BRIEF ANSWERS FOR TEST 1

1. T or F?
   1. T. The real numbers 0 and 1 under multiplication form a monoid that is not a group (with 1 as unit element).
   2. F. This is a consequence of Lagrange's Theorem: |G| = [G:H]*|H|.
   3. F. The Cauchy number of this permutation is 15, so it is odd.
   4. T. If z is any complex number (other than 1) which satisfies z^3 = 1, then z generates a 3-element subgroup of C-{0}. We need a root of z^3 - 1 = 0 different from 1. Since z^3 - 1 = (z - 1)(z^2 + z + 1), it is enough to pick z so that it satisfies z^2 + z + 1 = 0. Using the quadratic formula we find that both (-1 +/- square_root(-3))/2 are suitable choices for z.

2. The basis step, testing n = 3, is easy since A_3 = {1, (1 2 3), (1 3 2)}. (Every element is a product of zero or one 3-cycles.)

   Assume, by induction, that every even permutation of {1, 2, ..., n} is a product of 3-cycles. Let's show that every even permutation of {1, 2, ..., n, n+1} is a product of 3-cycles. Let alpha be an even permutation {1, 2, ..., n+1}. If alpha(n+1) = n+1, then alpha may be thought of as a permutation of {1, 2, ..., n}, so it is a product of 3-cycles. Suppose instead that alpha(n+1) = j and that j is not equal to n+1. Since n is at least three, there is an i different from j and n+1. Set beta = (i j n+1)*alpha. Since beta(n+1) = n+1, our earlier conclusion yields that beta is a product of 3-cycles. Thus

   alpha = (i j n+1)^(-1)*beta = (j i n+1)*beta

   is a product of 3-cycles.

3. Since alpha is not 1, there exist distinct i and j such that alpha(i) = j. Since n is greater than 3, there is at least one k between 1 and n which is different from both i and j. Let beta = (jk). Now alpha*beta(i) = j which is not equal to beta*alpha(i) = k. Therefore alpha*beta is not equal to beta*alpha.

4. Apply Lagrange's Theorem to the additive subgroups of K and H. We obtain that |K| = q^m is a divisor of |F| = p^n. Therefore we must have q = p (and so |K| = p^m). Now apply Lagrange's Theorem to the multiplicative group of nonzero elements of K and H. We have that |K-{0}| = p^m-1 divides |F-{0}| = p^n-1. A short calculation shows that m must divide n.

   (Here is the "short calculation". By the division algorithm there are q and r such that n = qm + r and r is a nonegative number less than m. Substituting x = p^m into the identity

   (x - 1)(x^(q-1) + ... + 1) = x^q - 1

   and multiplying by p^r we get

   p^r(p^m - 1)((p^m)^(q-1) + ... + 1) = p^r((p^m)^q - 1) = p^n - p^r,

   so p^n - p^r is divisible by p^m - 1. Now, if p^n - 1 is divisible by p^m - 1 as well, then the difference (p^n - 1) - (p^n - p^r) = p^r - 1 is also divisible by p^m - 1. Since r is less than m, p^r - 1 is less than p^m -1, so the only way this can happen is if p^r - 1 = 0. Thus r = 0. It follows that m divides n.)

5. Assume that H and K are proper subgroups of G and that G is the union of H and K. We cannot have H contained in K, since then the union of H and K would be K and also G. But K is not equal to G. Thus we can find some x on H-K, and for a similar reason we can find some y in K-H. The element xy is in G = H union K. If xy is in H, then since both x and xy are in H we have y = x^(-1)(xy) in H. This contradicts the choice of y. Similarly if xy is in K, then we contradict the choice of x. Either way, we have a contradiction to the assumption that G is the union of H and K.

6. If G is nontrivial, then it contains an element g not equal to 1. The cyclic subgroup generated by g is nontrivial, so it cannot be proper, hence G is the cyclic subgroup generated by g. If the order of G is n and n is not prime, then there is a divisor d of n which is between 1 and n. The elements {1, g^d, g^(2d), ...} form a proper nontrivial subgroup (of size n/d). The hypotheses rule out the possibility of such a subgroup, so |G| is prime.