please solve 2(a),2(b),2(c)and2(d) In this problem, we show that any group of or

Question

please solve 2(a),2(b),2(c)and2(d)

In this problem, we show that any group of order 3 must be cyclic. Let G be a group of order 3, and let e. a, b be the three elements of G, where e is the identity of the group. Prove that b must be the inverse of a. [Hint: explain why ab must equal e.] Using part (a), prove that G must be cyclic. Consider the group G = {e, a, b, c, d, f, g, h} whose Cayley table is given below. What is the center of the group? What is |a|? What is |b|? (Please show in some way how you found these.) Find a cyclic subgroup H of G such that |H | = 4. Find a noncyclic subgroup K of G such that |K| = 4.

Explanation / Answer

I already explained part (b) sufficiently, and parts (a) and (c) were quite self-explanatory, yet you are having doubts, so let me try again, with better explanations:

Looking at the table, we observe that:

(a) d commutes with all other elements in G. No other element of G(except e) has this property. Since the center of a group is the set of elements which commute with all elements of the group, hence e and d are the only elements in the center of G. Thus, center(G) = {d.e}

(b)For an element x in G, |x| is defined to be the smallest natural number k, for which xk = e

      |a| = 2, as a2 = a·a = e but a e, i.e., 2 is the smallest value of k for which ak = e.

      From the table, b e

                             b·b = d

                             (b·b)·b = d·b = g

                             ((b·b)·b)·b = g·b = e

     So, b4 = e, and 4 is the smallest value of k for which bk = e

(as we just calculated above that b, b2 and b3 are all different from e).

     Thus, |b| = 4

(c) The cyclic subgroup generated by an element x in G is the subgroup containing all powers og x, i.e. it is the set {xk; k is a natural number}. Clearly it has order equal to |x|, as by definition, x|x| = e, while, x|x|+k = xk, so that this subgroup has elements {x, x2, ..., x|x| = e}

Here, |b| = 4. So the cyclic subgroup generated by b must have order 4, i.e., H = {b, b2, b3, b4 = e} = {b, d, g, e} is a cyclic subgroupof order 4

I couldn't solve part (d) right now. Will try later. Please rate my ansewr for the other parts though.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.