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

3 Answers
Aug 25, 2016

Answer:

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#)

Aug 25, 2016

Answer:

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...

Aug 25, 2016

Answer:

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.