\[ \newcommand{\GL}{\text{GL}} \]
A non-empty set \(S\) with its combinative binary operation called SemiGroup. If there’s identical element in semigroup, it’s a Monoid. Monoid with all the elements reversible is a Group.
The subset \(M^{\times}\) which contains all the reversible elements of a monoid \(M\) construct a group, which is defined as the Unit Group of \(M\).
Formally, a group, with binary operation \(\cdot\), is restricted by four conditions:
Notice that by definition the identical \(e\) and reverse element is left-operated with element \(x\) traditionally, though we’ll prove that the opposite side is equivlent for group.
Furthermore, if the operation equiped is also commutative, the group is called additive or Abelian, and additive notation is proposed if the context is proper:
Particularly in the study of ring, the one operation that is commutative almos t always use the additive notation.
With the four restrictions listed above, a group is such a well-structed structure with nice properties:
Left Elimination Property of left operation. If \(x\cdot a = x\cdot b\) , then \(a = b\).
Proof. Left multiple \(x^{-1}\) for both sides. \[ \begin{align} x\cdot a &= x\cdot b \\ x^{-1}\cdot x \cdot a &= x^{-1}\cdot x\cdot b \\ (x^{-1}\cdot x) \cdot a &= (x^{-1}\cdot x)\cdot b \\ e\cdot a &= e\cdot b \\ a &= b \end{align} \]
As noticed, this is not a simple property, the proof uses all of the restrictions of a group to make the simple elimination possible, hence the elimination law is not held for the general monoid.
If \(x\cdot x = x\), then \(x = e\). This is the specical case of the elimination property, you can multiple \(x^{-1}\) to prove it.
This property gives an important rule to tell if an element is the identical: just simply square it, and check if the result is still itself. No more element involved.
Symmetric reverse element: If \(x^{-1}\) is the reversed element of \(x\), so \(x\) is the reverse of \(x^{-1}\), which indicates \(x^{-1}\cdot x = x\cdot x^{-1} = e\)
Proof. To prove that something is an element, we try to use property(2), that is, square it and check if it returns the input.
\[ (x\cdot x^{-1})\cdot(x\cdot x^{-1}) = x\cdot (x^{-1}\cdot x)\cdot x^{-1} = x\cdot x^{-1} \]
Hence we have \(x\cdot x^{-1} = e\)
By definition, if we consider \(x^{-1}\) as the “original element”, \(x\cdot x^{-1}\) indicates that \(x\) is exactly the reverse elementof $ x^{-1} $, hence $ (x{-1}){-1} = x$.
Since the reverse can be multiply either to the left side or right side, the elimination property now can be generalized to the right multiple case:
\[ a\cdot x = b\cdot x \Rightarrow a = b \]
Identical element commutativity. \(x\cdot e = e\cdot x = x\)
Proof.
\[ x\cdot e = x\cdot ( x^{-1} \cdot x) = (x\cdot x^{-1})\cdot x = e\cdot x = x \]
There’s only one reverse element of any \(x\in G\).
Proof. Assume there’s another reverse element \(x'\) for \(x\).
\[ \begin{align} x'\cdot x = e &= x^{-1}\cdot x \\ x'\cdot x\cdot x^{-1} &= x^{-1}\cdot x \cdot x^{-1} \\ x'\cdot (x\cdot x^{-1}) &= x^{-1}\cdot (x \cdot x^{-1}) \\ x'\cdot e &= x^{-1}\cdot e \\ x' &= x^{-1} \end{align} \\ \]
There’s only one identical element of any \(x\in G\).
Proof. Assume there’s another identical element notes as \(e'\) that statifies
\[ e'\cdot x = e\cdot x \]
Simply use the right-version elimination law, \(e' = e\).
$ (ab)^{-1} = b^{-1} a^{-1}$.
Proof.
\[ (a\cdot b)\cdot (b^{-1}\cdot a^{-1}) = a\cdot (b\cdot b^{-1})\cdot a^{-1} = a \cdot a^{-1} = e \]
Generally, for any \(n\) elements in the group:
\[ \left( \prod_{i=1}^{n} a_i \right)^{-1} = \prod_{i=n}^{1} a_{i}^{-1} \]
All permutations on a set, equiped with the permutation composition, construct a symmetric group.
The set \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}\) equiped with addition operation constructs abelian group.
