Chapter 5.2, Problem 12E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

1. Claim. If G is a group with identity e, then G is abelian.
   "Proof." Let a and b be elements of G. Then
   $\begin{array}{l}ab=aeb\\ =a\left(ab\right){\left(ab\right)}^{-1}b\\ =a\left(ab\right)\left(b^{-1}{a}^{-1}\right)b\\ =\left(aa\right)\left(b{b}^{-1}\right)a^{-1}b\\ =\left(aa\right)a^{-1}b\\ =\left(aa\right){\left(b^{-1}a\right)}^{-1}\\ =a\left(a{\left(b^{-1}a\right)}^{-1}\right)\\ =a\left({\left(b^{-1}a\right)}^{-1}a\right)\\ =a\left(\left(a^{-1}b\right)a\right)\\ =\left(a{a}^{-1}\right)\left(ba\right)\\ =e\left(ba\right)\\ =ba.\end{array}$

Therefore, $ab=ba$ and G is abelian.

Claim. If G is a group with elements x, y, and z and if $xz=xy,$ then $x=y.$
"Proof." If $z=e,$ then $xz=yz$ implies that $xe=ye,$ so $x=y.$ If $z\ne e,$ then the inverse of z exists, and $xz=yz$ implies

$\frac{xz}{z}=\frac{yz}{z}$ and $x=y.$ Hence, in all cases, if $xz=yz,$ then $x=y.$

Claim. The set $\mathbb{Q}+$ of positive rationals with the operation of multiplication is a group.
"Proof." The product of two positive rationals is a positive rational, so $\mathbb{Q}+$ is closed under multiplication. Because $1\cdot r=r=r\cdot 1$ for every $r\in \mathbb{Q},1$ is the identity. The inverse of the positive rational g is the positive rational $\frac{b}{a}.$ The rationals are associative under multiplication because the reals are associative under multiplication.

Claim. If m is prime, then $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ has no divisors of zero.
"proof." Suppose that a is a divisor of zero in $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right).$ Then $a\ne 0,$ and there exists $b\ne 0$ in ${\mathbb{Z}}_m$ such that $ab=0.$ Then $ab=m$ (mod m), so $ab=m.$ This contradicts the assumption that m is prime.

Claim. If m is prime, then $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ has no divisors of zero.
"proof." Suppose that a is a divisor of zero in $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right).$ Then $a\ne 0,$ and there exists $b\ne 0$ in ${\mathbb{Z}}_m$ such that $ab=0.$ Then $ab=0$ (mod m), so m divides ab. Because m is prime, m divides a, or m divides b. But because a and b are elements of ${\mathbb{Z}}_m,$ both are less than m. This is impossible.

Claim. For every natural number m, $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ is a group.
"Proof." We know that $\left(\mathbb{Z},\cdot \right)$ is associative with identity element 1. Therefore, $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ is associative with identity element 1. It remains to show every element has an inverse. For $x\in {\mathbb{Z}}_m-\left\{0\right\},x\ne 0.$ Therefore, $1/x\in {\mathbb{Z}}_m-\left\{0\right\}$ and $x\cdot 1/x=x\left(1/x\right)=1.$ Therefore, every element of ${\mathbb{Z}}_m-\left\{0\right\}$ has an inverse.

Claim. If $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ is a group, then m is prime.
"Proof." Assume that $\left({\mathbb{Z}}_m-\left\{0\right\},\cdot \right)$ is a group. Suppose that m is not prime. Let $m=rs,$ where r and s are integers greater than 1 and less than m. Then $r\cdot s=m=0$ (mod m). Because r has an inverse t in ${\mathbb{Z}}_m-\left\{0\right\},t\cdot r=1.$ Then $s=1\cdot s=\left(t\cdot r\right)\cdot s=t\cdot \left(r\cdot s\right)=t\cdot 0=0.$ That is, $s=0$ (mod m). This is impossible because $1<s<m.$