- We shall use the fundamental theorem for finitely generated
abelian groups. We may write
*A*= ⊕_{i=1}^{n}(**Z**/*q*_{i}**Z**)^{ai},*B*= ⊕_{i=1}^{n}(**Z**/*q*_{i}**Z**)^{bi},*C*= ⊕_{i=1}^{n}(**Z**/*q*_{i}**Z**)^{ci}*a*_{i},*b*_{i},*c*_{i},*n*are nonnegative integers, and the*q*_{i}are distinct prime powers. Then*A*⊕*B*≌*A*⊕*C*yields⊕The fundamental theorem now shows that_{i=1}^{n}(**Z**/*q*_{i}**Z**)^{ai+bi}≌⊕_{i=1}^{n}(**Z**/*q*_{i}**Z**)^{ai+ci}*a*_{i}+*b*_{i}=*a*_{i}+*c*_{i}and hence*b*_{i}=*c*_{i}for all*i*. It follows that*B*≌*C*as required. - Suppose by way of contradiction
*G*is a simple group of order 56. Sylow's theorem for the prime 7 shows that the number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 8, hence the number of Sylow 7-subgroups is 1 or 8. If there is one Sylow 7-subgroup, x then this subgroup must be normal in*G*which contradicts the hypothesis that*G*is simple, consequently*G*has 8 Sylow 7-subgroups.Let

*A*,*B*be two distinct Sylow 7-subgroups. Then*A*∩*B*is a subgroup of*A*, so by Lagrange's theorem |*A*∩*B*| divides |*A*|, hence |*A*∩*B*| = 1 or 7. Now*A*and*B*both have order 7, so we cannot have |*A*∩*B*| = 7. Therefore we must have |*A*∩*B*| = 1. Since every nonidentity element of a Sylow 7-subgroup has order 7, we deduce that*G*has at least 6×8 elements of order 7.Now every Sylow 2-subgroup has order 8, and every element of a Sylow 2-subgroup has order a power of 2 by Lagrange's theorem, so if

*G*has at least two Sylow 2-subgroups, then*G*has at least 9 elements of order a power of 2. Since we have already shown that*G*has at least 48 elements of order 7, we now have that*G*has at least 9 + 48 = 57 elements, which is not possible because |*G*| = 56. Therefore*G*has exactly one Sylow 2-subgroup, and so this Sylow subgroup must be normal which contradicts the hypothesis that*G*is simple. - Since we are working over
**C**, an algebraically closed field, we may use the Jordan canonical form. Here every matrix has a unique Jordan canonical form, and two canonical forms are in the same equivalence class of*T*if and only if they are equal. Since the eigenvalues of a matrix in*T*are 4,4,17,17,17, the Jordan canonical form of such a matrix must look like( 4 a 0 0 0 0 4 0 0 0 0 0 17 *b*0 0 0 0 17 *c*0 0 0 0 17 ) where

*a*,*b*,*c*are 1 or 0, and*b*= 0 if*c*= 0. Therefore there are**6**equivalence classes in*T*. - There are many answers to this problem; perhaps the simplest
example of a UFD which is not a PID is
**Z**[*X*]. This is a UFD because**Z**is a UFD, and a polynomial ring over UFD is again a UFD. We now establish that**Z**[*X*] is not a PID by showing that the ideal (2,*X*) (the ideal generated by 2 and*X*) is not principal.Suppose on the contrary that (2,

*X*) =*f*for some*f*∈**Z**[*X*]. Then there exist*g*,*h*∈**Z**[*X*] such that*fg*= 2 and*fh*=*X*. The equation*fg*= 2 shows that*f*must have degree 0, in other words*f*∈**Z**, and then*fh*=*X*shows that*f*= ±1. Thus we must have (2,*X*) =**Z**[*X*]. Now the general element of (2,*X*) is 2*a*+*Xb*where*a*,*b*∈**Z**[*X*]. The constant term of 2*a*must be divisible by 2, and the constant term of*Xb*must be zero, hence the constant term of 2*a*+*Xb*must be divisible by 2. In particular we cannot have 2*a*+*Xb*= 1, so 1 ∉ (2,*X*) and we have the required contradiction. - Let
*G*denote the Galois group of*x*^{3}- 10 over**Q**. By Eisenstein's criterion for the prime 2 (or otherwise), we see that*x*^{3}- 10 is irreducible over**Q**, hence*G*≌*S*_{3}or*A*_{3}. Let ω = (-1 +*i*√3)/2, a primitive cube root of unity. Then the roots of*x*^{3}- 10 are^{3}√10, ω^{3}√10, and^{3}√10 ( = (-1 -*i*√3)/2), hence*x*^{3}- 10 has one real root and two complex roots, so complex conjugation is an element of*G*. Therefore*G*has an element of order 2, which rules out the possibility*G*≌*A*_{3}. Therefore*G*≌*S*_{3}.A splitting field of

*x*^{3}- 10 is**Q**(*i*√3,^{3}√10). The normal subfields of the splitting field correspond to normal subgroups of*G*. The normal subfields corresponding to the subgroups 1 and*G*are**Q**(*i*√3,^{3}√10) and**Q**respectively.*G*has exactly one other normal subgroup, namely*A*_{3}, so there is exactly one other normal subfield. Since subfields of degree two over**Q**are always normal, this other normal subfield must be**Q**(*i*√3). - (a) How to prove this depends on how much field theory one is
allowed to assume. Also the result is true without the hypothesis
that
*f*is separable. Here is one way to proceed. Suppose*K*_{1},*K*_{2}are fields, θ :*K*_{1}->*K*_{2}is an isomorphism,*g*∈*K*_{1}[*X*] is an irreducible polynomial, α_{1}is a root of*g*in a splitting field*L*_{1}for*g*, and α_{2}is a root of the irreducible polynomial θ*g*in a splitting field*L*_{2}for θ*g*(recall if*g*=*a*_{0}+*a*_{1}*X*+ ... +*a*_{n}*X*^{n}with*a*_{i}∈*K*_{1}, then θ*g*= θ*a*_{0}+ θ*a*_{1}*X*+ ... + θ*a*_{n}*X*^{n}∈*K*_{2}[*X*]). Then θ extends to an isomorphism φ :*K*_{1}(α_{1}) ->*K*_{2}(α_{2}) such that φ(α_{1}) = φ(α_{2}). Using induction on the degree of*g*, we deduce that φ in turn extends to an isomorphism of*L*_{1}onto*L*_{2}. This is what is required.(b) Note that

*x*^{4}- 2 is indeed irreducible over**Q**, by Eisenstein's criterion for the prime 2. The roots of*x*^{4}- 2 are ±^{4}√2 and ±*i*^{4}√2. Consider the permutation of the roots^{4}√2 ->^{4}√2, -^{4}√2 ->*i*^{4}√2 -> -*i*^{4}√2 -> -^{4}√2. This cannot be induced by an element θ of the Galois group of*x*^{4}- 2 over**Q**, because if θ^{4}√2 =^{4}√2, then we must have θ(-^{4}√2) = -^{4}√2. - Since
Hom
_{A}(*M*,*k*) = 0, we must have*M*= M*M*. Then Nakayama's lemma yields the result.

Peter Linnell 2011-08-15