Tag Archives: groups

The set of all commutators in a group need not be a subgroup

I here provide a simple example of a group whose set of commutators is not a subgroup. The counterexample is due to P.J. Cassidy (1979).

Description of the group \(G\)

Let \(k[x,y]\) denote the ring of all polynomials in two variables over a field \(k\), and let \(k[x]\) and \(k[y]\) denote the subrings of all polynomials in \(x\) and in \(y\) respectively. \(G\) is the set of all upper unitriangular matrices of the form
\[A=\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)\] where \(f(x) \in k[x]\), \(g(y) \in k[y]\), and \(h(x,y) \in k[x,y]\). The matrix \(A\) will also be denoted \((f,g,h)\).
Let’s verify that \(G\) is a group. The products of two elements \((f,g,h)\) and \((f^\prime,g^\prime,h^\prime)\) is
\[\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)
\left(\begin{array}{ccc}
1 & f^\prime(x) & h^\prime(x,y) \\
0 & 1 & g^\prime(y) \\
0 & 0 & 1 \end{array}\right)\]
\[=\left(\begin{array}{ccc}
1 & f(x)+f^\prime(x) & h(x,y)+h^\prime(x,y)+f(x)g^\prime(y) \\
0 & 1 & g(y)+g^\prime(y) \\
0 & 0 & 1 \end{array}\right)\] which is an element of \(G\). We also have:
\[\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)^{-1} =
\left(\begin{array}{ccc}
1 & -f(x) & f(x)g(y) – h(x,y) \\
0 & 1 & -g(y) \\
0 & 0 & 1 \end{array}\right)\] proving that the inverse of an element of \(G\) is also an element of \(G\). Continue reading The set of all commutators in a group need not be a subgroup

Generating the symmetric group with a transposition and a maximal length cycle

Can the symmetric group \(\mathcal{S}_n\) be generated by any transposition and any \(n\)-cycle for \(n \ge 2\) integer? is the question we deal with.

We first recall some terminology:

Symmetric group
The symmetric group \(\mathcal{S}_n\) on a finite set of \(n\) symbols is the group whose elements are all the permutations of the \(n\) symbols. We’ll denote by \(\{1,\dots,n\}\) those \(n\) symbols.
Cycle
A cycle of length \(k\) (with \(k \ge 2\)) is a cyclic permutation \(\sigma\) for which there exists an element \(i \in \{1,\dots,n\}\) such that \(i, \sigma(i), \sigma^2(i), \dots, \sigma^k(i)=i\) are the only elements moved by \(\sigma\). We’ll denote the cycle \(\sigma\) by \((s_0 \ s_1 \dots \ s_{k-1})\) where \(s_0=i, s_1=\sigma(i),\dots,s_{k-1}=\sigma^{k-1}(i)\).
Transposition
A transposition is a cycle of length \(2\). We denote below the transposition of elements \(a \neq b\) by \((a \ b)\) or \(\tau_{a,b}\).

Continue reading Generating the symmetric group with a transposition and a maximal length cycle

Two subgroups whose product is not a subgroup

In this article, we consider a group \(G\) and two subgroups \(H\) and \(K\). Let \(HK=\{hk \text{ | } h \in H, k \in K\}\).

\(HK\) is a subgroup of \(G\) if and only if \(HK=KH\). For the proof we first notice that if \(HK\) is a subgroup of \(G\) then it’s closed under inverses so \(HK = (HK)^{-1} = K^{-1}H^{-1} = KH\). Conversely if \(HK = KH\) then take \(hk\), \(h^\prime k^\prime \in HK\). Then \((hk)(h^\prime k^\prime)^{-1} = hk(k^\prime)^{-1}(h^\prime)^{-1}\). Since \(HK = KH\) we can rewrite \(k(k^\prime)^{-1}(h^\prime)^{-1}\) as \(h^{\prime \prime}k^{\prime \prime}\) for some new \(h^{\prime \prime} \in H\), \(k^{\prime \prime} \in K\). So \((hk)(h^\prime k^\prime)^{-1}=hh^{\prime \prime}k^{\prime \prime}\) which is in \(HK\). This verifies that \(HK\) is a subgroup. Continue reading Two subgroups whose product is not a subgroup

A finitely generated soluble group isomorphic to a proper quotient group

Let \(\mathbb{Q}_2\) be the ring of rational numbers of the form \(m2^n\) with \(m, n \in \mathbb{Z}\) and \(N = U(3, \mathbb{Q}_2)\) the group of unitriangular matrices of dimension \(3\) over \(\mathbb{Q}_2\). Let \(t\) be the diagonal matrix with diagonal entries: \(1, 2, 1\) and put \(H = \langle t, N \rangle\). We will prove that \(H\) is finitely generated and that one of its quotient group \(G\) is isomorphic to a proper quotient group of \(G\). Continue reading A finitely generated soluble group isomorphic to a proper quotient group

A (not finitely generated) group isomorphic to a proper quotient group

The basic question that we raise here is the following one: given a group \(G\) and a proper subgroup \(H\) (i.e. \(H \notin \{\{1\},G\}\), can \(G/H\) be isomorphic to \(G\)? A group \(G\) is said to be hopfian (after Heinz Hopf) if it is not isomorphic with a proper quotient group.

All finite groups are hopfian as \(|G/H| = |G| \div |H|\). Also, all simple groups are hopfian as a simple group doesn’t have proper subgroups.

So we need to turn ourselves to infinite groups to uncover non hopfian groups. Continue reading A (not finitely generated) group isomorphic to a proper quotient group

Converse of Lagrange’s theorem does not hold

Lagrange’s theorem, states that for any finite group \(G\), the order (number of elements) of every subgroup \(H\) of \(G\) divides the order of \(G\) (denoted by \(\vert G \vert\)).

Lagrange’s theorem raises the converse question as to whether every divisor \(d\) of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when \(d\) is a prime.

However, this does not hold in general: given a finite group \(G\) and a divisor \(d\) of \(\vert G \vert\), there does not necessarily exist a subgroup of \(G\) with order \(d\). The alternating group \(G = A_4\), which has \(12\) elements has no subgroup of order \(6\). We prove it below. Continue reading Converse of Lagrange’s theorem does not hold