# 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 $$\tau_{a,b}$$.

# 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