 2\ \dots\ q)$ is in this subgroup. And I'm blanking thereafter. Attempt : $q$ being the larger prime, any subgroup of order $q$ is normal and hall. A normal hall subgroup is unique. $Z(G)$ is $1$ or non abelian is contradicted. The class equation for $G$ is $1 + p + q(\dots)$. If a subgroup of index $q$ is normal, it would again be hall and then unique, contradicting $pq = 1 + p + q(\dots)$. Let this be $H \le G$ of index $q$. $G$ acting on the cosets of $H$ has kernel the largest normal subgroup of $H$. Since $H$ not normal, and $|H|$ prime, the kernel is $1$ and $G/1 = G \cong$ a subgroup of $S_q$. The unique normal subgroup of size $q$ cycles the cosets $1H, qH, \dots$ so the left regular presentation of $G$ has the cycle $(12\dots q)$ in it. I'm stuck, can't see further. Pls help.

I am considering a cubic charge in Cartesian Coordinate system whose vertices are $(-20,-20,-20)$,$(-20,-20,20)$,$(-20,20,-20)$,$(-20,20,20)$,$(20,-20,-20)$,$(20,-20,20)$,$(20,20,-20)$,$(20,20,20)$. The charge density inside the cube is uniform and is unity. Let $r$ be distance between "a point inside the cube $(x,y,z)$" and "a point $(a,b,c)$ in $\mathbb{R}^3$" The electric field's $x$-component at $(a,b,c)$ is given by the integral : \begin{align} E_x (a,b,c) & = \int_{V_{cube}} \dfrac{a-x}{r^3}\ dV\\ & = \int_{-20}^{20}\int_{-20}^{20}\int_{-20}^{20} \dfrac{a-x}{[(a-x)^2+(b-y)^2+(c-z)^2]^{3/2}}\ dx\ dy\ dz\\ & = \int_{-20}^{20}\int_{-20}^{20} \dfrac{1}{[(a-x)^2+(b-y)^2+(c-z)^2]^{1/2}}\Biggl|_{x=-20}^{x=20} \ dy\ dz\\ & = \int_{-20}^{20}\int_{-20}^{20} \left[ \dfrac{1}{[(a-20)^2+(b-y)^2+(c-z)^2]^{1/2}}-\dfrac{1}{[(a+20^2+(b-y)^2+(c-z)^2]^{1/2}} \right] \ dy\ dz\\ & = \int_{-20}^{20} \left[ \int_{-20}^{20} \dfrac{1}{[(a-20)^2+(b-y)^2+(c-z)^2]^{1/2}}\ dy -\int_{-20}^{20} \dfrac{1}{[(a+20)^2+(b-y)^2+(c-z)^2]^{1/2}}\ dy \right]\ dz\\ & = \int_{-20}^{20} \left[ \ln \biggl| \sqrt{(a-20)^2+(b-y)^2+(c-z)^2} +y-b \biggl| \Biggl|_{y=-20}^{y=20} -\ln \biggl| \sqrt{(a+20)^2+(b-y)^2+(c-z)^2} +y-b \biggl| \Biggl|_{y=-20}^{y=20} \right]\ dz\\ & = \int_{-20}^{20}\left( \frac{ }{ } \right.\\ & + \ln \biggl| \sqrt{(a-20)^2+(b-20)^2+(c-z)^2} +20-b \biggl| \\ & - \ln \biggl| \sqrt{(a-20)^2+(b+20)^2+(c-z)^2} -20-b \biggl| \\ & - \ln \biggl| \sqrt{(a+20)^2+(b-20)^2+(c-z)^2} +20-b \biggl| \\ & + \ln \biggl| \sqrt{(a+20)^2+(b+20)^2+(c-z)^2} -20-b \biggl| \left. \frac{ }{ } \right) dz\\ & = T_1-T_2-T_3+T_4\Biggl|_{z=-20}^{z=20}\\ \end{align} where, according to online integral calculator : \begin{equation} T_1=\\ \left(20 - b\right) \left(\ln{\left|\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} } + z - c \right|}\\ - \ln{\left|\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} } - z + c \right|}\right)\\ + \left(z - c\right) \ln\left|\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2}} - b + 20\right|\\ + {2 \left(a - 20\right)\\ \tan^{-1}\left(\frac{\sqrt{\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2}} + b - 20} \left(\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2}} - \sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2}}\right)}{\left(z - c\right)\sqrt{\sqrt{\left(a - 20\right)^{2} + \left(b - 20\right)^{2}} - b + 20} }\right)} \end{equation} $ $ $ $ \begin{equation} T_2=\\ \left(-20 - b\right) \left(\ln{\left|\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} } + z - c \right|}\\ - \ln{\left|\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} } - z + c \right|}\right)\\ + \left(z - c\right) \ln\left|\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2}} - b - 20\right|\\ + {2 \left(a - 20\right)\\ \tan^{-1}\left(\frac{\sqrt{\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2}} + b + 20} \left(\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2}} - \sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2}}\right)}{\left(z - c\right)\sqrt{\sqrt{\left(a - 20\right)^{2} + \left(b + 20\right)^{2}} - b - 20} }\right)} \end{equation} $ $ $ $ \begin{equation} T_3=\\ \left(20 - b\right) \left(\ln{\left|\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} } + z - c \right|}\\ - \ln{\left|\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} } - z + c \right|}\right)\\ + \left(z - c\right) \ln\left|\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2}} - b + 20\right|\\ + {2 \left(a + 20\right)\\ \tan^{-1}\left(\frac{\sqrt{\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2}} + b - 20} \left(\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2} + \left(c - z\right)^{2}} - \sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2}}\right)}{\left(z - c\right)\sqrt{\sqrt{\left(a + 20\right)^{2} + \left(b - 20\right)^{2}} - b + 20} }\right)} \end{equation} $ $ $ $ \begin{equation} T_4=\\ \left(20 - b\right) \left(\ln{\left|\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} } + z - c \right|}\\ - \ln{\left|\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2} } - \sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} } - z + c \right|}\right)\\ + \left(z - c\right) \ln\left|\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2}} - b - 20\right|\\ + {2 \left(a + 20\right)\\ \tan^{-1}\left(\frac{\sqrt{\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2}} + b + 20} \left(\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2} + \left(c - z\right)^{2}} - \sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2}}\right)}{\left(z - c\right)\sqrt{\sqrt{\left(a + 20\right)^{2} + \left(b + 20\right)^{2}} - b - 20} }\right)} \end{equation} provided $c \neq 20$ and $c \neq -20$. Otherwise, in each of $T_1$,$T_2$,$T_3$,$T_4$; the denominator of tangent term containing $z-c$ will equal $z-20$ or $z-(-20)$ creating singularity when applying limits $z=20$ or $z=-20$. So to avoid this situation we must eliminate two planes $(c=20$ and $c=-20)$ from the domain of $E_x(a,b,c)$ Upto here, I see two planes $(c=20$ and $c=-20)$ which are not in the domain of $E_x(a,b,c)$ Are there any other points, lines, curves, planes or surfaces which are not in the domain of $E_x(a,b,c)$ which I may be not seeing ?

