A finite group of order $n$ has exponent $n$.

Definitions: The order $ |G|$ of a group $ G$ is the number of elements of $ G$ . The exponent of a group is an integer $ n$ such that $ x^n = e$ for all $ x\in G$ ($ e$ is the neutral element).

Particularly, the definition doesn’t say it’s the least integer with the property. So in this definition the 4-group $ V$ has exponent $ 4$ as well as exponent $ 2$ .

But I’m having a hard time proving the following statement:

Lagrange’s Theorem shows that a finite group $ G$ of order $ n$ has exponent $ n$ .

I know it must be very simple because it’s not even stated as a corollary of Lagrange’s.

If $(N,\|.\|)$ is a non trivial vector space with norm, then the diameter of the unit sphere is $\ge2$

It is easy to see that $ \delta(B_1(0))\le 2$ because $ $ \|x-y\|\le\|x\|+\|-y\|=\|x\|+|-1|\|y\|=\|x\|+\|y\|<1+1=2$ $ so $ \sup\{\|x-y\|:x,y\in B_1(0)\}\le 2$

But to show that $ \delta(B_1(0))\ge 2$ we chose $ a\in B_1(0)\backslash\{0\}$ and $ s\in[0,1)$ and observe that $ \|s\frac{a}{\|a\|}\|=s$ $ \implies s\frac{a}{\|a\|},-s\frac{a}{\|a\|}\in B_1(0)$ and $ \|s\frac{a}{\|a\|}-(-s\frac{a}{\|a\|})\|=2s$

But why does that imply that $ \delta(B_1(0))\ge2s$ ?

This is how I think I understand it: the diameter of $ B_1(0)$ is greater than that of $ B_s(0)$ for all $ s\in[0,1)$ because we it contains it. And the diameter of $ B_s(0)$ is $ 2s$ for each $ s$ , which is true because there exists a sequence $ (\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})$ of pairs of points, with $ 0\le\epsilon_n<s~\forall n$ and $ \lim\limits_{n\rightarrow\infty}\epsilon_n=s$ such that it is always contained in $ B_s(0)$ and $ \lim\limits_{n\rightarrow\infty}d(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})=2s$

I’m not sure if I got it right…

Simple way to generate (or characterize) real $N \times N$ matrices $K$ satisfying $(-1)^{|A|}\det(K-I_A) \ge 0,\;\forall A \subseteq [\![N]\!]$

Let $ N$ be a large positive integer.


What is a simple way to generate (or characterize) real $ N \times N$ matrices $ K$ such that $ $ (-1)^{|A|}\det(K-I_A) \ge 0,\;\forall A \subseteq [\![N]\!], $ $ where $ A$ is masked version of the $ N$ -by-$ N$ identity matrix defined by $ (I_A)_{i,j}=1$ if $ i=j \in A$ and $ 0$ otherwise.