# Solutions to Assignment 10

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1 Solutions to Assignment 10 Math 412, Winter Prove that each element of a finite group G appears exactly once in each row and exactly once in each column of the operation table. It is not difficult to show that each element appears at least once in each row and column. In fact, this is equivalent to the equations ax = b and xa = b always having a solution, which is easy to demonstrate. The required solutions are x = gb and x = bg where g G is such that ga = ag = e. So suppose an element a G appears twice in some column of the operation table. Then for some element b G, the equation xb = a has at least two solutions, which we will call c and d. Note that cb = a = db. Because there is an element f G such that bf = e, we get that c = ce = cbf = dbf = de = d, a contradiction. The argument is roughly the same if an element a G appears twice in some row of the operation table. It must then be the case that for some element b G, the equation bx = a has at least two solutions, which we will call c and d. Note that bc = a = bd. Then because there is an element f G such that fb = e, we get that c = ec = fbc = fbd = ed = d, a contradiction A partial operation table for a group G = {e, a, b, c, d, f} is shown below. Complete the table. a a b e d d d f f This is going to be a little painful! We proceed with the following steps: (a) We know by the one per row and column rule that ad = f and af = c. 1

2 d d f f (b) We know that ca = f. So da = (ac)a = aca = af = c. d d c f f (c) We know that cf = a. Thus df = (ac)f = acf = a 2 = b. d d c b f f (d) We know that da = c so db = d(aa) = daa = ca = f. f f (e) We know that df = b, so ff = (ad)f = adf = ab = e. f f e (f) By the one in each row and column trick we know that bf = d, and hence that ba = e and fa = d. 2

3 e d f f d e (g) We know that db = f, so fb = (ad)b = adb = af = c. e d f f d c e (h) We know that fb = c, so cb = (af)b = afb = ac = d. e d c c f d a f f d c e (i) Again by the one per row and column trick, we see that bb = a, and then bc = f and bd = c. e a f c d c c f d a f f d c e (j) We know that cf = a, so cc = c(fb) = cfb = ab = e. e a f c d c c f d e a f f d c e (k) Again by the one per row and column trick, we see that cd = b, dc = a, dd = e, fc = b, and fd = a. 3

4 e a f c d c c f d e b a d d c f a e b f f d c b a e Let G = {0, 1, 2, 3, 4, 5, 6, 7} and assume G is a group under an operation with these properties: (a) a b a + b for all a, b G; (b) a a = 0 for all a G. Write out the operation table for G. Again, fun to do, painful to type up! (a) We get for free that all of the entries on the main diagonal are (b) Now note that because , it must be that 0 0 = 0. Then due to the one per row and column rule, and the fact that , it must be that 0 1 = 1. Continuing in this manner allows us to fill in the first row and column of our table (and note that this makes zero the identity) (c) Now = 3, and we have already used 0, 1, and 2 in the second row and third column, so 2 1 = 3. Similarly, we may say that 1 2 = 3. At this point I ll note be so diligent about saying each time that we have already used such and so numbers in the column and row we are interested in. 4

5 (d) Now = 4, so 3 1 is either 2 or 4. If 3 1 = 4, then 3 = 3 (1 1) = = 4 1. This cannot be the case because we have already used 3 in the second column. It must be that 3 1 = 2. A similar argument demonstrates that 1 3 = (e) Note that = 5, and we have already used 4 in the fifth row, so 4 1 = 5. A similar argument demonstrates that 1 4 = (f) Now = 6, so that 5 1 is either 4 or 6. If 5 1 = 6, however, we get that 5 = 5 (1 1) = = 6 1, which cannot happen (we have already used 5 in the second column), so we conclude that 5 1 = 4. A similar argument show that 1 5 =

