Solutions to Problem Set 1

Math 345/504

1-26-2018

1. (a) Write the group element $x^5 y^{-2} z^3$ using additive notation. (Assume the operation is commutative.)

(b) Write the group element $-2 x
   + 11 y - 4 z$ using multiplicative notation. (Assume the operation is commutative.)

(a) In additive notation, this is $5 x - 2 y + 3 z$ .

(b) In multiplicative notation, this is $x^{-2} y^{11} z^{-4}$ .


2. Let G be a group (with the operation written multiplicatively), and let a and b be arbitrary elements of G.

(a) Simplify $b^3 a^{-1} (a
   b^{-2})^3 b^2 a^{-1} b^2$ as much as possible.

(b) Solve for x in terms of a and b, simplifying your answer as much as possible:

$$a^{-3} x a^2 b = a^{-1} b a b.$$

(a)

$$b^3 a^{-1} (a b^{-2})^3 b^2 a^{-1} b^2 = b^3 a^{-1} (a b^{-2})(a b^{-2})(a b^{-2}) b^2 a^{-1} b^2 = b a.\quad\halmos$$

(b)

$$\eqalign{ a^{-3} x a^2 b & = a^{-1} b a b \cr a^3 a^{-3} x a^2 b b^{-1} a^{-2} & = a^3 a^{-1} b a b b^{-1} a^{-2} \cr x & = a^2 b a^{-1} \quad\halmos \cr}$$


3. This is the operation table for a group G, written using multiplicative notation. The identity is 1.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\cdot$ & & 1 & & a & & b & & c & & d & & e & & f & & g & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & 1 & & 1 & & a & & b & & c & & d & & e & & f & & g & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & a & & a & & d & & c & & f & & e & & 1 & & g & & b & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & b & & b & & g & & d & & a & & f & & c & & 1 & & e & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & c & & c & & b & & e & & d & & g & & f & & a & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & d & & d & & e & & f & & g & & 1 & & a & & b & & c & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & e & & e & & 1 & & g & & b & & a & & d & & c & & f & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & f & & f & & c & & 1 & & e & & b & & g & & d & & a & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & g & & g & & f & & a & & 1 & & c & & b & & e & & d & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

(a) Compute $e \cdot f$ .

(b) Compute $b^2$ .

(c) Compute $b^{-1}$ .

(d) Give a specific example to prove that G is not abelian.

(a) $e \cdot f = c$ .

(b) $b^2 = d$ .

(c) $b^{-1} = f$ .

(d) $e \cdot f = c$ , but $f \cdot e = g$ , so $e \cdot f \ne f \cdot e$ .


4. The following set is a group under multiplication mod 20:

$$G = \{1, 3, 7, 9, 11, 13, 17, 19\}.$$

So, for example, $17 \cdot 11 =
   187 = 7$ .

(a) Compute $11 \cdot 9$ in G.

(b) Find $7^{-1}$ in G.

(c) When the operation in a group G is written using multiplicative notation, the order of an element $g \in G$ is the smallest positive power n such that $g^n = 1$ . The identity 1 is the only element with order 1.

For example, $11^2 = 1$ , so the order of 11 is 2.

Find the order of 13 in G.

[Compute powers of 13, stopping at the first power of 13 which is equal to 1.]

(a) $11 \cdot 9 = 99 = 19$ .

(b) Since $7 \cdot 3 = 1$ , I have $7^{-1} = 3$ .

(c)

$$13^2 = 9, \quad 13^3 = 7, \quad 13^4 = 1.$$

The order of 13 is 4.


5. The following set is a group under addition mod 12:

$$\integer_{10} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}.$$

For example,

$$3 + 7 = 10, \quad\hbox{but}\quad 5 + 9 = 14 = 2.$$

(a) Compute $9 + 11$ in $\integer_{12}$ .

(b) Find -8 in $\integer_{12}$ .

(c) When the operation in a group G is written using additive notation, the order of an element $g \in G$ is the smallest positive multiple n such that $n \cdot g = 0$ . The identity 0 is the only element with order 1.

(Remember that "$n \cdot g$ " is shorthand for $\underbrace{g + g + \cdots + g}_{n
   \rm\; times}$ . It is not multiplication in the group, since the operation is addition.)

For example,

$$3 \cdot 4 = 4 + 4 + 4 = 0 \quad\hbox{in}\quad \integer_{12}.$$

Hence, the order of 4 is 3.

Find the order of 8 in $\integer_{12}$ .

[Compute multiples of 8 and stop at the first multiple which equals 0.]

(a) $9 + 11 = 20 = 8$ in $\integer_{12}$ .

(b) $8 + 4 = 0$ , so $-8 = 4$ in $\integer_{12}$ .

(c)

$$2 \cdot 8 = 4, \quad 3 \cdot 8 = 0.$$

Hence, the order of 8 is 3.


[Math 504]

6. Let G be a group. Prove that G is abelian if and only if $(a b)^{-1} = a^{-1} b^{-1}$ for all $a, b \in G$ .

Suppose G is abelian. Let $a, b
   \in G$ . Then

$$\matrix{ (a b)^{-1} & = & b^{-1} a^{-1} & \hbox{(Formula for inverse of a product)} \cr & = & a^{-1} b^{-1} & \hbox{(G is abelian)} \cr}$$

Conversely, suppose that $(a
   b)^{-1} = a^{-1} b^{-1}$ for all $a, b \in G$ . Let $a, b \in
   G$ . I must show that $a b = b a$ . I have

$$\matrix{ (a b)^{-1} & = & a^{-1} b^{-1} & \hbox{(Given)} \cr b^{-1} a^{-1} & = & a^{-1} b^{-1} & \hbox{(Formula for inverse of a product)} \cr (b^{-1} a^{-1})^{-1} & = & (a^{-1} b^{-1})^{-1} & \cr a b & = & b a & \hbox{(Formula for inverse of a product)} \cr}$$

Therefore, G is abelian.


Courage consists of the power of self-recovery. - Ralph Waldo Emerson


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