Tag Archives: groups

Algebra

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

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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 →

Algebra

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

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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 →

Algebra

Two subgroups whose product is not a subgroup

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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 →

Algebra

A finitely generated soluble group isomorphic to a proper quotient group

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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 →

Algebra

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

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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 →

Algebra

Converse of Lagrange’s theorem does not hold

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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 →