1. Define subgroup.

A subgroup of a group G is a subset H of G which is closed under the law of composition, contains the identity element, and is closed under inverses.

2. Suppose that A is an n ×n matrix; let adj A denote the adjoint of A. What is the product A(adj A) equal to? (No proof required.)

It is equal to (detA) I_n.

3. Let G be a group, and let x be an arbitrary element of G. How does the order of x compare with the order of x^-1? Give a brief explanation.

They have the same order: the cyclic subgroup generated by x is the same as the cyclic subgroup generated by x^-1.

4. Let G be a group with law of composition $, G' a group with law of composition @. What formula must a homomorphism f from G to G' satisfy?

f (x $ y) = f(x) @ f(y)

5. Write down the multiplication table for the symmetric group S₃. (Recall that this is the group of permutations of the set { 1,2,3 }, or equivalently, it is the group of 3 ×3 permutation matrices. It has order 6.)

Let C be either of the two cyclic permutations, and let T be any of the three permutations that interchange two numbers. Then C³ = 1, T² = 1, and CT = TC². You can derive the rest of the table from this. See your class notes for details. (In other words, I'm too lazy to type in the table.)

6. True or false: if A and B are n ×n matrices, then det(AB) = det(A) det(B). (If true, give a brief explanation. If false, give a counterexample.)

This is true. We proved it in class like this: first we noticed that it was true when A was elementary. Thus it's true when A is a product of elementary matrices, i.e., when A is invertible. When A is not invertible, then we showed that both sides of the equation are zero.

7(a). How many 4 ×4 permutation matrices are there? (No proof required, just give me a number.)

24

(b) Give me three examples of 4 ×4 permutation matrices, and compute the sign of each one.

(There is no single correct answer here.)

8. Let G be a group. (To get full credit for parts (a) and (b), don't use Lagrange's theorem or any related result; instead, just use the definition of index.)

(a). Can there exist a subgroup H of G such that [G:H] = 0? If so, what can you say about H? If not, why not? Justify your answers.

No. 1H = H is always a coset, so the index is always at least 1.

(b). Answer the same questions for the case [G:H] = 1.

Yes. In this case, 1H is the only coset; since the cosets partition G, we conclude that if [G:H] = 1, then G = H.

(c). Answer the same questions for the case [G:H] = 2.

Yes. In this case, there are two cosets, so the order of H is half that of G. By a homework problem, if the index is 2, then H is a normal subgroup of G.

9. Let (R,+) be the group of real numbers under addition. Let (P,×) be the group of positive real numbers under multiplication. Prove that these groups are isomorphic.

[If you can't do this completely, then at least try to find a (non-trivial) homomorphism from one of these groups to the other.]

Define f: R --> P by f(x) = e^x. Note that e^x is always positive, so the codomain is indeed P. This function is a homomorphism: e^x+y = e^x e^y. Also, this function has an inverse function, log; therefore, it is a bijection.

10. Let n be a positive integer. The center Z of the group GL_n(R) is the subset of GL_n(R) consisting of matrices A which commute with every matrix in the group:

[Hint: Assume that A is in the center, and look at the equation AE = EA when E is elementary. If you can't do this problem in general, you can get one or two points if you do it when n = 1; you can get up to 7 for proving the n = 2 case; and you can get up to 12 points for the n = 3 case.]

Let H = { aI : in R^× }. We are supposed to show that Z = H. First, notice that Z contains H: if aI is in H, then for any matrix B, B(aI) = a(BI) = aB = (aI) B, by basic properties of scalar multiplication of matrices.

Now I want to show that Z is contained in H, so let A be in Z. I want to show that A is in H. I know that AB = BA for every matrix B. Let E be the (elementary) matrix which looks just like the identity matrix, except that there is a 2 in the (i,i)-entry. Consider the equation EA = AE. EA is obtained from A by multiplying the ith row by 2. AE is obtained from A by multiplying the ith column by 2. These are equal, so if I look at the (i,j)-entry of both sides, I find that 2a_i,j = a_i,j when i is not equal to j. Therefore a_i,j = 0 when i is not equal to j. In other words, if A is in the center, then A must be a diagonal matrix.

Now do the same thing but with the elementary matrix E that switches rows (or columns) i and j: EA = AE, EA is A with rows i and j switched, and AE is A with columns i and j switched. The (i,j)-entry of EA is a_jj. The (i,j)-entry of AE is a_ii. We conclude that a_jj = a_ii-all of the diagonal entries are the same. Therefore A is a scalar multiple of the identity matrix.