MTH3003 Weekly Problems 7

Original Documents: mth3003 weekly problem sheet 7.pdf / mth3003 weekly problem sheet 7 handwritten solutions.pdf / mth3003 weekly problem sheet 7 solutions.pdf

Vibes: All five problems are exercises in the Signature / Alternating group. Once you have the recipe - the signature of a permutation in disjoint-cycle form is $(-1{)}^{\sum (r_i-1)}$ - everything is mechanical. Builds toward systematically listing the 60 elements of $A_5$.

Used Techniques:

• Signature recipe: write $g=c_1{c}_2\cdots c_m$ in disjoint cycles, then $\sigma (g)={\prod }_i(-1{)}^{r_i-1}$ where $r_i=|c_i|$.
• Homomorphism property: $\sigma (gh)=\sigma (g)\sigma (h)$, so signs of products multiply.
• Inverse property: $\sigma (g^{-1})=\sigma (g)$ (cycles are reversed but lengths unchanged).
• $A_n$ membership: $g\in A_n\iff \sigma (g)=+1$.

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7.1. Signatures in $S_{10}$ and Membership of $A_{10}$

Question

Recall from lectures that the signature function $\sigma :S_n\to \{1,-1\}$ is used to define the alternating group $A_n$.

Calculate the signatures of the following permutations in $S_{10}$, and hence determine whether or not they lie in the alternating group $A_{10}$.

1. $g=(614)(5327)(8910)$
2. $h=(12)(34)(56)(78)$
3. $g^{-1}$
4. $gh$
5. $g^{-1}hg$

Solutions.

(1) $g$ is in disjoint-cycle form with cycle lengths $(3,4,3)$. So $\sigma (g)=(-1{)}^2(-1{)}^3(-1{)}^2=(+1)(-1)(+1)=-1$. So $\boxed{g\notin A_{10}}$.

(2) $h$ has cycle lengths $(2,2,2,2)$. $\sigma (h)=(-1{)}^4=+1$. So $\boxed{h\in A_{10}}$.

(3) $g^{-1}$ is $g$ with cycles reversed, so the cycle lengths are unchanged. $\sigma (g^{-1})=\sigma (g)=-1$. So $\boxed{g^{-1}\notin A_{10}}$.

(4) $\sigma (gh)=\sigma (g)\sigma (h)=(-1)(+1)=-1$. So $\boxed{gh\notin A_{10}}$.

(5) $\sigma (g^{-1}hg)=\sigma (g^{-1})\sigma (h)\sigma (g)=(-1)(+1)(-1)=+1$. So $\boxed{g^{-1}hg\in A_{10}}$.

Note

Conjugation $g^{-1}hg$ always preserves signature (the two factors of $\sigma (g{)}^{\pm 1}$ cancel). This is consistent with conjugation preserving cycle shape.

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7.2. Alternating Group Membership from Cycle Shapes

Question

Determine whether or not permutations with each of the following cycle shapes lie in some alternating group $A_n$.

1. Cycle shape $(3,2,2,2)$
2. Cycle shape $(7,3,3,2,2)$
3. Cycle shape $(5,5,5)$
4. Cycle shape $(2,2,2,2,2)$

Solutions. Compute $\sigma =\prod (-1{)}^{r_i-1}$ for each shape.

(1) $(3,2,2,2)$: $\sigma =(-1{)}^2(-1{)}^1(-1{)}^1(-1{)}^1=(+1)(-1)(-1)(-1)=-1$. Not in $A_n$.

(2) $(7,3,3,2,2)$: $\sigma =(-1{)}^6(-1{)}^2(-1{)}^2(-1{)}^1(-1{)}^1=(+1)(+1)(+1)(-1)(-1)=+1$. In $A_n$.

(3) $(5,5,5)$: $\sigma =(-1{)}^4(-1{)}^4(-1{)}^4=+1$. In $A_n$.

(4) $(2,2,2,2,2)$: $\sigma =(-1{)}^5=-1$. Not in $A_n$.

