BRIEF ANSWERS FOR TEST 1

1. A mistake was made. Watson's alpha * beta has a fixed point d while Holmes's alpha * beta has no fixed points.

2. • We are given that [M,N] is a subgroup, so we need only show that it is normal. It is enough to show that a conjugate of a generator is a generator:

     g(m^(-1)n^(-1)mn)g^(-1) = (gmg^(-1))^(-1)(gng^(-1))^(-1)(gmg^(-1))(gng^(-1)).

   • It is enough to show that any generator is in M and also in N. The element m^(-1)n^(-1)mn = m^(-1)(n^(-1)mn) is the product of two elements of M, so is in M, and a similar argument shows that it is in N.
   • Since (m^(-1)n^(-1)mn)^(-1) = n^(-1)m^(-1)nm, we get that the generators of [N,M] are contained in [M,N]. This gives [N,M] contained in [M,N], and symmetry gives the other inclusion.

3. The projection homomorphisms pi_i: G_1 x G_2 --> G_i composed with the natural homomorphisms nu_i: G_i --> G_i/N_i produce two homomorphisms G_1 x G_2 --> G_i/N_i, hence one homomorphism G_1 x G_2 --> (G_1/N_1) x (G_2/N_2). Check that this homomorphism is onto with kernel N_1 x N_2, and then use the First Isomorphism Theorem.

4. Label the picture with N at the bottom, G at the top, H on the left, and K < L on the right.

   (Solution 1) Assign indices to each interval. Use the Second Isomorphism Theorem to prove that [L:K] = 1, so K = L. Thus the interval is not actually a pentagon. (This solution works for finite groups only.)

   (Solution 2) Choose g in L-K. Since g is in G = HK we can write g as hk with h in H and k in K. But then h is in H and also L (since h = g*k^(-1), which is in L). Thus h is in the intersection of H and L, which is N. Therefore g = hk with h in N and k in K. Since N is a subset of K, and K is a subgroup, this shows that g is in K. But we chose g in L-K! Contradiction. (This solution works for infinite groups.)