# A group G isomorph to the product group G x G

Let’s provide an example of a nontrivial group $$G$$ such that $$G \cong G \times G$$. For a finite group $$G$$ of order $$\vert G \vert =n > 1$$, the order of $$G \times G$$ is equal to $$n^2$$. Hence we have to look at infinite groups in order to get the example we’re seeking for.

We take for $$G$$ the infinite direct product $G = \prod_{n \in \mathbb N} \mathbb Z_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \dots,$ where $$\mathbb Z_2$$ is endowed with the addition. Now let’s consider the map $\begin{array}{l|rcl} \phi : & G & \longrightarrow & G \times G \\ & (g_1,g_2,g_3, \dots) & \longmapsto & ((g_1,g_3, \dots ),(g_2, g_4, \dots)) \end{array}$

From the definition of the addition in $$G$$ it follows that $$\phi$$ is a group homomorphism. $$\phi$$ is onto as for any element $$\overline{g}=((g_1, g_2, g_3, \dots),(g_1^\prime, g_2^\prime, g_3^\prime, \dots))$$ in $$G \times G$$, $$g = (g_1, g_1^\prime, g_2, g_2^\prime, \dots)$$ is an inverse image of $$\overline{g}$$ under $$\phi$$. Also the identity element $$e=(\overline{0},\overline{0}, \dots)$$ of $$G$$ is the only element of the kernel of $$G$$. Hence $$\phi$$ is also one-to-one. Finally $$\phi$$ is a group isomorphism between $$G$$ and $$G \times G$$.

# A Commutative Ring with Infinitely Many Units

In a ring $$R$$ a unit is any element $$u$$ that has a multiplicative inverse $$v$$, i.e. an element $$v$$ such that $uv=vu=1,$ where $$1$$ is the multiplicative identity.

The only units of the commutative ring $$\mathbb Z$$ are $$-1$$ and $$1$$. For a field $$\mathbb F$$ the units of the ring $$\mathrm M_n(\mathbb F)$$ of the square matrices of dimension $$n \times n$$ is the general linear group $$\mathrm{GL}_n(\mathbb F)$$ of the invertible matrices. The group $$\mathrm{GL}_n(\mathbb F)$$ is infinite if $$\mathbb F$$ is infinite, but the ring $$\mathrm M_n(\mathbb F)$$ is not commutative for $$n \ge 2$$.

The commutative ring $$\mathbb Z[\sqrt{2}] = \{a + b\sqrt{2} \ ; \ (a,b) \in \mathbb Z^2\}$$ is not a field. However it has infinitely many units.

### $$a + b\sqrt{2}$$ is a unit if and only if $$a^2-2b^2 = \pm 1$$

For $$u = a + b\sqrt{2} \in \mathbb Z[\sqrt{2}]$$ we denote $$\mathrm N(u) = a^2- 2b^2 \in \mathbb Z$$. For any $$u,v \in \mathbb Z[\sqrt{2}]$$ we have $$\mathrm N(uv) = \mathrm N(u) \mathrm N(v)$$. Therefore for a unit $$u \in \mathbb Z[\sqrt{2}]$$ with $$v$$ as multiplicative inverse, we have $$\mathrm N(u) \mathrm N(v) = 1$$ and $$\mathrm N(u) =a^2-2b^2 \in \{-1,1\}$$.

### The elements $$(1+\sqrt{2})^n$$ for $$n \in \mathbb N$$ are unit elements

The proof is simple as for $$n \in \mathbb N$$ $(1+\sqrt{2})^n (-1 + \sqrt{2})^n = \left((1+\sqrt{2})(-1 + \sqrt{2})\right)^n=1$

One can prove (by induction on $$b$$) that the elements $$(1+\sqrt{2})^n$$ are the only units $$u \in \mathbb Z[\sqrt{2}]$$ for $$u \gt 1$$.

# A normal extension of a normal extension may not be normal

An algebraic field extension $$K \subset L$$ is said to be normal if every irreducible polynomial, either has no root in $$L$$ or splits into linear factors in $$L$$.

One can prove that if $$L$$ is a normal extension of $$K$$ and if $$E$$ is an intermediate extension (i.e., $$K \subset E \subset L$$), then $$L$$ is a normal extension of $$E$$.

However a normal extension of a normal extension may not be normal and the extensions $$\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension $$k \subset K$$ , i.e. an extension of degree two is normal. Suppose that $$P$$ is an irreducible polynomial of $$k[x]$$ with a root $$a \in K$$. If $$a \in k$$ then the degree of $$P$$ is equal to $$1$$ and we’re done. Otherwise $$(1, a)$$ is a basis of $$K$$ over $$k$$ and there exist $$\lambda, \mu \in k$$ such that $$a^2 = \lambda a +\mu$$. As $$a \notin k$$, $$Q(x)= x^2 – \lambda x -\mu$$ is the minimal polynomial of $$a$$ over $$k$$. As $$P$$ is supposed to be irreducible, we get $$Q = P$$. And we can conclude as $Q(x) = (x-a)(x- \lambda +a).$

The entensions $$\mathbb Q \subset \mathbb Q(\sqrt{2})$$ and $$\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ are quadratic, hence normal according to previous lemma and $$\sqrt[4]{2}$$ is a root of the polynomial $$P(x)= x^4-2$$ of $$\mathbb Q[x]$$. According to Eisenstein’s criterion $$P$$ is irreducible over $$\mathbb Q$$. However $$\mathbb Q(\sqrt[4]{2}) \subset \mathbb R$$ while the roots of $$P$$ are $$\pm \sqrt[4]{2}, \pm i \sqrt[4]{2}$$ and therefore not all real. We can conclude that $$\mathbb Q \subset \mathbb Q(\sqrt[4]{2})$$ is not normal.

