Question

What is an example of a non-abelian infinite group with infinitely many infinite non-abelian subgroups?

Answer 1

The group of invertible `2xx2` matrices with Real coefficients under matrix multiplication is such a group.

Explanation:

Let `G` be the group of invertible `2xx2` matrices with coefficients in `RR`.

Note that `G` satisfies:

• Has an identity element `((1,0),(0,1))`
• Is closed under multiplication, since if `A, B in G` then `AB` has Real coefficients and is invertible with inverse `B^(-1)A^(-1)`.
• Multiplication is associative since matrix multiplication is associative.

If `F` is a subfield of `RR` (see below) then the group of invertible matrices with coefficients in `F` is a subgroup of `G`. There are infinitely many subfields of `RR` and therefore are infinitely many subgroups of `G`, which are all non-abelian and infinite.

A subfield is a subset satisfying:

• Includes `0` and `1`.
• Closed under addition and subtraction.
• Closed under multiplication and under division by non-zero.

(Note that any subfield of `RR` will include all rational numbers `QQ`)

Answer 2

The group of non-zero quaternions under multiplication.

Explanation:

Hamilton's Quaternions `bb(H)` are an extension of the Complex numbers, dropping the requirement that multiplication be commutative, with three imaginary units: `i`, `j` and `k` which satisfy:

• `i^2=j^2=k^2=-1`
• `ij=k`, `jk=i`, `ki=j`
• `ji=-k`, `kj=-i`, `ik=-j`

Multiplication by Real numbers commutes with any quaternion.

Any quaternion can be expressed in the form:

`a+bi+cj+dk`

where `a, b, c, d in RR`

So we can multiply:

`(a_1+b_1i+c_1j+d_1k)(a_2+b_2i+c_2j+d_2k)`

`=a_1(a_2+b_2i+c_2j+d_2k)+b_1i(a_2+b_2i+c_2j+d_2k)+c_1j(a_2+b_2i+c_2j+d_2k)+d_1k(a_2+b_2i+c_2j+d_2k)`

`=a_1a_2+(a_1b_2)i+(a_1c_2)j+(a_1d_2)k+(b_1a_2)i+(b_1b_2)i^2+(b_1c_2)ij+(b_1d_2)ik+(c_1a_2)j+(c_1b_2)ji+(c_1c_2)j^2+(c_1d_2)jk+(d_1a_2)k+(d_1b_2)ki+(d_1c_2)kj+(d_1d_2)k^2`

`=a_1a_2+(a_1b_2)i+(a_1c_2)j+(a_1d_2)k+(b_1a_2)i-(b_1b_2)+(b_1c_2)k-(b_1d_2)j+(c_1a_2)j-(c_1b_2)k-(c_1c_2)+(c_1d_2)i+(d_1a_2)k+(d_1b_2)j-(d_1c_2)i-(d_1d_2)`

`=(a_1a_2-b_1b_2-c_1c_2-d_1d_2)+(a_1b_2+b_1a_2+c_1d_2-d_1c_2)i+(a_1c_2-b_1d_2+c_1a_2+d_1b_2)j+(a_1d_2+b_1c_2-c_1b_2+d_1a_2)k`

If this looks somewhat like matrix multiplication written out in full it is no coincidence. Quaternions have natural representations as `4xx4` matrices with Real coefficients or as `2xx2` matrices with Complex coefficients.

Quaternion multiplication is associative.

The subset of elements of the form `a+bi` is identical to the field of Complex numbers `CC`.

So is the subset of elements of the form `a+bj` or that of the form `a+bk`.

But multiplication in `bb(H)` as a whole is non-commutative.

`bb(H)` is called a skew field or a division algebra.

The set of non-zero elements form an infinite non-abelian group under multiplication.

If `F` is any subfield of `RR`, then the subset of elements of `bb(H)` of the form `a+bi+cj+dk` with `a, b, c, d in F` is a sub-algebra of `bb(H)`, closed under addition, multiplication, subtraction and division by non-zero elements.

There are infinitely many subfields of `RR` and therefore infinitely many infinite non-abelian subgroups of the group of non-zero elements of `bb(H)`.

`color(white)()`
For more information...

Here's an interesting talk on vectors and quaternions by one of my favourite maths lecturers...

Answer 3

The set of bijections (one to one onto functions) from any infinite set to itself (e.g. `NN -> NN`), is a suitable group under function composition.

Explanation:

Let `A` be any infinite set (e.g. `NN` or `RR`) and `B` the set of bijections from `A` onto `A`.

Consider function composition as a candidate for group multiplication:

• There is an identity, namely the identity function `I:A->A` defined as:

  `I(a) = a` for all `a in A`.

• `B` is closed under function composition since if `f` and `g` are bijections then so is `f@g`

• Function composition is associative:

  `(f@g)@h(a) = f(g(h(a))) = f@(g@h)(a)` for all `a`.

• Function composition has an inverse for every element of `B`, since every bijection has an inverse which is also a bijection.

So `B` is a group under function composition.

In addition note that:

• `B` is infinite

• Function composition is non-abelian. For example, if `a_1, a_2, a_3` are distinct elements of `A` then define:

`f(a) = { (a_2, " if " a = a_1), (a_1, " if " a = a_2), (a, " otherwise") :}`

`g(a) = { (a_2, " if " a = a_1), (a_3, " if " a = a_2), (a_1, " if " a = a_3), (a, " otherwise") :}`

Then we find:

`f@g(a_3) = a_2 " "` but `" " g@f(a_3) = a_1`

• If `C` is any finite subset of `A`, then the subset of `B` containing bijections that leave points in `C` fixed is closed under function composition and inversion, so is a subgroup, Since `A "\" C` is infinite, it is an infinite subgroup. Since there are infinitely many finite subsets of `A`, there are infinitely many such subgroups.