Tag Archives: algebra

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