Quick rule

A cycle of length $r$ is even iff $r$ is odd. So $\sigma (g)=+1$ iff the number of even-length cycles is even. Easier than computing each factor.

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7.3. Elements of $A_3$ via Cycle Shapes

Question

Without looking at your notes, try to list all the elements in the alternating group $A_3$ by first listing all the cycle shapes that can occur in $A_3$.

Cycle shapes in $S_3$.

Shape	Example	Signature	In $A_3$?
$\varnothing$	$e$	$+1$	✓
$(2)$	$(12)$	$-1$	✗
$(3)$	$(123)$	$+1$	✓

So $A_3$ consists of all permutations with shapes $\varnothing$ or $(3)$:

$$\boxed{A_3=\{e,(123),(132)\}\cong C_3.}$$

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7.4. Cycle Shapes in $S_5$ and Which Lie in $A_5$

Question

List all the cycle shapes that can occur in $S_5$ and determine which of these lie in the alternating group $A_5$.

Systematic enumeration by the length of the largest cycle:

Largest	Possibilities	Cycle shape
1	$(\ast )(\ast )(\ast )(\ast )(\ast )$	$\varnothing$ (identity)
2	$(\ast \ast )(\ast )(\ast )(\ast )$ or $(\ast \ast )(\ast \ast )(\ast )$	$(2)$, $(2,2)$
3	$(\ast \ast \ast )(\ast )(\ast )$ or $(\ast \ast \ast )(\ast \ast )$	$(3)$, $(3,2)$
4	$(\ast \ast \ast \ast )(\ast )$	$(4)$
5	$(\ast \ast \ast \ast \ast )$	$(5)$

Cycle shape	Example	Signature	In $A_5$?
$\varnothing$	$e$	$+1$	✓
$(2)$	$(12)$	$-1$	✗
$(2,2)$	$(12)(34)$	$+1$	✓
$(3)$	$(123)$	$+1$	✓
$(3,2)$	$(123)(45)$	$-1$	✗
$(4)$	$(1234)$	$-1$	✗
$(5)$	$(12345)$	$+1$	✓

So the cycle shapes in $A_5$ are: $\varnothing ,(2,2),(3),(5)$.

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7.5. Listing All Elements of $A_5$

Question

Use your answer to Question 7.4 to list all elements of the alternating group $A_5$.

Counts by cycle shape. Total $|A_5|=|S_5|/2=60$.

Shape $\varnothing$ (1 element): $e$.

Shape $(2,2)$ (15 elements). Choose the fixed point ($5$ choices), then partition the remaining $4$ into two pairs ($3$ choices), giving $5\cdot 3=15$:

• 1 fixed: $(23)(45),(24)(35),(25)(34)$
• 2 fixed: $(13)(45),(14)(35),(15)(34)$
• 3 fixed: $(12)(45),(14)(25),(15)(24)$
• 4 fixed: $(12)(35),(13)(25),(15)(23)$
• 5 fixed: $(12)(34),(13)(24),(14)(23)$

Shape $(3)$ (20 elements). Choose 3 of 5 elements ($\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$), then 2 cyclic orientations per choice: $10\cdot 2=20$. Listed by selected triple:

$\{1,2,3\}:(123),(132)$; $\{1,2,4\}:(124),(142)$; $\{1,2,5\}:(125),(152)$; $\{1,3,4\}:(134),(143)$; $\{1,3,5\}:(135),(153)$; $\{1,4,5\}:(145),(154)$; $\{2,3,4\}:(234),(243)$; $\{2,3,5\}:(235),(253)$; $\{2,4,5\}:(245),(254)$; $\{3,4,5\}:(345),(354)$.

Shape $(5)$ (24 elements). $5!/5=24$ (divide by $5$ for rotations of the same cycle). Listed by writing each $5$-cycle starting at $1$:

$(12345),(12354),(12435),(12453),(12534),(12543)$; $(13245),(13254),(13425),(13452),(13524),(13542)$; $(14235),(14253),(14325),(14352),(14523),(14532)$; $(15234),(15243),(15324),(15342),(15423),(15432)$.

Total. $1+15+20+24=60=|A_5|$. ✓