MAT102 Tutorial W11B

• 1
  • ${\mathbb{R}}^2$
  • $\oplus$ as $(a,b)\oplus (c,d)=(a+c,b+d)$
  • Show this is a group
  • Associativity
    • Fix $(a,b),(c,d),(e,f)\in {\mathbb{R}}^2$
    • $((a,b)\oplus (c,d))\oplus (e,f)=(a,b)\oplus ((c,d)\oplus (e,f))$
    • $(a+c,b+d)\oplus (e,f)$
    • $(a+c+e,b+d+f)$
    • By definition of $\oplus$
    • $(a,b)\oplus (c+e,d+f)$
    • $(a,b)\oplus ((c,d)\oplus (e,f))$
    • So it's associative
  • Identity
    • $(a,b)\oplus (e,e)=(a,b)$
    • Claim $e=(0,0)\in {\mathbb{R}}^2$ is the identity
    • $(a,b)\oplus (0,0)=(a+0,b+0)=(a,b)$
    • $(0,0)\oplus (a,b)=(0+a,0+b)=(a,b)$
    • So $e$ is the identity
  • Inverse
    • Let $(a,b)\in {\mathbb{R}}^2$
    • Find what element turns $(a,b)\to e$
    • Claim the inverse for $(a,b)$ is $(-a,-b)$
• 2
  • $H_1=\{(x,y):x^2+y^2=1\}$ is not a subgroup of $({\mathbb{R}}^2,\oplus )$
    • $\oplus$ as $(a,b)\oplus (c,d)=(a+c,b+d)$
    • Show that it violates at least one axiom
    • Inverse:
      • $(a,b)$
      • Claim that $(-a,-b)$ is not the inverse
      • $(a,b)\oplus (-a,-b)=(0,0)$
      • But see that $0^2+0^2\ne 1$
    • Solution:
      • 1
        • See that $(1,0),(0,1)\in H_1$
        • See that $b^{-1}=(0,-1)$
        • $a\oplus b^{-1}=(1,0)\oplus (0,-1)=(1,-1)$
        • See that $1^2+(-1{)}^2\ne 1$
      • 2
        • If something is a subgroup, the identity element must be in it.
        • $H_1\subseteq {\mathbb{R}}^2$
        • #tk, the identity element in the group should be unique
        • $(0,0)\notin H_1$
  • Show that $H_2=\{(x,y):y=17x\}$ is a subgroup of $({\mathbb{R}}^2,\oplus )$
    • Show that it has all axioms
    • Associativity:
      • Fix $(a,b),(c,d),(e,f)\in {\mathbb{R}}^2$
    • $H_2=\{(x,y):y=17x\}$
    • $(a,b)=(a,17a)$
    • $(c,d)=(c,17c)$
    • Find the inverse
    • $(c,d{)}^{-1}=(-c,-17c)$
    • $(a,b)\oplus (c,d{)}^{-1}$
    • $=(a,17a)\oplus (-c,-17c)$
    • $=(a-c,17a-17c)$
    • $=(a-c,17(a-c))\in H_2$
• 3
  • Suppose $\left|G\right|<\mathrm{\infty }$
  • prove that $\left|g\right|<\mathrm{\infty }$ for all $g\in G$
  • Show that $g^a=g^b⟹a=b$
    • $\left|g\right|=\mathrm{\infty }$
    • So then $g^a=g^b⟹a=b$
    • To get the identity element we need $0$
    • $(g^a{)}^{-1}(g^a)=(g^a{)}^{-1}(g^b)$
    • $e=g^{b-a}$
  • Contradiction
    • Suppose some $\left|g\right|=\mathrm{\infty }$ and $\left|G\right|<\mathrm{\infty }$
    • Which means that there is a bijection with a function $f:g\to \mathbb{R}$
    • Because $g\in G$, we count elements which are uncountable, which is a contradiction.
  • Solution:
    • Let $\left|G\right|=n<\mathrm{\infty }$
    • By contradiction, assume there is some $g\in G$ such that $\left|g\right|=\mathrm{\infty }$
    • $g^0=e,g^1,\dots ,g^n$
    • We have $n+1$ elements
    • There has to be some repetition in this set
    • $|\{g^0,g^1,\dots ,g^n\}|=n+1$
    • So there is $0<a<b\le n$ where $g^a=g^b$
    • By the lemma $g^a=g^b$ means that $(g^a{)}^{-1}{g}^a=(g^a{)}^{-1}{g}^b⟹g^{b-a}=e$
    • Since $b>a$ then $b-a=k\in \mathbb{N}$
    • But $k\ne 0$, which is a contradiction, meaning $g$ has to be of a finite order.
    • If $\left|g\right|=\mathrm{\infty }$ then the only way we get the element is $g^0=e$.