(10/17/07)
Math Kuperberg 150a: Modern Algebra First Midterm Solutions
1. Show that every finite group G has an even number of elements of order 3. Solution: In general, for any element g of any group, the orders of g and g-1 are the same. This comes from the fact that gn = e g-n = e. Now, if g G has order 3, then g-1 also has order 3, and we can begin to count off such elements (of order 3) in pairs. There won't be any double counting, since g = g-1 would imply that g has order 2. Since G is a finite group, we are also guaranteed to be finished counting at some point. Thus, we will have an even number of elements of order 3. 2. Consider a product a = xyzw of four transpositions in some symmetric group Sn . Can this product be a 4-cycle? No. A k-cycle must have signature (-1)k-1 . So an even permutation, such as a, cannot be a cycle of even length. Can it be a 5-cycle? Yes. Here's an example: (1 5)(1 4)(1 3)(1 2) = (1 2 3 4 5). Can it be a 9-cycle? No. Each transposition affects two numbers, while every other number is sent to itself. So the product of four transpositions can, at most, act nontrivially on 8 numbers. But a 9-cycle acts nontrivially on 9 numbers. 3. Is the dihedral group D4 isomorphic to the group C4 C2 ? (long): Solution No. D4 C4 C2 . Although the order of both groups is 8 (|D4 | = |C4 C2 | = 8), and both groups have elements of orders 2 and 4; D4 only has two elements of order 4, while C4 C2 has 4.
If D4 = {a, a2 , a3 , 1, b, ba, ba2 , ba3 }, only a and a3 are of order 4. On the other hand, C4 C2 = {(1, 1), (x, 1), (x2 , 1)(x3 , 1), (1, y), (x, y), (x2 , y), (x3 , y)} (x, 1), (x3 , 1), (x, y), (x3 , y) (which have order 4). Solution (short): C4 C2 is abelian, but D4 is not. This is a complete solution, but we may also
verify these facts: g, h C4 C2 , i, j, k, l N such that g = (xi , y j ) and h = (xk , yl ). Then gh = (xi , y j )(xk , yl ) = (xi+k , y j+k ) = (xk , yl )(xi , y j ) = hg. But in D4 , ab = ba-1 = ba. 4. In the dihedral group D6 (drawn on the board), let f be one of the reflections. Find an element which is conjugate to f (other than f itself) and find an element which is not conjugate to f (also other than f itself, and not equal to 1).
Solution: Let r be the rotation of the plane by /3. Then r D6 , and r has order 6. The reflection f has order 2, so it cannot be conjugate to r. On the other hand, conjugating by r (or by a power of r) gives another reflection (through a different line of symmetry).