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  • MAT102 Prep 11

Group Theory

• $G$ is a set, we have binary operators on it as: $+$ and $\cdot$, etc.
• Define a group as $(G,\cdot )$, the set and a binary operator.
• It's a group when it's: Associative, identity and inverse
  • Similar to fields
  • Associativity:
    • For every $a,b,c\in G$
    • $a\cdot (b\cdot c)=(a\cdot b)\cdot c$
    • $\cdot$ under $\mathbb{R}$ is a binary operator:
      • See that $a\cdot (b\cdot c)=(a\cdot b)\cdot c$
      • This is an axiom for $\mathbb{R}$
      • But see for something like mods it's not as well-defined
  • Identity:
    • Same as with fields, we need that there is an identity.
    • There exists $e\in G$ where $e\cdot a=a\cdot e=a$ for all $a\in G$
    • See that with mods, we must be mod prime $n$, in order to have a multiplicative inverse for all elements in the set.
      • ${\mathbb{Z}}_4$ has no multiplicative inverse for an element, meaning it violates this property.
  • Inverses:
    • For every $a\in G$ there is a $b\in G$ where $a\cdot b=b\cdot a=e$
    • See how the identity is linked to the inverse.
    • Notionally as $a^{-1}$
• Examples:
  • See that $\mathbb{R}$ are a group under $\cdot$
  • See that $\cancel{({\mathbb{Z}}^{+},+)}$
    • Let's find a counterexample:
    • Associativity:
      • $a+(b+c)=(a+b)+c$
      • This holds
    • Identity:
      • $e+a=a+e=a$
      • $e=0$, however $0\notin {\mathbb{Z}}^{+}$
      • This means that this doesn't have an identity
      • The next possible number could be $e=1$
      • But see that $1+a\ne a\ne a+1$
    • Inverse:
      • $a+b=b+a=e$
      • See that there is no inverse as $e$ is not well-defined.
      • So this is not a group.