Let $f = 4x^6 - 24x^5 - 2113x^4 + 8612x^3 + (-39114t - 12978)x^2 + (78228t + 8668)x + (58671t^2 + 18848602t - 2168)$ (I will explain its provenance if necessary, but there may be no need). Is there a value of the parameter $t \in \mathbb{Z}$ for which $f$ has a rational root? It is possible $t$ is very large, or even that there is no such $t$, e.g. if $f$ is always irreducible. It seems $f$ is irreducible for all $t$ from $0$ up to $10^7$, I didn't search any higher. It would be especially interesting if there is a very large $t$ which works and is computable.

Let $F$ be the free group with generators $x_1,\dots,x_n$. The basic commutators are elements of $F$ defined as follows: The basic commutators of weight $1$ are the generators $x_1,\dots,x_n$; the basic commutators of weight $w>1$ are the elements $[x,y]$ where $x$ and $y$ are basic commutators whose weights sum to $w$, $x>y$ and if $x=[u,v]$ for basic commutators $u$ and $v$ then $v\leq y$. Basic commutators are ordered so that $x>y$ if the weight of $x$ is greater than the weight of $y$ and for commutators of any fixed weight some total ordering is chosen. Denote by $F_k$ the $k$-th term of its lower central series. It is well known that $F_k/F_{k+1}$ is a free $\mathbb{Z}$-module with $\mathbb{Z}$-basis given by the basic commutators of weight $k$. Is it true also in the pro-$p$ case? More specifically: Let $F$ be the free pro-$p$ group on $x_1,\dots,x_n$ and denote by $F_k$ the $k$-th term of its lower central series. Is it true that $F_k/F_{k+1}$ is a free $\mathbb{Z}_p$-module with $\mathbb{Z}_p$-basis given by the basic commutators of weight $k$? References appreciated.

I calculated are of the sector by two methods, but they are giving different answers. Where is my mistake? Actually, I think it's when I am calculating arc(A,B), but it was true as shown in the second circle. Thank you in advance!