- Let
*G*be a simple group of order 2256 = 16*3*47. The number of Sylow 47-subgroups is congruent to 1 mod 47 and divides 48. Since*G*is simple, the number of Sylow 47-subgroups is not 1 (otherwise*G*would have a normal Sylow 47-subgroup), consequently*G*has 48 Sylow 47-subgroups. Next the number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 16*47. This number cannot be 1 because*G*is simple. Nor can it be 4, because then*G*would be isomorphic to a subgroup of*A*_{4}which only has order 12. We conclude that*G*has at least 16 Sylow 3-subgroups. Finally*G*has at least two Sylow 2-subgroups. Note that two distinct Sylow 47-subgroups*P*,*Q*intersect trivially, because |*P*/\*Q*| divides 47 and is not 47. Similarly two distinct Sylow 3-subgroups intersect trivially. On the other hand we cannot assert that two distinct Sylow 2-subgroups intersect trivially.We now count elements. The number of elements of order 47 is 46*48; the number of elements of order 3 is at least 2*16, and the number of elements of order a power of two is at least 17. We now have at least 17 + 2*16 + 46*48 = 2257 elements in

*G*, which is too many. This proves that there is no simple group of order 2256. - (i)
- Every element of
*G*can be written in the form*x*^{a1}*y*^{b1}...*x*^{an}*y*^{bn}, where*a*_{i},*b*_{i}**Z**. Using*x*^{7}=*y*^{3}= 1, we may assume that*a*_{i},*b*_{i}__>__0 for all*i*. Now whenever there is a subword*yx*, we can replace it with*xy*^{2}, because*yxy*^{-1}=*x*^{2}, so we may push all the*x*'s to the left and all the*y*'s to the right. This proves the result. - (ii)
- Using (i) and
*x*^{7}=*y*^{3}= 1, we may write every element of*G*in the form*x*^{i}*y*^{j}where 0__<__*i*__<__6 and 0__<__*j*__<__2. It follows that |*G*|__<__7*3 = 21. - (iii)
- This is because
q respects the relations. Specifically,
(1 3 4 5 6 7)
^{7}= 1, ((2 3 5)(4 7 6))^{3}= 1,((2 3 5)(4 7 6))(1 2 3 4 5 6 7)((2 3 5)(4 7 6))^{-1}= (1 3 5 7 2 4 6). - (iv)
- We see from (iii) that
imq has elements of order 7 and 3.
Therefore 21 divides
| imq| and hence 21 divides |
*G*|. Since |*G*|__<__21 from (ii), we deduce that |*G*| = 21 as required.

- We may assume that
*a*,*b*=/= 0. Let*d*be the greatest common divisor of*a*,*b*. We claim that*a*/*d*,*b*/*d*are coprime. If*r*divides*a*/*d*,*b*/*d*, then*rd*divides*a*,*b*and hence*rd*divides*d*, say*srd*=*d*for some*s**R*. Therefore*sr*= 1, which means that*r*is a unit and the claim is established. By hypothesis (*a*/*d*)*R*+ (*b*/*d*)*R*=*xR*for some*x**R*. Then*aR*+*bR*=*xdR*and the result follows. - Note that
*M*@**Z**/2**Z**as*R*-modules, so we need to prove**Z**/2**Z****Z**/2**Z**@**Z**/2**Z**. Define q :**Z**/2**Z****Z**/2**Z**- >**Z**/2**Z**by q(*x*,*y*) =*xy*. It is easily verified that q is an*R*-balanced map and therefore q induces an*R*-module map q' :**Z**/2**Z****Z**/2**Z**- >**Z**/2**Z**satisfying q(*x**y*) =*xy*. Now define an*R*-map f :**Z**/2**Z**- >**Z**/2**Z****Z**/2**Z**by f*x*=*x*1. Then q'f*x*= q'(*x*1) =*x*, so q'f is the identity. Also fq'(*x**y*) = f(*xy*) =*xy*1 =*x*1. Therefore fq' is the identity on a set of generators of**Z**/2**Z****Z**/2**Z**(in this case these generators*x**y*are the whole of**Z**/2**Z****Z**/2**Z**, but that won't always be the case) and we deduce that fq' is also the identity. This proves that**Z**/2**Z****Z**/2**Z**@**Z**/2**Z**as required. - Note that
*N*is finitely generated. This is because*M*is finitely generated implies*M**M*is finitely generated, which implies (*M**M*)/(*N*0) @*N*is finitely generated.We now apply the structure theorem for finitely generated modules over a PID. We may write