6 (g) It is now clear that 6 1 = 7, 7 1 = 6, 1 6 = 7, and 1 7 = (h) Note that 3 2 5, so 3 2 can equal 5, 4, or 1. But if 3 2 = 5, then 3 = 3 (2 2) = = 5 2, a contradiction as we have already used 3 in the third column. If 3 2 = 4, we get the same contradiction. Thus 3 2 = 1. A similar argument shows that 2 3 = (i) Note that 4 2 6, and that we have used 0, 1, 2, 3, 4, and 5 already in the corresponding column and row. So 4 2 = 6. A similar argument works for (j) Now 5 2 7, so 5 2 = 7 (by the one per column and row trick yet again!). The same can be said for

7 (k) Now 6 2 can equal 4 or 5. If 6 2 = 4, then 6 = 6 (2 2) = = 4 2 = 5, a contradiction. So 6 2 = 5, and we may then conclude that 7 2 = 4. The same goes for 2 6 and (l) We can now see that 4 3 = 7 = 3 4, 5 3 = 6 = 3 5, 6 3 = 4 = 3 6, and 7 3 = 5 = 3 7 (all using the one per column/row trick) (m) Note that 5 6 is either 1, 2, or 3. If 5 6 = 3, then 5 = 5 (6 6) = = 3 6 = 4, a contradiction. Similarly, if 5 6 = 1, then 5 = 5 (6 6) = = 1 6 = 7, a contradiction. Thus 5 5 = 2. We may conclude the same for (n) Note that 5 7 is either 1 or 3. If 5 7 = 3, then 5 = 5 (7 7) = = 7

8 3 6 = 4, a contradiction, so 5 7 = 1. The same may be concluded for 7 5. We can then see that 4 5 = 3, 4 6 = 1, 4 7 = 2, 5 4 = 3, 6 4 = 1, 6 7 = 3, 7 4 = 2, and 7 6 = (a) If a G and a 12 = e, find the orders of each of the elements a, a 2,..., a 11. According to theorem 7.8, the order of a i where i is a divisor of 12 is 12/i. Thus a 2 = 6, a 3 = 4, a 4 = 3, and a 6 = 2. We also know that a k = e if and only if 12 k. In this case we want to find the smallest t such that (a k ) t = a kt = e where k = 5, 7, 8, 9, 10, 11, so we need t such that 12 tk. If k is relatively prime to 12, then this is easy enough. In that case it must be that t = 12. Thus the orders of a 5, a 7, and a 11 are 60, 84, and 131. On the other hand, if k and 12 are not relatively prime, we can use a smaller t such that 12 will divide tk. For a 8, for instance, we note that if we take t = 3, then (a 8 ) 3 = (a 12 ) 2 = e. Similarly, for a 9, taking t = 4 will suffice (because this is the smallest integer such that 12 9t). Finally, for a 10, taking t = 6 works. So the orders are: a i order (b) Based on the evidence in part (a), make a conjecture about the order of a k when a = n. 8

9 It seems that we are always looking for a t such that 12 tk (note that this holds even when k 12). So the answer seems to be that the order of a k is 12/(k, 12) Prove that every nonabelian group G has order at least 6; hence, every group of order 2,3,4, or 5, is abelian. Suppose that G is a nonabelian group of order at least 2. We will show that G contains at least six elements. If G is of order exactly two, then because one of the elements must be the identity, all products are forced to commute (if we call the two elements of G e and a, then the only possible multiplications are e e, a a, e a, and a e, and these all commute). Thus we may suppose that G contains at least three elements, which we call e, a, and b. We can choose a and b in G such that ab ba because G is nonabelian. Of course, ab, ba G, and by cancellation, ab a ba and ab b ba. Also, because ab ba, we have that ab e ba. We have now identified five distinct elements in G: e, a, b, ab, and ba. So it is enough to show that {e, a, b, ab, ba} is not closed (because then we will have demonstrated that G contains at least six elements). First we show that if one of a 2 or b 2 is equal to e, then we are done. Suppose that a 2 = e. Then we know that aba ba, and aba ab (by cancellation), aba a (because ab e), aba e (because ba a), and aba b (because ab ba), and thus aba {e, a, b, ab, ba} so that G has at least six elements. This argument is symmetric in a and b, so we get the same result if b 2 = e. If neither a 2 = e or b 2 = e, then we will demonstrate that either a 2 {e, a, b, ab, ba} or a 1 {e, a, b, ab, ba}, again completing the proof. It is easy to see, again using cancellation, that a 2 a, a 2 ab, and a 2 ba. Suppose therefore that a 2 = b. We note that unless a 1 {e, a, b, ab, ba}, it must be the case that a(ab) = e or a(ba) = e. The first case, however, implies that b 2 = e, a contradiction. In the second case, we may assume that ba = a 1 and thus that baa = b 2 = e, again a contradiction. Thus if G is abelian, we have shown that it must have at least six elements Let G be a nonempty set equipped with an associative operation with these properties: (a) There is an element e G such that ea = a for every a G. This is a little tricky to show, until you see the trick, of course! We will let d be such that da = e and q be such that qd = e (these things exist by part (b) of the question). Now simply note that because ea = a for all a G, we have that ae = (ea)e = eae = (qd)ae = qdae = q(da)e = qee = q(ee) = qe = qda = ea as required. (b) For each a G, there exists d G such that da = e. 9

