Math 5246, Solutions to Final Exam

1. To say H is normal means that if x is in H, then so is gxg' for any g (where g' denotes the inverse). Take x=ab, g=b, so gxg'=ba. Another way to think about this is: if H is a normal subgroup of G, then saying ab in H is the same as saying that abH=H is the identity in the quotient group G/H. That is, (aH)(bH)=H, so aH and bH are inverses in G/H. This clearly happens iff (bH)(aH)=H, since inverses in a group are both left- and right-inverses.

For a counterexample when H is not normal, let G=Dn (n>2) and H={e,m} where m is any mirror reflection. If we write m=ab where b is a nontrivial rotation and a is some other reflection, then since a and b do not commute, ba is not m, and thus is not in H. (If you want a specific case, look at the tables in the book for D4.)

2. In a PID like Z[i], the concepts of prime and maximal and irreducible coincide. Here <3> is, but <2> and <4> and <5> are not. We can show 2=(1+i)(1-i) and 4=(2)(2) and 5=(2+i)(2-i) are reducible. To check 3 is irreducible, use the norm N(a+bi)=a^2+b^2, and check there are no elements of norm 3.

3. (a) False (try D=Z).

(b) False (factor (2)(5)=10=(2+sqrt(-6))(2-sqrt(-6))).

(c) False (try Z mapping to Z/6). Note however, that if the quotient of a PID is a domain at all, then it is a field and hence trivially a PID. (Since all prime ideals are maximal.)

(d) True. Let V={e,(12)(34),(13)(24),(14)(23)}. This is a normal subgroup and A4/V=Z3 since that is the only group of order 3.

4. For (a) use Eisenstein (p=3); for (b) reduce mod 2 and check that there are no roots and that it is not a multiple of x^2+x+1; for (c) use Eisenstein (p=2); for (d) reduce mod 3 and check there are no roots. Of course other approaches are possible.

5. Let K=Ker(phi), an ideal in F[x] consisting of all f such that f(a)=0 in E. Since F[x] is a PID, K=<g> for any nonzero polynomial g of lowest degree in K. Certainly the polynomial p is in K, so that means p=hg is a multiple of g. But p is irreducible, and g is not a unit, so h is a unit (a nonzero constant in F), and <p>=<g>=K.

Alternatively, we could note that <p> is a maximal ideal contained in K. But 1 is not in K, so K is not the whole ring, so K=<p>.

6. The characteristic of Z[i]/<a+bi> is the least positive integer n such that n times 1 is 0 in the quotient, ie, n is in the ideal <a+bi>. Clearly (a+bi)(a-bi)=a^2+b^2 is an integer in this ideal, so n is a factor of a^2+b^2.

7. (a) Z2+Z2+Z3=Z2+Z6 (not Z4+Z3=Z12).

(b) Z2+Z3+Z9=Z6+Z9=Z18+Z3 (not Z54=Z2+Z27 or Z3+Z3+Z6=Z3+Z3+Z3+Z2).

(c) Z5+Z5+Z5 (since the characteristic is 5).

(d) Z2+Z2 (since this is the only noncyclic group of order 4).