Finite Groups and Subgroups
Table of Contents
1. Order of Group & Finite Groups
We first define the order of a group, as well as the order of an element of a group.
The order of the group is defined as the number of elements in a group, denoted by \( |G| \) or \( o(G) \).
The order of an element \( a \) in a group \( G \) is the least positive integer \( n \) such that \( a^{n}=e \), where \( e\in G \) is the identity of the group. We denote it as \( o(a)=n \), the order of an element is denoted by \( |a| \) or \( o(a) \).
If the group is defined under addition, then the order of an element is defined as the least positive integer to make \( na=0 \). Moverover, the order of an element might be infinite.
Some useful results:
- \( o(a) = o(a^{-1}), \quad \forall a\in G \)
- \( o(x^{-1}ax) = o(a), \quad \forall a\in G \)
- \( o(ab) = o(ba),\quad\forall a,b\in G \)
- \( a^{m} = e, m\in \mathbb{Z}^{+} \implies o(a)\mid m \)
- \( o(a^{k}) = \frac{o(a)}{\gcd(o(a), k)} \)
- \( o(ab) = o(a)o(b) \) if \( \gcd(o(a), o(b))=1 \) and \( ab=ba \)
Proof of (1).
If in a group \( G \), we have \( a^{5}=e, aba^{-1}=b^{2} \) for all \( a,b\in G \), then show that \( o(b)=31 \)
Since the group is not abelian, we cannot use the property that \( (ab)^{n}=a^{n}b^{n} \).
Prove that if a group \( G \) has order \(4\), then it must be abelian.
2. Subgroups
- Subgroup
- A non-empty subset \( H \) of a group \( G \) is said to be a subgroup of \( G \) if \( H \) itself is a group under the operations of \( G \). We write \( H\le G \)
- Proper Subgroup
- If \( H \) is a subgroup of \( G \) but not equal to \( G \) itself, then \( H \) is called a proper subgroup of \( G \), written as \( H\lt G \)
Trivial subgroups are \( \{e\} \) and \( G \). All other subgroups are called non-trivial subgroups.
When checking subgroups, we mainly focus on closure, identity and inverse existence, because associativity is already guaranteed.
2.1. Subgroup Tests
The following tests can simplify the procedure of checking whether a subset is a subgroup.
Subgroup Test
Two Step Subgroup Test. A non-empty subset \( H \) of a group \( G \) is a subgroup of \( G \) if and only if
- \( a,b\in H \implies ab\in H \)
- \( a\in H \implies a^{-1}\in H \)
One Step Subgroup Test. A non-empty subset \( H \) of a group \( G \) is a subgroup of \( G \) if and only if \( a,b\in H\implies ab^{-1}\in H \)
Finite Subgroup Test. A non-empty finite subset \( H \) of a group \( G \) is a subgroup of \( G \) if and only if \( H \) is closed under the operation of \( G \), i.e., \( ab\in H, \forall a,b\in H \). Note that, the group \( G \) does not need to be finite. Only the subset.
Prove: A group of order \( 6 \) cannot have a subgroup of order \( 4 \).
Let \( o(G) = 6 \) and \( H\lt G, o(H) = 4 \). Suppose \( H = \{ h_{1}, h_{2}, h_{3}, h_{4} \} \) and \( x\in G, x\not\in H \).
Then \( xH = \{ xh_{1}, xh_{2}, xh_{3}, xh_{4} \} \). Let’s prove that there’s no common element between \( H, xH \).
If there exists, suppose \( h_{i} = xh_{j} \), where \( h_{i},h_{j} \in H \). Then we know \( x = h_{i}h_{j}^{-1} \). However, \( h_{i},h_{j}\in H \implies h_{i}h_{j}^{-1}\in H \), which is a contradiction.
Thus, there’s no common element between \( H, xH \). So \( o(G) \ge 8 \), but \( o(G) = 6 \), which is contradiction. Thus QED.
2.2. Special Class of Subgroups
2.2.1. Center of Group
- Center of Group
The center of a group \( G \) is the set of all those elements in \( G \) that commute with every element of \( G \), denoted by \( Z(G) \). i.e.
\[ Z(G) = \{ a\in G : ax=xa, \forall x\in G \} \]
The center of a group \( G \) is a subgroup of \( G \).
2.2.2. Centralizer of Element
- Centralizer
Let \( a \) be a fixed element of a group \( G \). The centralizer of \( a \) in \( G \) is the set of all those elements in \( G \) that commute with \( a \), denoted by \( C_{G}(a) \)
\[ C_{G}(a) = \{ g\in G : ga=ag \} \]
For each \( a \) in a group \( G \), the centralizer of \( a \) is a subgroup of \( G \).
2.2.3. Centralizer of Subgroup
- Centralizer of Subgroup
Let \(G\) be a group and let \(H\) be a subgroup of \(G\). Then the centralizer of subgroup \(H\) is defined as the set of all those elements of \(G\) which commute with every element of \(H\).
\[ C_{G}(H) = \{ x \in G: xh=hx,\forall h \in H \} \]
For any subgroup \(H\) of a group \(G\), \(C_{G}(H)\) is a subgroup of \(G\).
2.2.4. Conjugate of Subgroup
- Conjugate of Subgroup
Let \( G \) be a group. Let \( H \) be a subgroup of \( G \) and \( x\in G \), then conjugate of the subgroup \( H \) is defined as
\[ xHx^{-1} = \{ xhx^{-1} : h \in H \} \]
Let \( H \) be a subgroup of group \( G \), \( x\in G \). Then, \( xHx^{-1} \) is a subgroup of \( G \).
2.2.5. Normalizer of Subgroup
- Normalizer of Subgroup
Let \(G\) be a group and let \(H\) be a subgroup of \(G\). Then, normalizer of subgroup \(H\) in \(G\) is defined as
\[ N_{G}(H) = \{ x\in G: xHx^{-1}=H \} \]
For any subgroup \(H\) of a group \(G\), \(N_{G}(H)\) is a subgroup of \(G\).
2.2.6. Cyclic Subgroup
Let \(G\) be a group and let \(a\in G\), Let \(\lang a \rang = \{ a^{n}:n\in \mathbb{Z} \}\). Then \(\lang a \rang\) is called cyclic subgroup of the group \(G\) generated by \(a\). \(a\) is called the generator.
2.3. Intersection and Union of Subgroups
The intersection of a finite class of subgroups is also a subgroup.
If \(A\) and \(B\) are subgroups of \(G\), then \(A \cup B\) is a subgroup of \(G\) if and only if \(A \sube B\) or \(B \sube A\).
2.4. Product of Two Subgroups
Let \(H,K\) be 2 subgroups of a group \(G\), we define
\[ HK = \{ hk : h \in H \land k \in K \} \]
and
\[ KH = \{ kh : k \in K \land h \in H \} \]
Let \(H,K\) be 2 subgroups of a group \(G\), then
\[ x \in HK \iff x^{-1} \in KH \]
Let \(H,K\) be subgroups of \(G\). \(HK\) is a subgroup of \(G\) if and only if \(HK=KH\).