MTH3003 Weekly Problems 2

Original Documents: Problem Sheet / Handwritten Solutions / Provided Solutions

Vibes: Weird, but easy - just feels too easy and a lot of my initial solutions didn’t feel very formal.

Used Techniques:

• Basic definitions/properties of groups.
• Permutation definition and the Product of Permutations.
• Definition of a cyclic group.

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2.1. Prove a Group Equality

Question

Let $G$ be a group with $g,h\in G$. Prove that, for all $n\in \mathbb{N}$, $(g^{-1}hg{)}^n=g^{-1}{h}^{n}g$.

$$(g^{-1}hg{)}^n=(g^{-1}hg)(g^{-1}hg)\cdots (g^{-1}hg)=g^{-1}h\underset{=e_G}{\underbrace{(g{g}^{-1})}}h\underset{=e_G}{\underbrace{(g{g}^{-1})}}h\cdots h\underset{=e_G}{\underbrace{(g{g}^{-1})}}hg=g^{-1}{h}^{n}g\boxed{}$$

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2.2. Prove Group Properties

Question

Let $G$ be a group and $H\le G$.

1. Prove that $|H|=1⟺H=⟨e_G⟩$.
2. Suppose that $|G|$ is finite. Prove that $|H|=|G|⟺H=G$.
3. Is $|H|=|G|⟺H=G$ true if $|G|$ is infinite? $(\Rightarrow )$ By the Quick Subgroup Test, $e_G\in H$. If $|H|=1$, the unique element must be $e_G$, so $H=\{e_G\}=⟨e_G⟩$. $(\Leftarrow )$ $H=⟨e_G⟩\Rightarrow H=\{e_G\}\Rightarrow |H|=1$.

1. By the Quick Subgroup Test, $e_G\in H$. If $|H|=1$, the unique element must be $e_G$, so $H=\{e_G\}=⟨e_G⟩$. Alternatively, $H=⟨e_G⟩\Rightarrow H=\{e_G\}\Rightarrow |H|=1$.
2. Write $G=H\cup S$ with $S\cap H=\varnothing$, so $|G|=|H|+|S|$. Hence, $|G|=|H|\Rightarrow |S|=0\Rightarrow S=\varnothing \Rightarrow G=H$, and obviously vice versa.
3. No - $(\mathbb{Z},+)⪇(\mathbb{Q},+)$ but $|\mathbb{Z}|=|\mathbb{Q}|$ (both countably infinite).

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2.3. Prove a Relation between Subgroups

Question

Let $G$ be a group with $H\le G$ and $K\le G$. Prove that $H\cap K\le G$.

Apply the Quick Subgroup Test to $H\cap K$:

• Identity: $e_G\in H$ and $e_G\in K$ (both pass Quick Subgroup Test) $\Rightarrow e_G\in H\cap K$. ✓
• Closure $h,k\in H\cap K\Rightarrow hk\in H$ and $hk\in K\Rightarrow hk\in H\cap K$. ✓
• Inverse $h\in H\cap K\Rightarrow h^{-1}\in H$ and $h^{-1}\in K\Rightarrow h^{-1}\in H\cap K$. ✓

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2.4. Prove a Property of Cyclic Groups

Question

Let $G$ be a finite group and $g\in G$. Prove that $|⟨g⟩|=o(g)$.

Let $n:=o(g)$.

$\{g^0,\dots ,g^{n-1}\}\subseteq ⟨g⟩$, so $n\le |⟨g⟩|$.

For any $g^m\in ⟨g⟩$, write $m=an+r$ with $0\le r<n$ (division algorithm):

$$g^m=g^{an+r}=(g^n{)}^a\cdot g^r=e_G^a\cdot g^r=g^r\in \{g^0,\dots ,g^{n-1}\}$$

So $|⟨g⟩|\le n$, hence $|⟨g⟩|=n=o(g)$. ​

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2.5. Find Elements of a Cyclic Group of Permutations

Question

Let $g\in S_{10}$ be the permutation g=(1\\\)(2\). Write down all the elements in…

1. $⟨g⟩$.
2. $⟨g^2⟩$

Compute powers of $g$ until a repeat (order is $\text{lcm}(4,2)=4$):

Power	Result
$g^0$	$e$
$g^1$	$(1357)(24)$
$g^2$	$(15)(37)$
$g^3$	$(1753)(24)$
$g^4$	$e$ → stop

1. $⟨g⟩=\{e,(1357)(24),(15)(37),(1753)(24)\}$

2. $(g^2{)}^1=(15)(37)$, $(g^2{)}^2=e$ → stop

$⟨g^2⟩=\{e,(15)(37)\}$

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2.6. Prove that a Group is Cyclic

Question

Show that for all $n\in \mathbb{N}$, the group ${\mathbb{Z}}_n$ is cyclic, with ${\mathbb{Z}}_n=⟨[1{]}_n⟩$.

Take any $[m{]}_n\in {\mathbb{Z}}_n$. Since $\oplus$ is the group operation:

$$([1{]}_n{)}^m=\underset{m\text{ times}}{\underbrace{[1{]}_n\oplus \cdots \oplus [1{]}_n}}=[1+\cdots +1{]}_n=[m{]}_n$$

Every element of ${\mathbb{Z}}_n$ lies in $⟨[1{]}_n⟩$, so ${\mathbb{Z}}_n\subseteq ⟨[1{]}_n⟩$. Since $⟨[1{]}_n⟩\le {\mathbb{Z}}_n$, we have ${\mathbb{Z}}_n=⟨[1{]}_n⟩$. ​