What is an example of a nonabelian infinite group with infinitely many infinite nonabelian subgroups?
3 Answers
Answer:
The group of invertible
Explanation:
Let
Note that
 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
A subfield is a subset satisfying:
 Includes
#0# and#1# .  Closed under addition and subtraction.
 Closed under multiplication and under division by nonzero.
(Note that any subfield of
Answer:
The group of nonzero quaternions under multiplication.
Explanation:
Hamilton's Quaternions
#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
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_2b_1b_2c_1c_2d_1d_2)+(a_1b_2+b_1a_2+c_1d_2d_1c_2)i+(a_1c_2b_1d_2+c_1a_2+d_1b_2)j+(a_1d_2+b_1c_2c_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
Quaternion multiplication is associative.
The subset of elements of the form
So is the subset of elements of the form
But multiplication in
The set of nonzero elements form an infinite nonabelian group under multiplication.
If
There are infinitely many subfields of
For more information...
Here's an interesting talk on vectors and quaternions by one of my favourite maths lecturers...
Answer:
The set of bijections (one to one onto functions) from any infinite set to itself (e.g.
Explanation:
Let
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
In addition note that:

#B# is infinite 
Function composition is nonabelian. 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.