Groups: Definitions & Examples
Groups: Definitions & Examples
A group $(G,*)$ is a set with a binary operation satisfying closure, associativity, identity, and invertibility. Groups arise everywhere: symmetries of geometric objects, permutations, integer arithmetic, and matrices. Abstract algebra identifies the common algebraic structure across these disparate examples, yielding results that apply universally.
Group Axioms
A group $(G,*)$ satisfies: (i) Closure: $a,b\in G\Rightarrow a*b\in G$. (ii) Associativity: $(a*b)*c=a*(b*c)$. (iii) Identity: $\exists e\in G$ with $a*e=e*a=a$ for all $a$. (iv) Inverses: $\forall a\in G,\exists a^{-1}\in G$ with $a*a^{-1}=e$. A group is abelian (commutative) if $a*b=b*a$ for all $a,b$. The order of $G$ is $|G|$; the order of element $a$ is $\text{ord}(a)=\min\{n\geq 1:a^n=e\}$.
Lagrange's Theorem
If $H$ is a subgroup of a finite group $G$, then $|H|$ divides $|G|$, and $[G:H]=|G|/|H|$ (the index) equals the number of left cosets $aH=\{ah:h\in H\}$ of $H$ in $G$. Corollary: the order of any element $a\in G$ divides $|G|$, since $\langle a\rangle=\{a^0,a^1,\ldots,a^{\text{ord}(a)-1}\}$ is a subgroup. Corollary: every group of prime order $p$ is cyclic (isomorphic to $\mathbb{Z}_p$).
Example 1
Verify $(\mathbb{Z}_6,+)$ is a group and find all subgroups.
Solution: Identity: 0. Inverses: $-n\pmod{6}$: $1\leftrightarrow 5$, $2\leftrightarrow 4$, $3\leftrightarrow 3$, $0\leftrightarrow 0$. Subgroups by Lagrange: orders divide 6, so orders 1,2,3,6. $\{0\}$ (order 1); $\{0,3\}$ (order 2, generated by 3); $\{0,2,4\}$ (order 3, generated by 2); $\mathbb{Z}_6$ (order 6). These are all subgroups.
Example 2
Show $S_3$ (permutations of $\{1,2,3\}$) is a non-abelian group of order 6.
Solution: $|S_3|=3!=6$. Elements: $e,(12),(13),(23),(123),(132)$. Non-abelian: $(12)(13)=(132)$ but $(13)(12)=(123)\neq(132)$. Order of $(12)$: 2 (transpositions). Order of $(123)$: 3 (3-cycles). $S_3\cong D_3$ (dihedral group of triangle symmetries).
Practice
- Prove the identity element of a group is unique.
- Prove that $(ab)^{-1}=b^{-1}a^{-1}$ in any group.
- Find all groups of order 4 up to isomorphism.
- Show every subgroup of an abelian group is normal.
Show Answer Key
1. Suppose $e$ and $e'$ are both identities. Then $e=e\cdot e'=e'$ (using $e'$ as identity on the left, $e$ on the right). So the identity is unique.
2. $(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aea^{-1}=aa^{-1}=e$. Similarly $(b^{-1}a^{-1})(ab)=e$. By uniqueness of inverses, $(ab)^{-1}=b^{-1}a^{-1}$.
3. Up to isomorphism: $\mathbb{Z}_4$ (cyclic) and $\mathbb{Z}_2\times\mathbb{Z}_2$ (Klein four-group). These are the only two because every element has order dividing 4; if there's an element of order 4, the group is cyclic; otherwise all non-identity elements have order 2, giving the Klein group.
4. Let $H\le G$ (abelian) and $g\in G$. For any $h\in H$: $ghg^{-1}=hgg^{-1}=h\in H$ (using commutativity). So $gHg^{-1}=H$, hence $H\trianglelefteq G$.