All posts by Jean-Pierre Merx

Algebra

The image of an ideal may not be an ideal

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If \(\phi : A \to B\) is a ring homomorphism then the image of a subring \(S \subset A\) is a subring \(\phi(A) \subset B\). Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take \(A=\mathbb Z\) the ring of the integers and for \(B\) the ring of the polynomials with integer coefficients \(\mathbb Z[x]\). The inclusion \(\phi : \mathbb Z \to \mathbb Z[x]\) is a ring homorphism. The subset \(2 \mathbb Z \subset \mathbb Z\) of even integers is an ideal. However \(2 \mathbb Z\) is not an ideal of \(\mathbb Z[x]\) as for example \(2x \notin 2\mathbb Z\).

Analysis

A function whose Maclaurin series converges only at zero

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Let’s describe a real function \(f\) whose Maclaurin series converges only at zero. For \(n \ge 0\) we denote \(f_n(x)= e^{-n} \cos n^2x\) and \[
f(x) = \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty e^{-n} \cos n^2 x.\] For \(k \ge 0\), the \(k\)th-derivative of \(f_n\) is \[
f_n^{(k)}(x) = e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right)\] and \[
\left\vert f_n^{(k)}(x) \right\vert \le e^{-n} n^{2k}\] for all \(x \in \mathbb R\). Therefore \(\displaystyle \sum_{n=0}^\infty f_n^{(k)}(x)\) is normally convergent and \(f\) is an indefinitely differentiable function with \[
f^{(k)}(x) = \sum_{n=0}^\infty e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right).\] Its Maclaurin series has only terms of even degree and the absolute value of the term of degree \(2k\) is \[
\left(\sum_{n=0}^\infty e^{-n} n^{4k}\right)\frac{x^{2k}}{(2k)!} > e^{-2k} (2k)^{4k}\frac{x^{2k}}{(2k)!} > \left(\frac{2kx}{e}\right)^{2k}.\] The right hand side of this inequality is greater than \(1\) for \(k \ge \frac{e}{2x}\). This means that for any nonzero \(x\) the Maclaurin series for \(f\) diverges.

Algebra

A group that is not a semi-direct product

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Given a group \(G\) with identity element \(e\), a subgroup \(H\), and a normal subgroup \(N \trianglelefteq G\); then we say that \(G\) is the semi-direct product of \(N\) and \(H\) (written \(G=N \rtimes H\)) if \(G\) is the product of subgroups, \(G = NH\) where the subgroups have trivial intersection \(N \cap H= \{e\}\).

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group \(D_{2n}\) with \(2n\) elements is isomorphic to a semidirect product of the cyclic groups \(\mathbb Z_n\) and \(\mathbb Z_2\). \(D_{2n}\) is the group of isometries preserving a regular polygon \(X\) with \(n\) edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group \(\mathbb H_8\) is the group consisting of the symbols \(\pm 1, \pm i, \pm j, \pm k\) where\[
-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.\] One can prove that \(\mathbb H_8\) endowed with the product operation above is indeed a group having \(8\) elements where \(1\) is the identity element.

\(\mathbb H_8\) is not abelian as \(ij = k \neq -k = ji\).

Let’s prove that \(\mathbb H_8\) is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup \(N\) and a subgroup \(H\) such that \(G=N \rtimes H\).

  • If \(\vert N \vert = 4\) then \(H = \{1,h\}\) where \(h\) is an element of order \(2\) in \(\mathbb H_8\). Therefore \(h=-1\) which is the only element of order \(2\). But \(-1 \in N\) as \(-1\) is the square of all elements in \(\mathbb H_8 \setminus \{\pm 1\}\). We get the contradiction \(N \cap H \neq \{1\}\).
  • If \(\vert N \vert = 2\) then \(\vert H \vert = 4\) and \(H\) is also normal in \(G\). Noting \(N=\{1,n\}\) we have for \(h \in H\) \(h^{-1}nh=n\) and therefore \(nh=hn\). This proves that the product \(G=NH\) is direct. Also \(N\) is abelian as a cyclic group of order \(2\). \(H\) is also cyclic as all groups of order \(p^2\) with \(p\) prime are abelian. Finally \(G\) would be abelian, again a contradiction.