# 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 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.

# 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 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.

# 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 $$G$$.

# Field not algebraic over an intersection but algebraic over each initial field

Let’s describe an example of a field $$K$$ which is of degree $$2$$ over two distinct subfields $$M$$ and $$N$$ respectively, but not algebraic over $$M \cap N$$.

Let $$K=F(x)$$ be the rational function field over a field $$F$$ of characteristic $$0$$, $$M=F(x^2)$$ and $$N=F(x^2+x)$$. I claim that those fields provide the example we’re looking for.

### $$K$$ is of degree $$2$$ over $$M$$ and $$N$$

The polynomial $$\mu_M(t)=t^2-x^2$$ belongs to $$M[t]$$ and $$x \in K$$ is a root of $$\mu_M$$. Also, $$\mu_M$$ is irreducible over $$M=F(x^2)$$. If that wasn’t the case, $$\mu_M$$ would have a root in $$F(x^2)$$ and there would exist two polynomials $$p,q \in F[t]$$ such that $p^2(x^2) = x^2 q^2(x^2)$ which cannot be, as can be seen considering the degrees of the polynomials of left and right hand sides. This proves that $$[K:M]=2$$. Considering the polynomial $$\mu_N(t)=t^2-t-(x^2+x)$$, one can prove that we also have $$[K:N]=2$$.

### We have $$M \cap N=F$$

The mapping $$\sigma_M : x \mapsto -x$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$M$$ are fixed under $$\sigma_M$$. Similarly, the mapping $$\sigma_N : x \mapsto -x-1$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$N$$ are fixed under $$\sigma_N$$. Also $(\sigma_N\circ\sigma_M)(x)=\sigma_N(\sigma_M(x))=\sigma_N(-x)=-(-x-1)=x+1.$ An element $$z=p(x)/q(x) \in M \cap N$$ where $$p(x),q(x)$$ are coprime polynomials of $$K=F(x)$$ is fixed under $$\sigma_M \circ \sigma_N$$. Therefore following equality holds $\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)},$ which is equivalent to $p(x)q(x+1)=p(x+1)q(x).$ By induction, we get for $$n \in \mathbb Z$$ $p(x)q(x+n)=p(x+n)q(x).$ Assume $$p(x)$$ is not a constant polynomial. Then it has a root $$\alpha$$ in some finite extension $$E$$ of $$F$$. As $$p(x),q(x)$$ are coprime polynomials, $$q(\alpha) \neq 0$$. Consequently $$p(\alpha+n)=0$$ for all $$n \in \mathbb Z$$ and the elements $$\alpha +n$$ are all distinct as the characteristic of $$F$$ is supposed to be non zero. This implies that $$p(x)$$ is the zero polynomial, in contradiction with our assumption. Therefore $$p(x)$$ is a constant polynomial and $$q(x)$$ also according to a similar proof. Hence $$z$$ is constant as was supposed to be proven.

Finally, $$K=F(x)$$ is not algebraic over $$F=M \cap N$$ as $$(1,x, x^2, \dots, x^n, \dots)$$ is independent over the field $$F$$ which concludes our claims on $$K, M$$ and $$N$$.

# Complex matrix without a square root

Consider for $$n \ge 2$$ the linear space $$\mathcal M_n(\mathbb C)$$ of complex matrices of dimension $$n \times n$$. Is a matrix $$T \in \mathcal M_n(\mathbb C)$$ always having a square root $$S \in \mathcal M_n(\mathbb C)$$, i.e. a matrix such that $$S^2=T$$? is the question we deal with.

First, one can note that if $$T$$ is similar to $$V$$ with $$T = P^{-1} V P$$ and $$V$$ has a square root $$U$$ then $$T$$ also has a square root as $$V=U^2$$ implies $$T=\left(P^{-1} U P\right)^2$$.

### Diagonalizable matrices

Suppose that $$T$$ is similar to a diagonal matrix $D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 & d_2 & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n \end{bmatrix}$ Any complex number has two square roots, except $$0$$ which has only one. Therefore, each $$d_i$$ has at least one square root $$d_i^\prime$$ and the matrix $D^\prime=\begin{bmatrix} d_1^\prime & 0 & \dots & 0 \\ 0 & d_2^\prime & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n^\prime \end{bmatrix}$ is a square root of $$D$$. Continue reading Complex matrix without a square root