Show that a group in which all the $m$th powers commute with each other and all the $n$th powers commute with each other ,$m$ and $n$ relatively prime, is abelian.  How to do this ? we know $1=mx+yn$ for $x,y \in \mathbb{Z}$  $ab=a^{mx}a^{ny}b^{yn}b^{mx}=a^{mx}b^{yn}a^{ny}b^{mx}$  but now how to switch $b^{yn}$ and $a^{mx}$ ??
Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$  Then how prove $G$ is an abelian group?  Some context: Some of these commutation relations often imply that $G$ is abelian, for example , or . This looks like another example of this phenomenon, but the same techniques do not apply.
Using processing.runandload in QGIS 3 Show that $S$ is compact. Let $A \in M_n(\bf C)$ and let   $S=\{ A \in M_n(\bf C) : A=A ^*$ and  $\rho (A) \leq 1 \}$, $\rho(A)$ denotes spectral radius of $A$.     Show that $S$ is compact subset of $M_n(\bf C) \cong \bf C^{n^2} $ with usual topology.    $S$ is subset of $\bf O_n(C)$ which is compact so I think its enough to show that $S$ is closed.Please help. Equation of a rectangle I need to graph a rectangle on the Cartesian coordinate system. Is there an equation for a rectangle? I can't find it anywhere. During Winter Bash, users can earn hats. But sometimes it seems that a hat was awarded even though its requirements weren't (yet) met, or vice versa.  What could be the reason for this? Limit of $\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} }$ Calling PHP functions within HEREDOC strings In PHP, the HEREDOC string declarations are really useful for outputting a block of html.  You can have it parse in variables just by prefixing them with $, but for more complicated syntax (like $var[2][3]), you have to put your expression inside {} braces.  In PHP 5, it is possible to actually make function calls within {} braces inside a HEREDOC string, but you have to go through a bit of work.  The function name itself has to be stored in a variable, and you have to call it like it is a dynamically-named function.  For example:  $fn = 'testfunction'; function testfunction() { return 'ok'; } $string = &lt;&lt;&lt; heredoc plain text and now a function: {$fn()} heredoc;   As you can see, this is a bit more messy than just:  $string = &lt;&lt;&lt; heredoc plain text and now a function: {testfunction()} heredoc;   There are other ways besides the first code example, such as breaking out of the HEREDOC to call the function, or reversing the issue and doing something like:  ?&gt; &lt;!-- directly output html and only breaking into php for the function --&gt; plain text and now a function: &lt;?PHP print testfunction(); ?&gt;   The latter has the disadvantage that the output is directly put into the output stream (unless I'm using output buffering), which might not be what I want.  So, the essence of my question is: is there a more elegant way to approach this?  Edit based on responses: It certainly does seem like some kind of template engine would make my life much easier, but it would require me basically invert my usual PHP style.  Not that that's a bad thing, but it explains my inertia..  I'm up for figuring out ways to make life easier though, so I'm looking into templates now. Is there a fast way (without looping over all edges/faces/verts) using ops.mesh or bmesh to get the lists of indices associated with mesh islands as shown below? These islands are part of one mesh but are disjoint from each other.    I know how to select the boundaries by running the operator Select boundary Loop or calling bpy.ops.mesh.region_to_loop() but can't find any API calls to return the vert/edge/face indices associated with the individual islands. Runtime's exec() method is not redirecting the output Idea behind proof that $\frac{21n+4}{14n+3}$ is irreducible for all $n$ I have a proof of $\frac{21n+4}{14n+3}$ being an irreducible fraction for any positive integer $n$. The proof starts with the following: Indeed, from the equality: $$2(21n+4)-3(14n+3)=-1$$ and so on..But how does this equality follow from the fraction? Factor analysis and regression XP Activation and Dual Booting Roots of a functionwith condition $\int_0^\pi f(x) \sin x dx = \int_0^\pi f(x) \cos x dx =0.$ Let $f:[0,\pi] \rightarrow \Bbb R$ be a continuous function which satisfies $\int_0^\pi f(x) \sin x dx = \int_0^\pi f(x) \cos x dx =0.$ Show that $f$ has at least two roots. How can I instantiate a generalized hash function? I've come across a bunch of "strange" hash function notations, such as the following ones and now I don't know how to choose / instantiate them.  Can you please explain me what this notation means and how I can actually instantiate such a hash function?  The notation in question: $$H_1:\mathcal M \rightarrow \mathbb F_p$$ $$H_2:\mathcal M \rightarrow \mathbb Z_n^*$$ $$H_3:\mathcal M \rightarrow \{0,1\}^l$$ $$H_4:\mathcal M_1\times ...\times \mathcal M_k\rightarrow \mathcal S$$  where $\mathcal M\in\{\mathbb F_p,\mathbb Z_n^*,\{0,1\}^*,\mathbb G,...\}$ where $\mathbb G$ is a generic (but there instantiated) group and $\mathcal S\in\{\mathbb F_p,\mathbb Z_n^*,\{0,1\}^l\}$ Determine group G as cyclic If $G$ has no proper subgroup, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.     I know that since $G $is a group with no proper subgroups, $g \in G$ is not just the identity. I don't know where to go from there. Let $\vec{k}$, $\vec{v}$, and $\vec{u}$ be vectors, where $\vec{u}$ is unknown and $\vec{k}$ and $\vec{v}$ are known vectors. Given:     $\vec{u}\cdot\vec{k}=c$  $\vec{u} \times \vec{k}= \vec{v}$   From this relations, how can i determine the vector $u$? I tried to construct orthogonal coordinate system from $(\vec{k}, \vec{v},\vec{k}\times \vec{v})$ but i couldn't proceed from there. Any idea? I can follow the proofs in which the asymptotic normal-distribution of a maximum likelihood estimator $\tilde{\theta}_n$ is derived.   however, does this already imply that the maximum likelihood estimates are asymptotically unbiased, i.e. do we have $$E(\tilde{\theta}_n) \to \theta \text{ as } n \to \infty?$$  Since I am aware that in general it is not true that convergence in distribution implies convergence in moments, an explanation would be nice.  The result is often stated e.g.  as "...it means that the bias of the maximum likelihood estimator is equal to zero up to the order $n^{-1/2}$")  1.) Are  some kind of regularity conditions from the mle theory used to establish this result?  2.) Or is a $\sqrt{n}$-convergence of an estimator (to a normal distribution) in general already enough to establish convergence of its moments?  Note: the Wikipedia article mentions Cox, David R.; Snell, E. Joyce (1968).  , as a source where the order of the bias is derived (formula (12) or (20)).   However in this paper I can't follow the arguments 100%, since their Taylor approximation of $L'(\widehat{\beta})$ is lacking the remainder term. What is the argument used here to discard it completely? At what pressure will be particles in a medium be unable to form a sound wave when disturbed? How can this pressure be described mathematically?  My guess is that this would correspond to the point at which the restoring force due to pressure is unable to create a transverse wave and the disturbed particles travel infinitely far away before the hypothetical wave reaches it's amplitude. But I have no idea how you would even begin to start finding a quantitative value for this point. In my experience, input type="text" onchange event usually occurs only after you leave (blur) the control.  Is there a way to force browser to trigger onchange every time textfield content changes? If not, what is the most elegant way to track this “manually”?  Using onkey* events is not reliable, since you can right-click the field and choose Paste, and this will change the field without any keyboard input.  Is setTimeout the only way?.. Ugly :-) How to write (La)TeX (with parentheses) [or any other TeX-related logo] Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.