Now we claim that for each i=2..k, the group S_1 S_2 ... S_i is just the direct product S_1 x S_2 x .. x S_i. The case i=k will give the theorem. We need to do this inductively because the theorem we have is only about products of two subgroups.

The base case is i=1, where the claim is that S_1 = S_1. Sure enough.

The induction step is to show that S_1 S_2 ... S_i is an internal direct product (S_1 S_2 ... S_{i-1}) x S_i. By induction, we know that the group inside the parentheses is of order p_1^{e_1} * ... * p_{i-1}^{e_{i-1}}. That's relatively prime to the order of S_i, so their intersection must be 1. Since they're both normal, the group they generate is the direct product of the two.

Then we stick into (S_1 S_2 ... S_{i-1}) x S_i our inductively gained knowledge of the first factor, and we get the result for i. Then use i=k.

5.4#13. Prove that D_{8n} is not isomorphic to D_{4n} x Z_2.

In D_{2m} in general, the orders of the elements in the rotation half of the group are divisors of m, whereas the orders of the reflections are all 2.

In a product GxH, the order of (g,h) is the LCM of the individual orders.

Therefore the largest order that shows up in D_{8n} is 4n, whereas the largest order that shows up in D_{4n} x Z_2 is LCM(2n,2) = 2n.

(Note that if n is odd, and this question were about D_{4n} vs. D_{2n} x Z_2, this argument wouldn't work, because LCM(n,2) = 2n.)

5.4#15. If G/A and G/B are abelian, show G/(A intersect B) is abelian.

G/A abelian <=> for all gA,hA in G/A, (gA)(hA) = (hA)(gA) <=> for all g,h in G, ghA = hgA <=> for all g,h in G, g^{-1} h^{-1} g h A = A <=> for all g,h in G, g^{-1} h^{-1} g h is in A.

G/A and G/B are both abelian <=> for all g,h in G, g^{-1} h^{-1} g h is in both A and B <=> for all g,h in G, g^{-1} h^{-1} g h is in A intersect B <=> G/(A intersect B) is abelian.

5.5#2. Let G = H semidirect K, H the normal one. Show C_H(K) = N_H(K).

C_H(K) = {those h such that h k h^{-1} = k for all k in K}

N_H(K) = {those h such that h k h^{-1} in K for all k in K}

The second requirement looks weaker, so obviously C_H(K) is inside N_H(K). We need the opposite.

If h in N_H(K), and k in K, then
h k h^{-1} k^{-1} = (h k h^{-1}) k^{-1} = h (k h^{-1} k^{-1})
The first one lets us see this commutator as a product of two elements of K, the second as a product of two elements of H (this uses H normal). So it's in both H and K and hence the identity. Therefore h was in the centralizer.

5.5#6. Let phi_1,phi_2 be two maps of K (cyclic) into H with conjugate images; if K is infinite assume they are injective.

Let sigma conjugate phi_1(K) to phi_2(K), as in the hint. Let g generate K. Then phi_2(g) generates phi_2(K) = sigma phi_1(K) sigma^-1, in which one can find sigma phi_1(g) sigma^-1, so the latter is some power of phi_2(g). Call this power a, as in the hint. Now we have
sigma phi_1(g) sigma^-1 = phi_2(g)^a
as hinted. Since every k is some power of g, we can take that power of each side and get the same equation for every k.

Now we consider the map psi(h k) = sigma(h) k^a. This is obviously a homomorphism if we restrict to just h=1 or k=1; the only real thing to check is that it satisfies psi(kh) = psi(k) psi(h).
psi(k h) = psi( psi_1(k)(h) k ) = sigma(psi_1(k)(h)) k^a, vs.
psi(k) psi(h) = k^a sigma(h) = psi_2(k^a)(sigma(h)) k^a
so we need sigma o psi_1(k) = psi_2(k^a) o sigma, which we saw above.

Finally, we need to show this homomorphism is 1:1 and onto. Let's look for an inverse of the form
rho(h k) = sigma^-1(h) k^b.
Then psi(rho(hk)) = h k^{ab}, likewise rho(psi(hk)) = h k^{ab}. (Careful: these look like the same map, but they're really defined as endomorphisms of different groups using the same set of elements!).

Restricting to K, we want psi to be 1:1 and onto, i.e. raising to the ath power should be an invertible operation. If K is finite then sigma phi_1(g) sigma^-1 = phi_2(g)^a
we know that phi_1(g) and phi_2(g)^a have the same finite order, so a must be relatively prime to that order, and hence there exists a b such that ab=1.

Whereas if K is infinite, then the assumption phi_1,phi_2 injective says that we can define an isomorphism k |-> phi_1(k) |-> sigma phi_1(k) sigma^{-1} |-> phi_2^{-1}( sigma phi_1(k) sigma^{-1} ) of K to itself, which we know to be k |-> k^a. Since it's an isomorphism a is +/-1, so let b=a.