10 Now that we know part (a), this is not difficult. Using the notation from above we have that a = ea = qda = qe = eq = q so that ad = qd = e = da as required If a is the only element of order 2 in a group G, prove that a Z(G). We need to show that for all g G, ag = ga. We do know that for all g G, (gag 1 ) 2 = gag 1 gag 1 = gaeag 1 = ga 2 g 1 = geg 1 = gg 1 = e. But a is the only order two element of G. So if gag 1 e (the only order 1 element), it must be that gag 1 = a, and then we will be done because multiplying both sides on the right by g we then obtain gag 1 g = ag, that is, ga = ag. Suppose then that gag 1 = e. This implies that ga = g, or that a = g 1 ga = g 1 (ga) = g 1 g = e. But this is a contradiction because a has order two and hence a e Let G be an abelian group and n a fixed positive integer. (a) Prove that H = {a G a n = e} is a subgroup of G. As we proved in class, it is enough to show that H is nonempty, and for all a, b H, ab 1 H. The first condition is quite simple, because e n = e, so e H. The second is similarly straight forward. If a, b H, that is, if a n = e and b n = e, then (ab 1 ) n = a n (b n ) 1 = ee = e, so that ab 1 H. (b) Show by example that part (a) may be false if G is nonabelian. Consider the nonabelian group S 3, and let n = 2. Then it is not difficult to see that the elements ρ S 3 such that ρ 2 = e are H = {( ), ( ), ( ), ( )}. But this set is not closed under composition, for example ( ) ( ) ( ) = H, and hence H is not a subgroup Let G = a be a cyclic group of order n. (a) Prove that the cyclic subgroup generated by a m is the same as the cyclic subgroup generated by a d where d = (m, n). It is enough to show that a m a d and a d a m. In fact, because a m and a d are the respective generators of a m and a d, it is enough to show that a m a d and a d a m. Note that because d = (m, n) is is clear that there is a t Z such that dt = m. Thus a m = (a d ) t a d. On the other hand, we know that there exist u, v Z such that un + vm = d. Thus a d = a un+vm = a un a vm = (a n ) u a vm = ea vm = a vm a m. (b) Prove that a m is a generator of G if and only if (m, n) = 1. Suppose that (m, n) = 1, then again there are u, v Z such that un + vm = 1. Thus for any a i a we have that i = iun + ivm so that 10

11 a i = a iun+ivm = a iun a ivm = (a n ) iu a ivm = e(a m ) iv = (a m ) iv, so that a m is a generator of a. On the other hand, if a m is a generator of a, then a a m, that is, there is a v such that (a m ) v = a. By theorem 7.8, this implies that mv 1 (mod n), and thus that there is an u Z such that mv 1 = un. Thus mv + ( u)n = 1 and we know that this implies that (m, n) = 1. 11

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