0002 Groups
Let $X$ be a nonmepty set and $\ast:X^2\to X$ an operation on $X$.
If
$e\in X$ is such that
$$ \forall x \in X, \quad x\ast e = e\ast x = x $$
then we say that $e$ is an
identity for $\ast$.
If $x\in X$ and
there exists an element $y\in X$ such that
$$ x\ast y=y\ast x=e $$
then we say that
$y$ is the
inverse of $x$.
Typically we denote the identity of an
operation by $e$ and the inverse of $x$ by $x^{-1}$.
If $G$ is a nonmepty
set with an operation $\ast:G^2\to G$ then we say that $G$ is a
group if
- $\ast$ is an associative operation.
- There is an
identity for $\ast$.
- Each $x\in G$ has an inverse element.
We abbreviate $x\ast y$ as $xy$ and we abbreviate $x y^{-1}$ as $x/y$.
For any $n\in\Bbb N$ we abbreviate
- $x^n = \overbrace{x\ast x\ast \cdots \ast x}^n$
- $x^0=e$
- $x^{-n} = (x^{-1})^n$
If $(G,\ast)$ is a group, as a shorthand we often write and say that $G$ is a group. The operation is left implicit by context.
If $\ast$ is a commutative operation, then we say that $G$ is a
commutative group.
In all that follows, assume that $(G,\ast)$ is a group and $x\in G$.
Theorem: Assoc. Op. Uniq. Id. & Inv.
The identity and inverse elements of an associative operation are unique.
Statement and proof.
Statement: Let $X$ be a nonempty set and $\ast: X^2\to X$ be an associative operation on $X$. Assume that $e\in X$ is an identity element for $\ast$. Let $a\in X$ have an inverse element $a^{-1}\in X$.
Then $e$ is the unique element with the identity property. Also $a^{-1}$ is the unique element with the property of the inverse of $a$.
Proof: Let $y\in X$ have the property of the identity. Then
$$ ey = e = y $$
Now let $b\in X$ be an inverse of $a$. Then
$$ a^{-1} = a^{-1}e = a^{-1}ab = eb = b $$
$\Box$
Theorem: Group Uniq. Id. & Inv.
In a group, the identity and inverses are unique.
Statement and proof.
Statement: Let $(G,\ast)$ be a group with identity $e$. Then $e$ is the unique identity element. Also, for each $x\in G$, the inverse $x^{-1}$ is unique.
Proof: Because the operation of any group is associative, therefore these elements are unique by the Assoc. Op. Uniq. Id. & Inv. theorem.
$\Box$
Theorem: Group Inv. Inv.
$(x^{-1})^{-1} = x$
Statement and proof.
Statement: $(x^{-1})^{-1} = x$.
Proof: $xx^{-1} = e$ by definition of the inverse.
Then $(x^{-1})^{-1} = x$ by the Group Uniq. Id. & Inv. theorem.
$\Box$
Theorem: Group Uniq. Sol.
Group equations have unique solutions.
Statement and proof.
Statement: Let $a,b\in G$ such that
$$ ax = b $$
Then this uniquely determines the element $x$.
Likewise $xa = b$ uniquely determines $x$.
Proof: From $ax=b$ we obtain
$$ x = a^{-1}b $$
Because inverses and products are uniquely determined, then this value is unique.
The case for $xa = b$ is similar.
$\Box$
Theorem: Group Prod. Inv.
The inverse of a product is the product of inverses, in reverse order.
Statement and proof.
Statement: Let $a,b\in G$. Then
$$ (ab)^{-1} = b^{-1}a^{-1} $$
Proof: By associativity of the operation,
$$ (ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = e $$
This shows, by the uniqueness of inverses, that $b^{-1}a^{-1}$ is the inverse of $ab$.
$$ b^{-1}a^{-1} = (ab)^{-1} $$
$\Box$
Theorem: Group Exp. Laws
Groups obey the familiar exponent laws.
Statement and proof.
Statement: Let $m,n\in\Bbb Z$. Then
$$ x^mx^n = x^{m+n} $$
and
$$ (x^m)^n = x^{mn} $$
Proof: Without loss of generality assume $m\le n$.
Case 1: $m$ or $n$ is 0.
Suppose, without loss of generality, that $m=0$. Then $x^m = e$ and therefore
$$ x^m x^n = x^n = x^{m+n} $$
Case 2: $0 < m$.
Suppose $0 < m$. Then by associativity,
$$
\begin{aligned}
x^{m}x^n &= \overbrace{x x \cdots x}^m \overbrace{ x x \cdots x}^n \\
&= \overbrace{x x \cdots x}^{m+n} \\
&= x^{m+n}
\end{aligned}$$
Case 3: $m < 0 < n$.
Suppose $m < 0 < n$. Then
$$\begin{aligned}
x^m x^n &= \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m} \overbrace{xx\cdots x}^n
\end{aligned}$$
If $-m \le n$ then $-m$ instances of $x^{-1}$ and $x$ will cancel. In that case, the above is
$$ \overbrace{xx\cdots x}^{n-(-m)} = x^{n+m} $$
Note that the above includes the case $-m=n$ in which case all instances of $x^{-1}$ and $x$ cancel, resulting in $e=x^{n+m}=x^0$.
Finally suppose $n < -m$, in which case $n$ instances of $x^{-1}$ and $x$ cancel. Then $x^mx^n$ is
$$\begin{aligned}
\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m-n} &= x^{m+n}
\end{aligned}$$
Case 4: $m \le n < 0$.
Suppose $m \le n < 0$. Then
$$\begin{alinged}
x^m x^n &= \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m} \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-n+(-m)} \\
&= x^{m+n}
\end{aligned}$$
Since the above exhausts all cases, we have now shown
$$ x^mx^n = x^{m+n}$$
Now to show $(x^m)^n$:
Case 1: $m$ or $n$ is 0.
If $m=0$ then $(x^m)^n = e^n = e$. If $n=0$ then immediately $(x^m)^n = e$.
Case 2: $0 < m$.
Suppose $0 < m$. Then
$$\begin{aligned}
(x^m)^n &= \overbrace{\overbrace{xx\cdots x}^m\overbrace{xx\cdots x}^m\cdots \overbrace{xx\cdots x}^m}^n \\
&= \overbrace{xx\cdots x}^{mn} \\
&= x^{mn}
\end{aligned}$$
Case 3: $m < 0 < n$.
Suppose $m < 0 < n$.
$$\begin{aligned}
(x^m)^n &= \overbrace{\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m}\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m}\cdots \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m}}^n \\
&= \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{n(-m)} \\
&= \overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-mn} \\
&= x^{mn}
\end{aligned}$$
Case 4: $m \le n < 0$.
Suppose $m \le n < 0$.
$$\begin{aligned}
(x^m)^n &= \overbrace{(\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m})^{-1}(\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m})^{-1}\cdots (\overbrace{x^{-1}x^{-1}\cdots x^{-1}}^{-m})^{-1}}^{-n}\\
&= \overbrace{xx\cdots x}^{(-m)(-n)} \\
&= x^{mn}
\end{aligned} $$
Therefore in all cases $(x^m)^n = x^{mn}$.
$\Box$