A simpler example for a group theory problem

[Boi]

The problem is from section 2.6 of Herstein's "Topics in algebra", in which the idea of normal subgroups and quotient groups is introduced.

The example I came up was this: the group [imath]G[/imath] is [imath] \{ (a, b) \vert a \in \mathbb{R}/ \{ 0 \}, b \in \mathbb{R} \} [/imath] ( with multiplication defined as [imath](a, b)(c, d) = (ac, ad+b)[/imath] ), subgroup [imath]H[/imath] is [imath] \{ (1, n) \vert n \in \mathbb{Z} \} [/imath] and element [imath]a[/imath] is [imath] (2, 0) [/imath]. To be frank, I wouldn't have thought of this example if I hadn't had some prior knowledge of group theory. That got me thinking that there must be a simpler example for this problem which I missed (after all, this is basically just the beginning of the book). Can anyone think of one?

 

[blamocur]

I couldn't come up with a simple alternative, but your example isn't as artificial as it looks at the first sight. If I am not mistaken,
[imath]G[/imath] is equivalent to the group of invertible 2x2 matrices of the form
[math]M = \begin{matrix} a & b \\ 0 & 1 \end{matrix}[/math]i.e., X-skew and X-scaling. It is easy to see that:

• [imath]H[/imath] is the group of all integer X-skews without scaling,
• [imath]a[/imath] is scaling by 2 along X
• [imath]aHa^{-1}[/imath] is the group of all X-skews by even integers.

 

[fresh_42]

We need an infinite, non-abelian group, since conjugation is a bijection and commutativity means [imath] gHg^{-1}=H. [/imath] Matrix groups are therefore a natural choice. Now, we are looking for [imath] gH\subsetneq Hg, [/imath] i.e., multiplication from left is a different shift than multiplication from right. Let's consider upper triangular matrices and a small subgroup:

[math] G=\left\{\left.\sum_{i\le j}g_{ij}E_{ij}\right| \; g_{11}\cdots g_{nn}\neq 0\right\}\quad , \quad H=I+hE_{1n}\,(h\in \mathbb{Z})[/math]
where [imath] E_{ij} [/imath] denote the matrices with zeros everywhere except a one at position [imath] (i,j). [/imath] Then



[math]\begin{array}{lll} gH&=\displaystyle{\sum_{i\le j}g_{ij}E_{ij}+\sum_{i\le j}g_{ij}E_{ij}E_{1n}\mathbb{Z}}=\displaystyle{\sum_{i\le j}g_{ij}E_{ij}+ g_{11}E_{1n}\mathbb{Z} }\\[16pt] Hg&=\displaystyle{\sum_{i\le j}g_{ij}E_{ij}+ \sum_{i\le j}g_{ij}E_{1n}E_{ij}\mathbb{Z}}=\displaystyle{\sum_{i\le j}g_{ij}E_{ij}+ g_{nn}E_{1n}\mathbb{Z}} \end{array}[/math]

and [imath] gH\subsetneq Hg [/imath] means [imath] g_{11}\mathbb{Z}\subsetneq g_{nn}\mathbb{Z}. [/imath] Now, we can set [imath] g_{11}=4 [/imath] and [imath] g_{nn}=2 [/imath] with [imath] n>2 [/imath], big enough not to open another loophole. Note that we only need a single element [imath] g\in G. [/imath]




(I hope I haven't overlooked a loophole.)

 

[blamocur]

We need an infinite, non-abelian group, since conjugation is a bijection and commutativity means gHg−1=H. gHg^{-1}=H. gHg−1=H.

Good point! And I was scratching my head trying to find an example with permutation groups

I wonder if you limit your statements to 2x2 matrix would the original example be the only one possible. Well, maybe more general case of [imath](m,0)[/imath] for [imath]m>1[/imath] instead of just [imath](2,0)[/imath].

 

[fresh_42]

Good point! And I was scratching my head trying to find an example with permutation groups

I wonder if you limit your statements to 2x2 matrix would the original example be the only one possible. Well, maybe more general case of [imath](m,0)[/imath] for [imath]m>1[/imath] instead of just [imath](2,0)[/imath].

I first hoped for S(3) (smallest non-abelian), too, or A(5) (smallest simple).