MAT102 Lecture 22B

• If we have $(G,\oplus )$ as a group, we say that $H\subseteq G$ is a subgroup if $(H,\oplus )$ is a group
• Properties:
  • Restricting $\oplus$ to a subset will not change associativity.
  • $H$ must have the same identity element as $G$
  • We need to check closure and inverse.
  • Closure
    • $\mathrm{\forall }a,b\in H,a\oplus b\in H$
    • for all $a$ and $b$, is the operator still in $H$
    • Example:
      • $(\mathbb{Z},+)$
      • ${\mathbb{Z}}^{+}$
        • These are closed
        • no identity
      • $2\mathbb{Z}=\{2n:n\in \mathbb{Z}\}$
        • If I take two evens, we get another even. Closed.
        • Identity $0$
        • and inverses
        • Subgroup
      • $2\mathbb{Z}+1\{2n+1:n\in \mathbb{Z}\}$
        • If I take two odds, we get an even, not closed.
        • Identity $0$
  • Identity
  • Inverses
    • $\mathrm{\forall }a\in H,a^{-1}\in H$
• One step subgroup test
  • If $(G,\oplus )$ is a group, and $H\subseteq G$
  • Then $H⊴G⟺\mathrm{\forall }a,b\in H,a{b}^{-1}\in H$
• If $(G,\oplus )$ is a group and $g\in G$ then the cyclic subgroup generated by $g$ is $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$
• $G={\mathbb{Z}}_{60}$
  • $g^n=\underset{h\text{ times}}{\underbrace{g\oplus g\oplus g\oplus \cdots \oplus g}}$
  • $⟨12⟩=\{0,12,24,36,48\}$
  • $⟨25⟩=\{0,25,50,15,40,5,30,55,20,45,10,35\}=⟨5⟩$
  • See that the $gcd(25,60)=5$ so $⟨25⟩=⟨5⟩$
  • #tk prove this
  • Is the union of these two cyclic sets a subgroup?
    • It wouldn't be closed
    • See that $12\oplus 25=37\notin G$
  • If we have ${\mathbb{Z}}_n$ and $⟨r⟩,⟨s⟩$
  • we take the intersection and get $\text{lcm}(r,s)$
• Prove that the intersection of two subgroups are still subgroups
  • If $(G,\oplus )$ is a group with $H,K⊴G$ then $H\cap K⊴G$
  • $⟹$
    • We have that $H,K$ are subgroups
    • Want to show that $H\cap K$ is a subgroup
    • If $H$ and $K$ are both, closed, associative and have inverses.
    • $H=\{h_0,h_1,\dots ,h_n\}$
    • $K=\{k_0,k_1,\dots ,k_n\}$
    • $H\cap K=\{h_i,k_i,\dots \}$ for some $i$
  • Solution:
    • Let $a,b\in H\cap K$
    • To apply the subgroup test we want to show that $a{b}^{-1}\in H\cap K$
    • Let's show that $a{b}^{-1}\in H$ and $a{b}^{-1}\in K$
    • Without loss of generality
    • Since $a$ and $b$ are in $H$
    • By one step subgroup test
    • Then $a{b}^{-1}\in H$
    • So we have $a{b}^{-1}\in K$
    • So it's done.
• Claim: Let $G$ be an abelian group with $H,K⊴G$ define a new set $HK=\{hk:h\in H,k\in K\}$
  • Show $HK⊴G$
  • We have that $H⊴G$ and $K⊴G$
  • To show that $HK⊴G$ we need that $a{b}^{-1}\in HK$
    • $a=h_1{k}_1$
    • $b=h_2{k}_2$
    • $(h_1{k}_1)(h_2{k}_2{)}^{-1}$
    • $h_1{k}_1{h}_2^{-1}{k}_2^{-1}$
    • $\underset{\in HK}{\underbrace{\underset{\in H}{\underbrace{h_1{h}_2^{-1}}}\underset{\in K}{\underbrace{k_1{k}_2^{-1}}}}}$
      11GroupTheoryI_Students.pdf
• $G={\mathbb{Z}}_2\times {\mathbb{Z}}_2$
  • $(a_1,b_1)\oplus (a_2,b_2)=(a_1+a_2,b_1+b_2)$
• 1
  • Show that $G$ is a group what is $\left|G\right|$?
