## Math 417 Practice Problems

• [1] Problem #4a on p. 89.

An element x of a group G is in the center if the row and column corresponding to x in the multiplication table for G are the same. Thus from the table on p. 89, we see that Z = {e, a2}.

• [2] Problem #6 on p. 89.

Suppose (ba)n = e for some positive integer n. Then we have e = (ba)n = (ba)(ba)...(ba), so ab = aeb = a(ba)...(ba)b = (ab)...(ab) = (ab)n+1, hence e = (ab)n. This means that |ba| >= |ab|. A similar argument reversing the roles of a and b shows that |ab| >= |ba|. Thus |ab| = |ba|.

• [3] Problem #10 on p. 89.

Let G be a group of order 4. Cyclic groups are abelian, so we are done if G is cyclic. So now assume G is not cyclic. Let g be an element of G of largest possible order. Thus 1 < |g| < 4. If |g| = 3, then there is an element x in G which is not a power of g. If x2 = x, then x = e, which is a contradiction, since e = g0, but x is not a power of g. Thus x2 is not x, so x is not its own inverse, so x-1 must be a power of g. But if x-1 = gi for some i, then x = g-i so x is in fact a power of g, contrary to hypothesis. Thus it is impossible for there to be an element g of order 3. Thus all elements of G but e have order 2. So now let g and h be any elements of G. If either g or h is e, then gh = hg. If neither g nor h is e, then g, h and gh all have order 2, so e = ghgh implies gh = geh = g(ghgh)h = gg(hg)hh = ehge = hg. Thus G is abelian.

• [4] Problem 14 on p. 89.

Let t denote the positive square root of two. We must show that G = {a + bt : a and b are rational, not both 0} is a group under multiplication. We know multiplication is associative. And a + bt = 1 + 0t is an identity. To check closure, consider a + bt and c + dt in G: (a + bt)(c + dt) = (ac + 2bd) + (ad + bc)t, and since ac + 2bd and ad + bc are rational, we see that (a + bt)(c + dt) is in G. (Note that if ac + 2bd and ad + bc were both 0, then (ac + 2bd) + (ad + bc)t and hence (a + bt)(c + dt) would be 0 even though neither factor is 0.) Finally, the multiplicative inverse of a + bt is 1/(a + bt); by rationalizing the denominator we get (a - bt)/(a2 - 2b2), which is in G. Thus G is a group. (We could also use the fact that the nonzero rationals form a group under multiplication. Then we only need to check that G is a subgroup.)

• [5] Problem 16 on p. 89.

Let G be an abelian group, and let T be the set of elements of G of finite order. Then T is nonempty since e is in T. And if x is in T, then since |x| = |x-1|, we know x-1 is in T also. Finally, let x and y be in T, and say |x| = m and |y| = n. Then (xy)mn = xmnymn = enem = e, so xy has finite order so xy is in T. Thus T is a subgroup. Note that we used commutativity by asserting (xy)mn = xmnymn. Since we can't do this if G is not abelian, this argument doesn't work when G is nonabelian. (In fact it is false, in general; look at the group G of symmetries of the infinite strip of H's for an explicit example. Let x be reflection across a vertical line through H number 1, and let y be reflection across a vertical line through H number 2. Then yx is a slide two H's to the right. Thus even though x and y have order 2 and hence are in T, yx has infinite order, and so is not in T.)

• [6] Problem 18 on p. 89.

Since |y| = 2, we know that y-1 = y, so yxy = x2 implies x = exe = yyxyy = yx2y = yxexy = yxyyxy = x4, so e = x3. Thus |x| <= 3. But x is not e, so we know that |x| > 1, and this means |x| is not 2, since otherwise e = x3 = x2x = ex = x. Thus |x| = 3.

• [7] Problem 36 on p. 92.

So say p is an odd prime and G is a group with exactly p elements of order p. Let g be one of those elements. Then we know that the cyclic subgroup generated by g has p - 1 generators, namely g, g2, ... , gp - 1, each of which has order p. Thus there is one more element of G of order p, say x, and it cannot be a power of g. But this means x, x2, ... , xp - 1 all have order p, and all generate the same subgroup. Thus the sets {g, g2, ... , gp - 1} and {x, x2, ... , xp - 1} are disjoint, so G must have at least 2p - 2 elements of order p. Since p is an odd prime, hence bigger than 2, 2p - 2 is bigger than p, which is a contradiction. (What this argument really shows is that the number of elements of G of order p is a multiple of p - 1. Since no odd prime is a multiple of p - 1, G can't have exactly p elements of order p.)

• [8] Problem 12 on p. 111.

Let b be in Sn for some n. If b = e, then b is its own inverse, so both b and its inverse are even. If b is not e, we can write b as a product of one or more 2-cycles: b = c1...ct. But b-1 = ct...c1, so both b and b-1 can be written using the same number of 2-cycles, so either both are even or both are odd.

• [9] Problem 24 on p. 111.

Let b be an element of S7 of order 5. Write b as a product of disjoint cycles. Since the order, which is 5, of b is the least common multiple of the lengths of these cycles, we know that every cycle in the product has length 1 or 5, and at least one of them has length 5. But we can ignore the length 1 cycles, and there is room for only one 5-cycle (since the sum of the lengths of the cycles is at most 7). Thus b must in fact be a 5-cycle. Thus the number of elements of S7 is just the number of 5-cycles. There are (7)(6)(5)(4)(3) = 2520 different ways strings abcde where a, b, c, d, e are chosen from the numbers 1 to 7. Each of these give a 5-cycle (abcde), and every 5-cycle is one of these. However, they are not all different, since (abcde) = (bcdea) = (cdeab) = etc. Since each 5-cycle can be written 5 different ways, there are only 2520/5 = 504 different elements of order 5 in S7.