Chapter 6.4, Problem 8E

a.

To determine

To prove the remaining parts of theorem 6.4.1.

Explanation of Solution

Consider $f:(A,\circ )\to (B,\times )$ an operation preserving map therefore,

  $f(x\circ y)=f(x)\times f(y)$

Consider $f:A\to B$ is onto on $B$ .

Consider $\circ$ is associative in $A$ therefore, for some $x,y,z\in A$

  $x\circ (y\circ z)=(x\circ y)\circ z$

Consider,

  $u,v,w\in B$

As $f:A\to B$ is onto on $B$ therefore their exists $x,y,z\in A$ such that

  $\begin{array}{l}u=f(x)\\ v=f(y)\\ w=f(z)\end{array}$

Consider,

  $\begin{array}{l}u\times (v\times w)=f(x)\times (f(y)\times f(z))\\ =f(x\circ (y\circ z))\\ =f((x\circ y)\circ z)\\ =f(x\circ y)\times f(z)\\ =(f(x)\times f(y))\times f(z)\\ u\times (v\times w)=(u\times v)\times w\end{array}$

Therefore $\times$ is associative on $B$ .

Hence the theorem is proved.

b.

To determine

To prove the remaining parts of theorem 6.4.1.

Explanation of Solution

Consider $f:(A,\circ )\to (B,\times )$ an operation preserving map therefore,

  $f(x\circ y)=f(x)\times f(y)$

Consider $f:A\to B$ is onto on $B$ .

Let $e$ be the identity in $A$ therefore,

  $e\circ e=e$ In $A$

Therefore,

  $\begin{array}{l}f(e\circ e)=f(e)\\ f(e)\times f(e)=f(e)\\ f(e)\times f(e)=f(e)\times e^{\text{'}},e^{\text{'}}\text{ is the identity in B}\end{array}$

By left cancellation law,

  $f(e)=e^{\text{'}}$

Therefore, $f(e)$ is the identity for $B$ .

Hence the theorem is proved.

c.

To determine

To prove the remaining parts of theorem 6.4.1.

Explanation of Solution

Consider $f:(A,\circ )\to (B,\times )$ an operation preserving map therefore,

  $f(x\circ y)=f(x)\times f(y)$

Consider $f:A\to B$ is onto on $B$ .

Consider,

  $x\in A$

Therefore,

  $\begin{array}{l}f(e)=f(x\circ x^{-1})\\ =f(x)\times f(x^{-1})\end{array}$

And

  $\begin{array}{l}f(e)=f(x^{-1}\circ x)\\ =f(x^{-1})\times f(x)\end{array}$

As $e^{\text{'}}=f(e)$ is the identity in $B$

  $f(x)\times f(x^{-1})=f(x^{-1})\times f(x)=e\text{'}$

Therefore, $f(x^{-1})$ is the inverse of $f(x)$ .

Thus,

  $(f{(x)}^{-1})=f(x^{-1})$

Hence the theorem is proved.