*M*@ *R*^{a}(*R*/*q*_{1}*R*)^{a1}... (*R*/*q*_{n}*R*)^{an}*N*@ *R*^{b}(*R*/*q*_{1}*R*)^{b1}... (*R*/*q*_{n}*R*)^{bn}

where*n*,*a*,*b*,*a*_{i},*b*_{i}are non-negative integers and the*q*_{i}are nonassociate prime powers. Since*M**M*@*N**N*, we see that*R*^{2a}(*R*/*q*_{1}*R*)^{2a1}... (*R*/*q*_{n}*R*)^{2an}@*R*^{2b}(*R*/*q*_{1}*R*)^{2b1}... (*R*/*q*_{n}*R*)^{2bn}.*a*= 2*b*, 2*a*_{1}= 2*b*_{1}, ..., 2*a*_{n}= 2*b*_{n}. Therefore*a*=*b*,*a*_{1}=*b*_{1}, ...,*a*_{n}=*b*_{n}and the result follows. - (i)
- Note that
*f*has degree 4. Since a_{i}satisfies*f*, we have [*K*(a_{i}) :*K*]__<__4. Therefore [*K*(a_{i},a_{j}) :*K*]__<__4*4 = 16, in particular*K*(a_{i},a_{j}) =/=*L*as required. - (ii)
- Assume that
*f*is monic. Then if*a**K*is the coefficient of*x*^{3}of*f*, we have a_{1}+ a_{2}+ a_{3}+ a_{4}= -*a*. Since*L*is the splitting field over*K*of the polynomial*f*and we are in characteristic zero,*L*is a Galois extension of*K*. Let*G*denote the Galois group of*L*over*K*. Then the subfields of*L*containing*K*are in a one-to-one correspondence with the subgroups of*G*, and*G*permutes the a_{i}faithfully. If*K*(a_{1}+ 2a_{2}+ 3a_{3}) =/=*L*, then*K*(a_{1}+ 2a_{2}+ 3a_{3}) corresponds to a nontrivial subgroup of*G*, in particular a_{1}+ 2a_{2}+ 3a_{3}is fixed by some nontrivial permutation of the a_{i}. In other words there is 1 =/= s*S*_{4}such thataBy inspection, this will yield a different linear relation from a_{1}+ 2a_{2}+ 3a_{3}= a_{s1}+ 2a_{s2}+ 3a_{s3}._{1}+ a_{2}+ a_{3}+ a_{4}= -*a*from above. (Clearly this relation is not 0 = 0. If we subtract the right hand side from the left hand side, then the sum of the coefficients of the a_{i}is 0.) Therefore we may express the a_{i}in terms of just two of the a_{i}; in other words,*L*=*K*(a_{i},a_{j}) for some*i*,*j*, which contradicts (i). Finally*K*[a_{1}+ 2a_{2}+ 3a_{3}] =*K*(a_{1}+ 2a_{2}+ 3a_{3}) because a_{1}+ 2a_{2}+ 3a_{3}is algebraic over*K*and the result follows.

- If the assertion is false, then
I(
*V*_{1}) + I(*V*_{2}) is contained in a maximal ideal*M*of*k*[*x*_{1},...,*x*_{n}]. By Hilbert's Nullstellensatz,*M*is of the form (*x*_{1}-*a*_{1},...,*x*_{n}-*a*_{n}), where*a*_{i}*k*for all*i*. Since*f*(*a*_{1},...,*a*_{n}) = 0 for all*f**M*, we see that*f*(*a*_{1},...,*a*_{n}) = 0 for all*f*I(*V*_{1}) and*f*I(*V*_{2}). Therefore (*a*_{1},...,*a*_{n})*V*_{1}/\*V*_{2}which is a contradiction, and the result follows.