MTH3003 Weekly Problems 8

Original Documents: mth3003 weekly problem sheet 8.pdf / mth3003 weekly problem sheet 8 handwritten solutions.pdf / mth3003 weekly problem sheet 8 solutions.pdf

Vibes: All eight problems are about Group actions - verifying actions, computing orbits and stabilisers in $S_3$, and proving that stabilisers are subgroups. Foundational drilling on the group-action axioms.

Used Techniques:

• Group action axioms (i) $\lambda (e)x=x$, (ii) $\lambda (gh)x=\lambda (g)(\lambda (h)x)$.
• Regular action: $\lambda (g)x=gx$ on $X=G$.
• Conjugation action: $\lambda (g)x=gx{g}^{-1}$ on $X=G$. Stabiliser is the centraliser.
• Quick Subgroup Test for showing ${\mathrm{Stab}}_G(x)\le G$.
• Closure check: an action requires $\lambda (g)x\in X$ - easy to forget!

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8.1. Verify Regular Action Is a Group Action

Question

Check that the regular action satisfies the definition of a group action by verifying that both parts (1) and (2) of the group action definition hold.

Solution. Take $X=G$, $\lambda (g)x=gx$.

(1) For all $x\in X=G$: $\lambda (e)x=ex=x$. ✓

(2) For all $f,g\in G$, $x\in X=G$: $\lambda (fg)x=(fg)x=f(gx)=\lambda (f)(\lambda (g)x)$. ✓

So the regular action is an action. $■$

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8.2. Verify Conjugation Action Is a Group Action

Question

Check that the conjugation action satisfies the definition of a group action.

Solution. Take $X=G$, $\lambda (g)x=gx{g}^{-1}$.

(1) $\lambda (e)x=ex{e}^{-1}=x$. ✓

(2) Using $(fg{)}^{-1}=g^{-1}{f}^{-1}$:

$$\lambda (fg)x=(fg)x(fg{)}^{-1}=fgx{g}^{-1}{f}^{-1}=f(gx{g}^{-1})f^{-1}=\lambda (f)(\lambda (g)x).✓$$

So conjugation is an action. $■$

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8.3. Action of $S_6$ on $2$-Subsets

Question

Let $G=S_6$ and let $X$ be the set of 2-element subsets of $\{1,\dots ,6\}$. Define $\lambda (g)\{i,j\}=\{gi,gj\}$. Prove $\lambda$ is an action.

Solution. Three checks.

(0) Closure. $\lambda (g)\{i,j\}=\{gi,gj\}\in X$ (still a 2-subset of $\{1,\dots ,6\}$ since $g$ is a bijection on $\{1,\dots ,6\}$).

(1) For all $\{i,j\}\in X$: $\lambda (e)\{i,j\}=\{ei,ej\}=\{i,j\}$. ✓

(2) For all $g,h\in S_6$, $\{i,j\}\in X$:

$$\lambda (gh)\{i,j\}=\{(gh)i,(gh)j\}=\{g(hi),g(hj)\}=\lambda (g)\{hi,hj\}=\lambda (g)(\lambda (h)\{i,j\}).✓$$

So $\lambda$ is an action and $X$ is a $G$-set. $■$

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8.4. Translation Action of $(\mathbb{Z},+)$ on $\mathbb{N}$

Question

Let $G=(\mathbb{Z},+)$, $X=\mathbb{N}$, $\lambda (z)n=z+n$. Decide whether $\lambda$ is an action.

Solution. Not an action. The closure condition $\lambda (z)n\in X$ fails.

Counterexample. Take $z=-5\in G$, $n=1\in X$. Then $\lambda (z)n=-5+1=-4\notin \mathbb{N}$. So $\lambda (z)n$ is not always in $X$ - the map fails to be defined on all of $G\times X$ with codomain in $X$.

Note

If we replaced $X=\mathbb{N}$ with $X=\mathbb{Z}$, the map would be an action (the shift action of $\mathbb{Z}$ on itself). Closure must be checked, even when the operations look natural.

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8.5. Orbit and Stabiliser of $(123)$ under Regular Action of $S_3$

Question

Let $G=S_3$ act on itself via the regular action. Compute the orbit and stabiliser of $(123)$.

