# The image of an ideal may not be an ideal

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 $$2x \notin 2\mathbb Z$$.

# A function whose Maclaurin series converges only at zero

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.

# A group that is not a semi-direct product

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.

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!

# A normal subgroup that is not a characteristic

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.

# A non complete normed vector space

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.

# Uniform continuous function but not Lipschitz continuous

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 $\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert$

# A nonabelian $$p$$-group

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.

# Raabe-Duhamel’s test

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.

# Subset of elements of finite order of a group

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 $$G$$.