Given A Is closed B Is commutative C The identity is D The

Given

A.) Is & closed?

B.) Is & commutative?

C.) The identity is?

D.) The inverse of P is?

E.) Determine if & is associative? Show why or why not

F.) Find if each element has an inverse

Solution

ANSWER

Part (A)

Definition: an operation * is closed over a set S, if for all a, b S, a * b S.

Now, in the given Cayley’s table, all entries in the body of the table are one of p, q, r and s, which actually form the set S. Since each entry in the body of the table represents the output of operation & acting on two elements of S, this implies that all outputs belong to S => & is closed.

Part (B)

Definition: an operation * is commutative over a set S, if for all a, b S, a * b = b * a.

Now, in the given Cayley’s table, p & r = r and r & p = s which is not equal to r.

Hence & is not commutative

Part (C)

Definition: an operation * has an identity element, say e, over a set S, if for all a, b S, a * e = e * a = a and e S.

Now, in the given Cayley’s table, it can be observed that for all of p, q, r and s, operation & with q yields the original element itself. So, q is the identity

Part (D)

Definition: If a^{- 1} is the inverse of an element a of set S with an operation *, then, a * a^{- 1} = a^{- 1} * a = e where e is identity element.

Now, in the given Cayley’s table, p & s = s & p = q. Since from Part (c), q is the identity element, s is the inverse of p   

Part (E)

Definition: an operation * is associative over a set S, if for all a, b, c S, a * (b * c) = (a * b) * c.

Now, in the given Cayley’s table, p & (q & r) = p & r = r and (p & q) & r) = p & r = r. Likewise, other combinations can also be checked to see that & is associative

Part (F)

By actual observation of the given Cayley’s table,inverse of p is s, inverse of s is p and q and r are their own inverses.