We can conclude that \(G\) is not a semi-direct product.

Analysis

Painter’s paradox

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Can you paint a surface with infinite area with a finite quantity of paint? For sure… let’s do it!

Consider the 3D surface given in cylindrical coordinates as \[
S(\rho,\varphi):\begin{cases}
x &= \rho \cos \varphi\\
y &= \rho \sin \varphi\\
z &= \frac{1}{\rho}\end{cases}\] for \((\rho,\varphi) \in [1,\infty) \times [0, 2 \pi)\). The surface is named Gabriel’s horn.

Volume of Gabriel’s horn

The volume of Gabriel’s horn is \[
V = \pi \int_1^\infty \left( \frac{1}{\rho^2} \right) \ d\rho = \pi\] which is finite.

Area of Gabriel’s horn

The area of Gabriel’s horn for \((\rho,\varphi) \in [1,a) \times [0, 2 \pi)\) with \(a > 1\) is: \[
A = 2 \pi \int_1^a \frac{1}{\rho} \sqrt{1+\left( -\frac{1}{\rho^2} \right)^2} \ d\rho \ge 2 \pi \int_1^a \frac{d \rho}{\rho} = 2 \pi \log a.\] As the right hand side of inequality above diverges to \(\infty\) as \(a \to \infty\), we can conclude that the area of Gabriel’s horn is infinite.

Conclusion

Gabriel’s horn could be filled with a finite quantity of paint… therefore painting a surface with infinite area. Unfortunately the thickness of the paint coat is converging to \(0\) as \(z\) goes to \(\infty\), leading to a paint which won’t be too visible!

Algebra

A normal subgroup that is not a characteristic

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Let’s \(G\) be a group. A characteristic subgroup is a subgroup \(H \subseteq G\) that is mapped to itself by every automorphism of \(G\).

An inner automorphism is an automorphism \(\varphi \in \mathrm{Aut}(G)\) defined by a formula \(\varphi : x \mapsto a^{-1}xa\) where \(a\) is an element of \(G\). An automorphism of a group which is not inner is called an outer automorphism. And a subgroup \(H \subseteq G\) that is mapped to itself by every inner automorphism of \(G\) is called a normal subgroup.

Obviously a characteristic subgroup is a normal subgroup. The converse is not true as we’ll see below.

Example of a direct product

Let \(K\) be a nontrivial group. Then consider the group \(G = K \times K\). The subgroups \(K_1=\{e\} \times K\) and \(K_2=K \times \{e\} \) are both normal in \(G\) as for \((e, k) \in K_1\) and \((a,b) \in G\) we have
\[(a,b)^{-1} (e,x) (a,b) = (a^{-1},b^{-1}) (e,x) (a,b) = (e,b^{-1}xb) \in K_1\] and \(b^{-1}K_1 b = K_1\). Similar relations hold for \(K_2\). As \(K\) is supposed to be nontrivial, we have \(K_1 \neq K_2\).

The exchange automorphism \(\psi : (x,y) \mapsto (y,x)\) exchanges the subgroup \(K_1\) and \(K_2\). Thus, neither \(K_1\) nor \(K_2\) is invariant under all the automorphisms, so neither is characteristic. Therefore, \(K_1\) and \(K_2\) are both normal subgroups of \(G\) that are not characteristic.

When \(K = \mathbb Z_2\) is the cyclic group of order two, \(G = \mathbb Z_2 \times \mathbb Z_2\) is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group.

Example on the additive group \(\mathbb Q\)

Consider the additive group \((\mathbb Q,+)\) of rational numbers. The map \(\varphi : x \mapsto x/2\) is an automorphism. As \((\mathbb Q,+)\) is abelian, all subgroups are normal. However, the subgroup \(\mathbb Z\) is not sent into itself by \(\varphi\) as \(\varphi(1) = 1/ 2 \notin \mathbb Z\). Hence \(\mathbb Z\) is not a characteristic subgroup.

