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

Aug 25, 2016

The group of invertible $2 \times 2$ matrices with Real coefficients under matrix multiplication is such a group.

#### Explanation:

Let $G$ be the group of invertible $2 \times 2$ matrices with coefficients in $\mathbb{R}$.

Note that $G$ satisfies:

• Has an identity element $\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$
• Is closed under multiplication, since if $A , B \in G$ then $A B$ 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 $\mathbb{R}$ (see below) then the group of invertible matrices with coefficients in $F$ is a subgroup of $G$. There are infinitely many subfields of $\mathbb{R}$ 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 $\mathbb{R}$ will include all rational numbers $\mathbb{Q}$)

Aug 25, 2016

The group of non-zero quaternions under multiplication.

#### Explanation:

Hamilton's Quaternions $\boldsymbol{H}$ are an extension of the Complex numbers, dropping the requirement that multiplication be commutative, with three imaginary units: $i$, $j$ and $k$ which satisfy:

• ${i}^{2} = {j}^{2} = {k}^{2} = - 1$
• $i j = k$, $j k = i$, $k i = j$
• $j i = - k$, $k j = - i$, $i k = - j$

Multiplication by Real numbers commutes with any quaternion.

Any quaternion can be expressed in the form:

$a + b i + c j + \mathrm{dk}$

where $a , b , c , d \in \mathbb{R}$

So we can multiply:

$\left({a}_{1} + {b}_{1} i + {c}_{1} j + {d}_{1} k\right) \left({a}_{2} + {b}_{2} i + {c}_{2} j + {d}_{2} k\right)$

$= {a}_{1} \left({a}_{2} + {b}_{2} i + {c}_{2} j + {d}_{2} k\right) + {b}_{1} i \left({a}_{2} + {b}_{2} i + {c}_{2} j + {d}_{2} k\right) + {c}_{1} j \left({a}_{2} + {b}_{2} i + {c}_{2} j + {d}_{2} k\right) + {d}_{1} k \left({a}_{2} + {b}_{2} i + {c}_{2} j + {d}_{2} k\right)$

$= {a}_{1} {a}_{2} + \left({a}_{1} {b}_{2}\right) i + \left({a}_{1} {c}_{2}\right) j + \left({a}_{1} {d}_{2}\right) k + \left({b}_{1} {a}_{2}\right) i + \left({b}_{1} {b}_{2}\right) {i}^{2} + \left({b}_{1} {c}_{2}\right) i j + \left({b}_{1} {d}_{2}\right) i k + \left({c}_{1} {a}_{2}\right) j + \left({c}_{1} {b}_{2}\right) j i + \left({c}_{1} {c}_{2}\right) {j}^{2} + \left({c}_{1} {d}_{2}\right) j k + \left({d}_{1} {a}_{2}\right) k + \left({d}_{1} {b}_{2}\right) k i + \left({d}_{1} {c}_{2}\right) k j + \left({d}_{1} {d}_{2}\right) {k}^{2}$

$= {a}_{1} {a}_{2} + \left({a}_{1} {b}_{2}\right) i + \left({a}_{1} {c}_{2}\right) j + \left({a}_{1} {d}_{2}\right) k + \left({b}_{1} {a}_{2}\right) i - \left({b}_{1} {b}_{2}\right) + \left({b}_{1} {c}_{2}\right) k - \left({b}_{1} {d}_{2}\right) j + \left({c}_{1} {a}_{2}\right) j - \left({c}_{1} {b}_{2}\right) k - \left({c}_{1} {c}_{2}\right) + \left({c}_{1} {d}_{2}\right) i + \left({d}_{1} {a}_{2}\right) k + \left({d}_{1} {b}_{2}\right) j - \left({d}_{1} {c}_{2}\right) i - \left({d}_{1} {d}_{2}\right)$

$= \left({a}_{1} {a}_{2} - {b}_{1} {b}_{2} - {c}_{1} {c}_{2} - {d}_{1} {d}_{2}\right) + \left({a}_{1} {b}_{2} + {b}_{1} {a}_{2} + {c}_{1} {d}_{2} - {d}_{1} {c}_{2}\right) i + \left({a}_{1} {c}_{2} - {b}_{1} {d}_{2} + {c}_{1} {a}_{2} + {d}_{1} {b}_{2}\right) j + \left({a}_{1} {d}_{2} + {b}_{1} {c}_{2} - {c}_{1} {b}_{2} + {d}_{1} {a}_{2}\right) k$

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

Quaternion multiplication is associative.

The subset of elements of the form $a + b i$ is identical to the field of Complex numbers $\mathbb{C}$.

So is the subset of elements of the form $a + b j$ or that of the form $a + b k$.

But multiplication in $\boldsymbol{H}$ as a whole is non-commutative.

$\boldsymbol{H}$ 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 $\mathbb{R}$, then the subset of elements of $\boldsymbol{H}$ of the form $a + b i + c j + \mathrm{dk}$ with $a , b , c , d \in F$ is a sub-algebra of $\boldsymbol{H}$, closed under addition, multiplication, subtraction and division by non-zero elements.

There are infinitely many subfields of $\mathbb{R}$ and therefore infinitely many infinite non-abelian subgroups of the group of non-zero elements of $\boldsymbol{H}$.

$\textcolor{w h i t e}{}$

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

Aug 25, 2016

The set of bijections (one to one onto functions) from any infinite set to itself (e.g. $\mathbb{N} \to \mathbb{N}$), is a suitable group under function composition.

#### Explanation:

Let $A$ be any infinite set (e.g. $\mathbb{N}$ or $\mathbb{R}$) 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 $I : A \to A$ defined as:

$I \left(a\right) = a$ for all $a \in A$.

• $B$ is closed under function composition since if $f$ and $g$ are bijections then so is $f \circ g$

• Function composition is associative:

$\left(f \circ g\right) \circ h \left(a\right) = f \left(g \left(h \left(a\right)\right)\right) = f \circ \left(g \circ h\right) \left(a\right)$ 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.

• $B$ is infinite

• Function composition is non-abelian. For example, if ${a}_{1} , {a}_{2} , {a}_{3}$ are distinct elements of $A$ then define:

$f \left(a\right) = \left\{\begin{matrix}{a}_{2} & \text{ if " a = a_1 \\ a_1 & " if " a = a_2 \\ a & " otherwise}\end{matrix}\right.$

$g \left(a\right) = \left\{\begin{matrix}{a}_{2} & \text{ if " a = a_1 \\ a_3 & " if " a = a_2 \\ a_1 & " if " a = a_3 \\ a & " otherwise}\end{matrix}\right.$

Then we find:

$f \circ g \left({a}_{3}\right) = {a}_{2} \text{ }$ but $\text{ } g \circ f \left({a}_{3}\right) = {a}_{1}$

• 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 $A \text{\} C$ is infinite, it is an infinite subgroup. Since there are infinitely many finite subsets of $A$, there are infinitely many such subgroups.