MTH3003 Weekly Problems 4

Original Documents: Problem Sheet / [[mth3003 weekly problem sheet 4 handwritten solutions.pdf|My Handwritten Solutions]] / Provided Solutions

Vibes: Pretty mechanical, meh.

Used Techniques:

• Quick Subgroup Test (identity, closure, inverses).
• Lagrange’s Theorem (order divisors; corollaries).
• Order Switching Lemma ($h^{-1}nh\in N$ for $n\in N$).
• Normality check: $g^{-1}Hg\le H$ for all $g$.

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4.1. Intersection of Subgroup and Normal Subgroup

Question

Let $G$ be a group with $H\le G$ and $N⊴G$. Prove that $H\cap N$ is a normal subgroup of $H$.

Hint: Show that for all $h\in H$ and all $k\in H\cap N$ we have $h^{-1}kh\in H\cap N$.

First, $H\cap N\le H$ by intersection (Quick Subgroup Test inherited).

For normality in $H$: fix $h\in H$, $k\in H\cap N$; then $h^{-1}kh\in H$ (as $h,k\in H$), and $h^{-1}kh\in N$ ($N⊴G$). Thus, $h^{-1}kh\in H\cap N$.

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4.2. Nonexistence of a Subgroup of Order 9 in $S_5$

Question

Prove that the symmetric group $S_5$ has no subgroup of order $9$.

Hint: Use Lagrange’s Theorem.

$|S_5|=5!=120=2^3\cdot 3\cdot 5$. By Lagrange’s Theorem, any subgroup order divides 120, but $9=3^2\nmid 120$. Thus, no subgroup of order 9.

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4.3. Equality of Products $HN$ and $NH$

Question

Let $G$ be a group with $H\le G$ and $N⊴G$. Prove that $HN=\{hk:h\in H,k\in N\}$ is equal to $NH=\{kh:k\in N,h\in H\}$.

Hint: Use the Order Switching Lemma to show $HN\subseteq NH$ and $NH\subseteq HN$ separately. Do not assume $G$ is Abelian.

$HN\subseteq NH$: fix $hk\in HN$; by Order Switching ($N⊴G$), $\exists k^{\prime }\in N$ s.t. $hk=k^{\prime }h$, so $hk\in NH$.

$NH\subseteq HN$: fix $kh\in NH$; $\exists k^{\prime \prime }\in N$ s.t. $kh=h{k}^{\prime \prime }$, so $kh\in HN$. Thus, $HN=NH$.

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4.4. Groups of Prime Order Are Cyclic

Question

Let $p$ be a prime and let $G$ be a group with $|G|=p$.

1. Prove that the only subgroups of $G$ are $⟨e_G⟩$ and $G$ itself.
2. Hence prove that $G$ is cyclic.

Hint: For part (1), use Lagrange’s Theorem. For part (2), use (1) to analyse what any cyclic subgroup $⟨g⟩\le G$ must look like.

1. By Lagrange’s Theorem, $|H|\mid p$ ($p$ prime) so $|H|=1$ ($⟨e_G⟩$) or $p$ ($H=G$).
2. Take nontrivial $g\in G$: $⟨g⟩\le G$, $⟨g⟩\ne ⟨e_G⟩$, so $⟨g⟩=G$ by (1); thus cyclic.

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4.5. The Subgroup Generated by $H$ and $N$

Question

Let $G$ be a group with $H\le G$ and $N⊴G$. Prove that $HN\le G$.

Hint: Use the Order Switching Lemma to verify the conditions of the Quick Subgroup Test for $HN$. Do not assume that $G$ is abelian.

Identity: $e_G=e_H{e}_N\in HN$.

Closure: $(h_1{k}_1)(h_2{k}_2)=h_1(k_1{h}_2)k_2$; by Order Switching, $\exists k_1^{\prime }\in N$ s.t. $k_1{h}_2=h_2{k}_1^{\prime }$, so $=h_1{h}_2{k}_1^{\prime }{k}_2\in HN$.

Inverse: $(hk{)}^{-1}=k^{-1}{h}^{-1}$; $\exists k^{\prime }\in N$ s.t. $k^{-1}{h}^{-1}=h^{-1}{k}^{\prime }$, so $\in HN$. Thus, $HN\le G$ (Quick Subgroup Test).

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4.6. Normality of $C_3$ in $S_3$

Question

Show that the cyclic subgroup $C_3$ is a normal subgroup of $S_3$.

Hint: Write out all the elements of $S_3$ and of $C_3$, then consider: – the product of a $2$-cycle with a $3$-cycle, – the product of two $2$-cycles, – the product of two $3$-cycles, and use this to prove normality.

$S_3=\{e,(123),(132),(12),(13),(23)\}$, $C_3=\{e,(123),(132)\}$.

In $S_3$: (2-cycle)(3-cycle)=(2-cycle); (2-cycle)^2=e or (3-cycle).

For $g\in S_3$, $c\in C_3$: if $g\in C_3$, $g^{-1}cg\in C_3$ (closure). If $g=$2-cycle, $g^{-1}=g$, so $g^{-1}c=gc=$2-cycle, then $g^{-1}cg=(2-cycle)(2-cycle)=e$ or 3-cycle $\in C_3$. Thus, $C_3⊴S_3$.

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4.7. Finitary Permutations of $\mathbb{Z}$

Question

Recall that $FS(\mathbb{Z})$ denotes the group of finitary permutations of $\mathbb{Z}$, consisting of those permutations of $\mathbb{Z}$ that move only finitely many elements.

1. Determine whether the permutation $(135)(6810)$ lies in $FS(\mathbb{Z})$.
2. Let $\tau$ be the permutation that maps each $n\in \mathbb{Z}$ to $-n$. Determine whether $\tau$ lies in $FS(\mathbb{Z})$.
3. Prove that $FS(\mathbb{Z})\le \mathrm{Sym}(\mathbb{Z})$.
4. Decide whether $FS(\mathbb{Z})$ is abelian.
5. Prove that $FS(\mathbb{Z})⊴\mathrm{Sym}(\mathbb{Z})$.

Note: For a general set $X$, the group $FS(X)$ is called the finitary symmetric group on $X$, and it is always a normal subgroup of $\mathrm{Sym}(X)$.

1. Moves only $\{1,3,5,6,8,10\}$ (finite), so $\in FS(\mathbb{Z})$.
2. Moves all $n\ne 0$ (infinite), so $\notin FS(\mathbb{Z})$.
3. $FS(\mathbb{Z})\subseteq \mathrm{Sym}(\mathbb{Z})$; Quick Subgroup Test: $e\in FS(\mathbb{Z})$; closure/inverses preserve finite support.
4. No: embeds $S_n$ (non-Abelian) via finite support.
5. For $g\in \mathrm{Sym}(\mathbb{Z})$, $h\in FS(\mathbb{Z})$: let $X=\mathrm{supp}(h)$ (finite); $g^{-1}hg$ fixes points outside $g^{-1}X$ (finite), so $\in FS(\mathbb{Z})$. Thus normal.