Topology

A non complete normed vector space

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Consider a real normed vector space \(V\). \(V\) is called complete if every Cauchy sequence in \(V\) converges in \(V\). A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like \((\ell_p, \Vert \cdot \Vert_p)\) the space of real sequences endowed with the norm \(\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}\) for \(p \ge 1\), the space \((C(X), \Vert \cdot \Vert)\) of real continuous functions on a compact Hausdorff space \(X\) endowed with the norm \(\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert\) or the Lebesgue space \((L^1(\mathbb R), \Vert \cdot \Vert_1)\) of Lebesgue real integrable functions endowed with the norm \(\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx\).

Let’s give an example of a non complete normed vector space. Let \((P, \Vert \cdot \Vert_\infty)\) be the normed vector space of real polynomials endowed with the norm \(\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert\). Consider the sequence of polynomials \((p_n)\) defined by
\[p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.\] For \(m < n \) and \(x \in [0,1]\), we have \[\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}\] which proves that \((p_n)\) is a Cauchy sequence. Also for \(x \in [0,1]\) \[ \lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.\] As uniform converge implies pointwise convergence, if \((p_n)\) was convergent in \(P\), it would be towards \(p\). But \(p\) is not a polynomial function as none of its \(n\)th-derivative always vanishes. Hence \((p_n)\) is a Cauchy sequence that doesn't converge in \((P, \Vert \cdot \Vert_\infty)\), proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

Analysis

Uniform continuous function but not Lipschitz continuous

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Consider the function \[
\begin{array}{l|rcl}
f : & [0,1] & \longrightarrow & [0,1] \\
& x & \longmapsto & \sqrt{x} \end{array}\]

\(f\) is continuous on the compact interval \([0,1]\). Hence \(f\) is uniform continuous on that interval according to Heine-Cantor theorem. For a direct proof, one can verify that for \(\epsilon > 0\), one have \(\vert \sqrt{x} – \sqrt{y} \vert \le \epsilon\) for \(\vert x – y \vert \le \epsilon^2\).

However \(f\) is not Lipschitz continuous. If \(f\) was Lipschitz continuous for a Lipschitz constant \(K > 0\), we would have \(\vert \sqrt{x} – \sqrt{y} \vert \le K \vert x – y \vert\) for all \(x,y \in [0,1]\). But we get a contradiction taking \(x=0\) and \(y=\frac{1}{4 K^2}\) as \[
\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert\]

Algebra

A nonabelian \(p\)-group

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Consider a prime number \(p\) and a finite p-group \(G\), i.e. a group of order \(p^n\) with \(n \ge 1\).

If \(n=1\) the group \(G\) is cyclic hence abelian.

For \(n=2\), \(G\) is also abelian. This is a consequence of the fact that the center \(Z(G)\) of a \(p\)-group is non-trivial. Indeed if \(\vert Z(G) \vert =p^2\) then \(G=Z(G)\) is abelian. We can’t have \(\vert Z(G) \vert =p\). If that would be the case, the order of \(H=G / Z(G) \) would be equal to \(p\) and \(H\) would be cyclic, generated by an element \(h\). For any two elements \(g_1,g_2 \in G\), we would be able to write \(g_1=h^{n_1} z_1\) and \(g_2=h^{n_1} z_1\) with \(z_1,z_2 \in Z(G)\). Hence \[
g_1 g_2 = h^{n_1} z_1 h^{n_2} z_2=h^{n_1 + n_2} z_1 z_2= h^{n_2} z_2 h^{n_1} z_1=g_2 g_1,\] proving that \(g_1,g_2\) commutes in contradiction with \(\vert Z(G) \vert < \vert G \vert\). However, all \(p\)-groups are not abelian. For example the unitriangular matrix group \[
U(3,\mathbb Z_p) = \left\{
\begin{pmatrix}
1 & a & b\\
0 & 1 & c\\
0 & 0 & 1\end{pmatrix} \ | \ a,b ,c \in \mathbb Z_p \right\}\] is a \(p\)-group of order \(p^3\). Its center \(Z(U(3,\mathbb Z_p))\) is \[
Z(U(3,\mathbb Z_p)) = \left\{
\begin{pmatrix}
1 & 0 & b\\
0 & 1 & 0\\
0 & 0 & 1\end{pmatrix} \ | \ b \in \mathbb Z_p \right\},\] which is of order \(p\). Therefore \(U(3,\mathbb Z_p)\) is not abelian.