Solution. Compute $\lambda (g)(123)=g\cdot (123)$ for each $g\in S_3$:

• $e\cdot (123)=(123)$
• $(12)\cdot (123)=(23)$
• $(13)\cdot (123)=(12)$
• $(23)\cdot (123)=(13)$
• $(123)\cdot (123)=(132)$
• $(132)\cdot (123)=e$

Orbit. $G\cdot (123)=\{(123),(23),(12),(13),(132),e\}=S_3=X$. The action is transitive. ✓

Stabiliser. Only $e$ fixes $(123)$, so ${\mathrm{Stab}}_G((123))=⟨e⟩$.

Note

This generalises: for the regular action of any group on itself, the action is always transitive and every stabiliser is trivial. Compatible with $|G\cdot x|=|G|/|{\mathrm{Stab}}_G(x)|=|G|/1=|G|=|X|$ in the Orbit-Stabiliser theorem.

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8.6. Orbit and Stabiliser of $(123)$ under Conjugation in $S_3$

Question

Let $G=S_3$ act on itself by conjugation. Compute the orbit and stabiliser of $(123)$.

Solution. Compute $\lambda (g)(123)=g(123)g^{-1}$:

• $e(123)e^{-1}=(123)$
• $(12)(123)(12{)}^{-1}=(132)$
• $(13)(123)(13{)}^{-1}=(132)$
• $(23)(123)(23{)}^{-1}=(132)$
• $(123)(123)(123{)}^{-1}=(123)$
• $(132)(123)(132{)}^{-1}=(123)$

Orbit. $G\cdot (123)=\{(123),(132)\}\ne X$. The action is not transitive.

Stabiliser. ${\mathrm{Stab}}_G((123))=\{e,(123),(132)\}=⟨(123)⟩\cong A_3$.

Note

Orbit-Stabiliser sanity check: $|G|=6$, $|G\cdot x|=2$, $|\mathrm{Stab}|=3$, and $6=2\cdot 3$. ✓

The stabiliser is the centraliser $C_G((123))$ - elements that commute with $(123)$. Sensibly, the cyclic subgroup it generates does commute with it, and the transpositions don’t.

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8.7. Stabiliser Is a Subgroup

Question

Let $X$ be a $G$-set and $x\in X$. Prove ${\mathrm{Stab}}_G(x)\le G$.

Solution. Quick Subgroup Test.

Identity. $\lambda (e)x=x$, so $e\in {\mathrm{Stab}}_G(x)$.

Closure. Let $g,h\in {\mathrm{Stab}}_G(x)$, so $\lambda (g)x=x$ and $\lambda (h)x=x$. Then

$$\lambda (gh)x=\lambda (g)(\lambda (h)x)=\lambda (g)x=x,$$

so $gh\in {\mathrm{Stab}}_G(x)$.

Inverse. Let $g\in {\mathrm{Stab}}_G(x)$. Then

$$x=\lambda (e)x=\lambda (g^{-1}g)x=\lambda (g^{-1})(\lambda (g)x)=\lambda (g^{-1})x,$$

so $g^{-1}\in {\mathrm{Stab}}_G(x)$. $■$

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8.8. Homomorphisms into $\mathrm{Sym}(X)$ as Actions

Question

Let $\lambda :G\to \mathrm{Sym}(X)$ be a homomorphism. Prove $\lambda$ is an action of $G$ on $X$.

Solution. Verify the two action axioms.

(0) Closure. Since $\lambda (g)\in \mathrm{Sym}(X)$ is a permutation of $X$, $\lambda (g)x\in X$ for any $x\in X$.

(1) Homomorphisms preserve identity: $\lambda (e_G)=e_{\mathrm{Sym}(X)}={\mathrm{id}}_X$, so $\lambda (e_G)x=x$ for all $x\in X$.

(2) Homomorphism property gives $\lambda (gh)=\lambda (g)\lambda (h)$ as permutations, hence $\lambda (gh)x=\lambda (g)(\lambda (h)x)$ for all $x\in X$.

So $\lambda$ is an action of $G$ on $X$. $■$

Note

Combined with the previous direction (every action is a homomorphism into $\mathrm{Sym}(X)$, lecture 8.1), we have a bijection:

$$\{\text{actions of }G\text{ on }X\}⟷\{\text{homomorphisms }G\to \mathrm{Sym}(X)\}.$$

Two equivalent ways of viewing the same data.