As mentioned in the previous post, here is the algebra qual with my solutions. To be honest, this is one of the easiest quals I’ve seen from them. The scores were quite high; I scored 24 out of 25 and there were a couple of perfect scores. I think my only mistake came on problem 5(a), I gave a pretty hand wavy argument. Oh well…

**1.** Let be a group.

**(a)** Let be defined by for all . Prove that is a homomorphism if and only if is abelian.

*Proof. *Suppose that is a homomorphism. Let . Then . Also, . Thus, . Multiplying on the left by and on the right by we have that .

Conversely, suppose that is abelian. Let . Then . Where the second equality holds because and commute.

**(b)** If is abelian and finite, show that is an automorphism if and only if has odd order.

*Proof. *Suppose that is an automorphism. Seeking a contradiction assume that does not have odd order. Then is a prime which divides , so by Cauchy’s theorem, has an element of order . Let be this element. Then , so is a nontrivial element in the kernel of . This, of course, is a contradiction because is injective and so must be the only element in the kernel of .

Conversely, suppose that has odd order. Since is abelian, by part (a) we know that is a homomorphism. Moreover, because and , being injective will imply being surjective. Thus, it suffices to show that is injective. We do this by showing that the kernel of is trivial. Suppose that is in the kernel of . Then . Thus must be or . But, the order of must divide the order of . Since has odd order, so . Thus, the kernel of is trivial.

**2. **Let be a group and let . Prove: If is a subgroup of and , then is a normal subgroup of .

*Proof. *Let and . We are given that is a subgroup so exists and it suffices to show that is normal. We have that . But so for some . Thus, because is closed.

**3. **Prove that in an integral domain every prime element is an irreducible.

*Proof. *Let be a prime element of . Then the ideal generated by , , is a prime ideal. Suppose that . Then and so or . Thus, or for some . By cancellation in an integral domain we see that or . In either case is a unit or is a unit which means that is an irreducible.

**4. **Find necessary and sufficient conditions on such that the matrix

is diagonalizable over .

*Solution. *Recall, an matrix is diagonalizable if and only if the sum of the dimensions of the eigenspaces is equal to . Let

First, we find the eigenvalues of . Computing the characteristic polynomial of , we obtain . The eigenvalues of are then and .

Next we find the corresponding eigenspaces.

For we solve the system given by and get that and . Thus,

is a basis for the eigenspace corresponding to .

For we solve the system given by and get that is free, , and .

If then and

would be our basis for the eigenspace corresponding to . In this case, would not be diagonalizable.

If then

would be our basis for the eigenspace corresponding to . In this case, would be diagonalizable.

In conclusion, is diagonalizable if and only if .

**5. **Let be the vector space of matrices with entries in and let and denote the set of real symmetric and skew-symmetric matrices, respectively.

**(a) **Show that the dimension of is . A brief justification is sufficient.

*Proof. *Let

be an element of . Then

.

Thus,

…and so on. By counting, we see that there must be elements in our basis for .

**(b) **Let be the linear transformation defined by for all . Prove that and .

*Proof. *First we show that . Let . Then and so . Thus, . Now, let . Then and so . Thus, .

Next we show that . Let . Then there exists such that . This implies that and so . Now, let . Then and so . Thus, .

**(c) **Compute the dimension of .

*Solution. *By the rank nullity theorem we have that

.

In particular,

.

Thus, .