Analysis

Raabe-Duhamel’s test

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The Raabe-Duhamel’s test (also named Raabe’s test) is a test for the convergence of a series \[
\sum_{n=1}^\infty a_n \] where each term is a real or complex number. The Raabe-Duhamel’s test was developed by Swiss mathematician Joseph Ludwig Raabe.

It states that if:

\[\displaystyle \lim _{n\to \infty }\left\vert{\frac {a_{n}}{a_{n+1}}}\right\vert=1 \text{ and } \lim _{{n\to \infty }} n \left(\left\vert{\frac {a_{n}}{a_{{n+1}}}}\right\vert-1 \right)=R,\]
then the series will be absolutely convergent if \(R > 1\) and divergent if \(R < 1\). First one can notice that Raabe-Duhamel's test maybe conclusive in cases where ratio test isn't. For instance, consider a real \(\alpha\) and the series \(u_n=\frac{1}{n^\alpha}\). We have \[ \lim _{n\to \infty } \frac{u_{n+1}}{u_n} = \lim _{n\to \infty } \left(\frac{n}{n+1} \right)^\alpha = 1\] and therefore the ratio test is inconclusive. However \[ \frac{u_n}{u_{n+1}} = \left(\frac{n+1}{n} \right)^\alpha = 1 + \frac{\alpha}{n} + o \left(\frac{1}{n}\right)\] for \(n\) around \(\infty\) and \[ \lim _{{n\to \infty }} n \left(\frac {u_{n}}{u_{{n+1}}}-1 \right)=\alpha.\] Raabe-Duhamel's test allows to conclude that the series \(\sum u_n\) diverges for \(\alpha <1\) and converges for \(\alpha > 1\) as well known.

When \(R=1\) in the Raabe’s test, the series can be convergent or divergent. For example, the series above \(u_n=\frac{1}{n^\alpha}\) with \(\alpha=1\) is the harmonic series which is divergent.

On the other hand, the series \(v_n=\frac{1}{n \log^2 n}\) is convergent as can be proved using the integral test. Namely \[
0 \le \frac{1}{n \log^2 n} \le \int_{n-1}^n \frac{dt}{t \log^2 t} \text{ for } n \ge 3\] and \[
\int_2^\infty \frac{dt}{t \log^2 t} = \left[-\frac{1}{\log t} \right]_2^\infty = \frac{1}{\log 2}\] is convergent, while \[
\frac{v_n}{v_{n+1}} = 1 + \frac{1}{n} +\frac{2}{n \log n} + o \left(\frac{1}{n \log n}\right)\] for \(n\) around \(\infty\) and therefore \(R=1\) in the Raabe-Duhamel’s test.

Algebra

Subset of elements of finite order of a group

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Consider a group \(G\) and have a look at the question: is the subset \(S\) of elements of finite order a subgroup of \(G\)?

The answer is positive when any two elements of \(S\) commute. For the proof, consider \(x,y \in S\) of order \(m,n\) respectively. Then \[
\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e\] where \(e\) is the identity element. Hence \(xy\) is of finite order (less or equal to \(mn\)) and belong to \(S\).

Example of a non abelian group

In that cas, \(S\) might not be subgroup of \(G\). Let’s take for \(G\) the general linear group over \(\mathbb Q\) (the set of rational numbers) of \(2 \times 2\) invertible matrices named \(\text{GL}_2(\mathbb Q)\). The matrices \[
A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}\] are of order \(2\). They don’t commute as \[
AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.\] Finally, \(AB\) is of infinite order and therefore doesn’t belong to \(S\) proving that \(S\) is not a subgroup of \(G\).