To make the ring equiped with multiplication operation also constructs group we need to substract \(0\) from them: \(\mathbb{Q}^{\times}=\mathbb{Q}-\{0\}, \mathbb{R}^{\times}=\mathbb{R}-\{0\}, \mathbb{C}^{\times}=\mathbb{C}-\{0\}\). Notice that \(\mathbb{Z}^{\times}=\mathbb{Z} - \{0\}\) still cannot constructs group since the reverse of those \(n\neq\pm1\) does not hold.
It’s obvious that to the general \(m\times n\) dimensional vector/matrix equiped with the normal entry-wise addition operation above any group also constructs a group.
Define \(M_n(\mathbb{R})\) as the set of all \(n\times n\) real matrix, and \(\GL(n, \mathbb{R})\) is a subset of \(M_n(\mathbb{R})\) which contains the reversible parts.
As mentioned above, the \(M_n(\mathbb{R})\) with the matrix addition constructs a group. Besides, \(M_n(\mathbb{R})\) with matrix multiplication is a monoid, and \(\GL(n,\mathbb{R})\) with matrix multiplication is a group. \(\GL(n,\mathbb{R})\) is the unit group of \(M_n(\mathbb{R})\). Generally, we can define \(M_n(F)\) and \(\GL(n, F)\) for any domain \(F\). \(\GL(n, F)\) is named as General Linear Group upon domain \(F\).
We define the \(n\in\mathbb{Z}\) exponent of element \(x\in G\) recursively:
$$ x^n = { \[\begin{array}{ll} \begin{align} &1, &n = 0 \\ &x^{n-1}\cdot x, &n\ge 1 \\ &(x^{-1})^{-n}, &n\lt 0 \end{align} \end{array}\]. $$
Here’re some simple properties about the exponent:
\((a^n)^m = a^{nm}\)
\(a^n\cdot a^m = a^{n+m}\)
if \(a, b\) commutes, then \((ab)^n = a^n b^n\)
If there’s one \(k\ge1\) that makes \(a^k = 1\), the smallest \(k\) is defined as the order of \(a\). otherwise we say \(a\) has infinite order. It’s easy to prove that, any \(n\) satisifies that \(a^n = 1\) must be the multiple of order \(k\).
Proof. If it is not, let \(n = qk + r\) where \(r\neq0\), \(e = a^n = a^{qk+r} = a^{qk}a^r = a^r\), since \(0<r<k\), this infers that we’ve found a smaller number \(r\) which makes \(a^r = 1\), which is contradict with the defintion of \(k\).
If the subset \(H\subset G\) meets
\(1\in H\).
H is closed under the operation \(\cdot\) of group \(G\).
for any \(x\in H\), \(x^{-1}\in H\).
We define the subset \(H\) is a subgroup of \(G\). Here we don’t mention about the associvity since it is ensured by the group \(G\). Notice that the condition (1) is necessary to avoid \(H=\varnothing\).
The proper subgroup is those \(H\neq G\), and nontrivial group is those \(H\neq\{1\}\).
As a shortcut for proving the subgroup, we assert that \(H\) is a subgroup iff it’s nonempty and for any \(x, y\in H\), \(xy^{-1}\in H\).
Proof. (\(\Rightarrow\)) If \(H\) is a subgroup, it’s obvious that \(y^{-1}\in H\), hence \(xy^{-1}\in H\).
(\(\Leftarrow\)) We’re required to prove the three conditions of subgroup.
Let \(y = x\), \(xy^{-1} = xx^{-1} = 1\), so \(1\in H\).
Based on (1), let \(x = 1\), \(xy^{-1} = y^{-1}\in G\).
Based on (3), let \(y = y^{-1}\), \(xy^{-1} = x(y^{-1})^{-1} = xy\in G\).
This concludes that \(H\) is the subgroup of \(G\).
Although its tedious and hard to prove the subgroup in general group, while for finite group \(G\), the subset \(H\subset G\) is only required closed to be a subgroup, due to its finity , which makes there must be at least one element whose exponent will eventually “falls into itself”.
Proof. Since \(H\neq \varnothing\), assume that \(a\in H\), hence \(a^n\in H\) for all the \(n\ge0\) according to the closed conditionof \(H\).
In the sequence of exponent of \(a\), there must be at least two integer \(i<j\) which makes \(a^j = a^i\), since the element number of \(H\subset G\) is finite.
This results in \(a^{j-i} = a^{i-i} = a^0 = 1 \in H\), which ensures that the identical element is in \(H\).
Besides, since \(j>i\), \(j-i-1\ge0\), \(a^{-1} = a^{j-i-1}\in H\).
Hence the set \(H\) is the subgroup of \(G\).