  • ${\mathbb{Z}}_2\times {\mathbb{Z}}_2=\left\{\begin{array}{c}(0,0),(0,1)\\ (1,0),(1,1)\end{array}\right\}$
  • So $\left|G\right|=4$
  • To show $G$ is a group
  • Associativity
    • fix $a,b,c\in G$
    • $(a\oplus b)\oplus c=a\oplus (b\oplus c)$
    • $((a_0,a_1)\oplus (b_0,b_1))\oplus (c_0,c_1)$
    • $((a_0+b_0,a_1+b_1))\oplus (c_0,c_1)$
    • $(a_0+b_0+c_0,a_1+b_1+c_1)$
    • By definition of $\oplus$
    • $(a_0,a_1)\oplus (b_0+c_0,b_1+c_1)$
    • By definition of $\oplus$
    • $(a_0,a_1)\oplus ((b_0,b_1)\oplus (c_0,c_1))$
    • Proven
  • Identity
    • Show that for $a,e$
    • $a\oplus e=a$
    • Claim $e=(0,0)$
    • $(a_0,a_1)\oplus (0,0)=(a_0,a_1)$
    • By definition of $\oplus$
    • $(a_0+0,a_1+0)=(a_0,a_1)$
  • Inverse
    • Show that for $a,b$ where $b=a^{-1}$, that $a\oplus a^{-1}=e$
    • Claim: $a^{-1}=a$
    • $a\oplus a=e$
    • $(a_0,a_1)\oplus (a_0,a_1)=e$
    • $(2a_0,2a_1)=e$
    • Without loss of generality, if $a_0,a_1$ are $0$ we get $0$
    • If $a_0,a_1$ are $1$ we get $2$, see that $2(\mathrm{mod}2)=0$
    • So we get that $(2a_0,2a_1)=e$
• 2
  • Determine the order of every element in this group
  • $a^n=\underset{n\text{ times}}{\underbrace{a\oplus a\oplus \cdots \oplus a}}$
  • ${\mathbb{Z}}_2\times {\mathbb{Z}}_2=\left\{\begin{array}{c}(0,0),(0,1)\\ (1,0),(1,1)\end{array}\right\}$
  • $a=(0,0)$
    • $n=1$ is our order, it's already the identity
  • $b=(0,1)$
    • $n=2$ is our order
    • $(0,1)\oplus (0,1)=(0,2)$
    • $(0,2)\oplus (0,1)=e$
  • $c=(1,0)$
    • $n=2$ is our order
  • $d=(1,1)$
    • $n=2$ is our order
• 3
  • Find all subgroups of $G$
  • $H_1=\{(0,0)\}$
    • Subgroup test
    • $a{b}^{-1}\in H_1$
    • $(a_0,a_1)\oplus (b_0,b_1{)}^{-1}$
    • $(a_0+b_0^{-1},a_1+b_1^{-1})$
    • The inverse of $b$ based on this set, the only thing it could be is $0$
    • So we just have the identity element and it's in $H_1$ so it's a subgroup.
  • $H_2=\{(0,0),(0,1)\}$
    • $a{b}^{-1}\in H_2$
  • $H_3=\{(0,0),(1,0)\}$
  • $H_4=\{(0,0),(1,1)\}$
  • $H_5=\{(0,0),(0,1),(1,0)\}$
    • $a{b}^{-1}\in H_5$
    • $a\oplus b^{-1}$
    • $(a_0+b_0^{-1},a_1+b_1^{-1})$
    • I might get $(1,1)$ as an output element from this, which isn't in $H_5$
  • $H_6=\{(0,0),(0,1),(1,1)\}$
    • Same argument for these others
  • $H_7=\{(0,0),(1,0),(1,1)\}$
  • $H_8=\{(0,0),(1,0),(0,1),(1,1)\}$
  • $\mathcal{P}(G)\cup \{(0,0)\}$ to take care of our identity.

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  • Cyclic subgroup
    • Lagrange theorem
      • Cardinality of subgroups must divide the cardinality of the group
      • This is why we have the cardinality of these subgroups are $1,2,4$
    • We have order $2$ so the elements of order $2$
  • $\{(0,0)\}$
  • $\{(0,0),(1,0)\}$
  • $\{(0,0),(0,1)\}$
  • $\{(0,0),(0,1),(1,0),(1,1)\}$