Group homomorphism with an infinite kernel

Let f:\mathbb{C}^* \rightarrow \mathbb{R}^* be a homomorphism. Prove that kernel of f is infinite.


If \mathcal{G} < \mathbb{C}^\ast is a finite subgroup then all z \in \mathcal{G} must have norm 1, i.e. z \in \mathbb{S}^1. ( Otherwise otherwise z, z^2, z^3, \dotsc is an infinite sequence of distinct elements in \mathcal{G}. )

Suppose , that \mathcal{G} is finite and is the kernel of f:\mathbb{C}^\ast \rightarrow \mathbb{R}^\ast. Then let z \in \mathcal{G} and

    \[f(2z) = f(2)f(z) = 2\cdot 0 = 0\]

So,  2z \in \ker f = G. But this is a contradiction since |2z| = 2 \neq 1.

Hence,  no finite subgroup \mathal{G} of \mathbb{C}^\ast can be a kernel.

Note: The homomorphisms that can easily be described are for the form f(z)=|z|^\alpha.

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Irreducible function on Q[z]

Let f(z) \in \mathbb{Q}[z] be irreducible with degree n>1. If f has a root on the unit circle then n is even and

    \[z^n f\left( \frac{1}{z} \right) = f(z)\]


Let \alpha be a root of f with |\alpha|=1. Since f has real coefficients \bar{\alpha}= \frac{1}{\alpha} is also a root of f. The product z^n f\left( \frac{1}{z} \right) is a polynomial in \mathbb{Q}[z] of degree n ( its leading coefficient is f(0) ) with root \alpha. By the irreducibility of f we have

(1)   \begin{equation*} z^n f\left( \frac{1}{z} \right) =  c f(z) \end{equation*}

for some non zero rational number c. Setting z=1 we have that f(1)=c f(1). Since f(1) \neq 0  , by our hypotheses , c=1 hence z^n f\left( \frac{1}{z} \right) = f(z) . Setting z=-1 we get that f(-1)=(-1)^n f(-1) and because f(-1) \neq 0 we deduce that n is even.

Note: The above tells us that f(z) can be expressed in terms of z + \frac{1}{z}.

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Inequality with roots

Let a, b, c be positive real numbers. Prove that



We apply the AM – GM inequality, thus:

    \begin{align*} \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} &=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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On permutation

Let x_1, x_2, \dots, x_k be vectors of m – dimensional Euclidean space such that x_1+x_2+\cdots+x_k=0. Prove that there exists a permutation \pi of the integers \{1, 2, \dots,k\} such that

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|\leq\left ( \sum_{i=1}^{k} \left \| x_i \right \|^2 \right )^{1/2}\]

for each n=1,2, \dots, k.


We define \pi inductively. Set \pi(1)=1.Assume \pi is defined for i=1, 2, \dots, n and also

(1)   \begin{equation*} \left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \end{equation*}

Note that (1) is true for n=1. We choose \pi(n+1) in a way that (1) is fulfilled for n+1 instead of n. Set y=\sum \limits_{i=1}^{n} x_{\pi(i)} and A=\{1,2, \dots, k\} \setminus \{\pi(i) : i=1,2, \dots, n\}. Assume that (y, x_r)>0 for all r \in A.  Then \left ( y , \sum \limits_{r \in A} x_r \right )>0 and in view of y + \sum \limits_{r \in A} x_r =0 ones gets -(y, y)>0 which is impossible. Hence , there is r\in A such that

(2)   \begin{equation*} (y, x_r)\leq 0\end{equation*}

Put \pi(n+1) = r . Then using (2) and (1) we have

    \begin{align*} \left \| \sum_{i=1}^{n+1} x_{\pi(i)} \right \|^2 &= \left \| y + x_r \right \|^2 \\ &=\left \| y \right \|^2 + 2\left ( y, x_r \right ) + \left \| x_r \right \|^2 \\ &\leq \left \| y \right \|^2 + \left \| x_r \right \|^2 \\ &\leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 + \left \| x_r \right \|^2\\ &= \sum_{i=1}^{n+1} \left \| x_{\pi(i)} \right \|^2 \end{align*}

which verifies (1) for n+1. Thus we define \pi for every n=1, 2, \dots, k. Finally from (1) we get

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{k} \left \| x_i \right \|^2\]

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On linear operators

Let \alpha \in \mathbb{R} \setminus \{0\} and suppose that F ,G are linear operators from \mathbb{R}^n into \mathbb{R}^n satisfying

(1)   \begin{equation*}F\circ G - G \circ F =\alpha F \end{equation*}

  1. Show that for all k \in \mathbb{N} one has

        \[F^k \circ G - G \circ F ^k= \alpha k F^k\]

  2. Show that there exists k \geq 1 such that F^k =0.


  1. Using the assumptions we have

    \begin{aligned} F^k \circ G - G \circ F^k &= \sum_{i=1}^{k} \bigg (F^{k-i+1} \circ G \circ F^{i-1} - F^{k-i} \circ G \circ F^i \bigg ) \\ &= \sum_{i=1}^{k} \bigg( F^{k-i} \circ \left ( F \circ G- G \circ F \right ) \circ F^{i-1} \bigg)\\ &= \sum_{i=1}^{k} F^{k-i} \circ \alpha F \circ F^{i-1} \\ &= \alpha k F^k \end{aligned}

  2. Consider the linear operator \mathcal{L}(F) = F\circ G - G \circ F acting over all n \times n matrices F. It may have at most n^2 different eigenvalues. Assuming that F^k \neq 0 for every k we get that \mathcal{L} has infinitely many different eigenvalues \alpha k in view of (i). This is a contradiction.

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