Abstract Algebra: Groups, Rings and Fields: Very Difficult Problems with Solutions

Solution:
For [tex]n=1[/tex] there is nothing to prove, so let [tex]n\ge 2[/tex]. For a divisor [tex]d[/tex] of [tex]n[/tex] let [tex]N(d)[/tex] be the number of elements of order [tex]d[/tex] in [tex]G[/tex]. By Lagrange's theorem the order of every element divides [tex]n[/tex], so
[tex]\sum_{d\mid n}N(d)=n[/tex]
Let [tex]d\mid n[/tex], [tex]d\lt n[/tex], and suppose that [tex]G[/tex] has an element [tex]x[/tex] of order [tex]d[/tex]. Every element [tex]y[/tex] of order [tex]d[/tex] generates a subgroup [tex]\langle y\rangle[/tex] of order [tex]d[/tex], which by hypothesis coincides with [tex]\langle x\rangle[/tex]. So the elements of order [tex]d[/tex] are exactly the generators of the cyclic group [tex]\langle x\rangle[/tex], and there are [tex]\varphi(d)[/tex] of them, where [tex]\varphi[/tex] is Euler's totient function. Thus [tex]N(d)=0[/tex] or [tex]N(d)=\varphi(d)[/tex], and in every case
[tex]N(d)\le\varphi(d)\qquad\left(d\mid n,\ d\lt n\right)[/tex]
Using Gauss's identity [tex]\sum_{d\mid n}\varphi(d)=n[/tex], we get
[tex]n=\sum_{d\mid n}N(d)\le\sum_{d\mid n,\ d\lt n}\varphi(d)+N(n)=n-\varphi(n)+N(n)[/tex]
Therefore [tex]N(n)\ge\varphi(n)\ge 1[/tex], i.e. [tex]G[/tex] contains an element [tex]g[/tex] of order [tex]n[/tex]. Then [tex]\langle g\rangle[/tex] has [tex]n=|G|[/tex] elements, so [tex]G=\langle g\rangle[/tex] is cyclic.

Solution:
First we show that [tex]x^3=x[/tex] for every [tex]x\in G[/tex]. Apply the property to the element [tex]ax[/tex]:
[tex](ax)^3=a(ax)a\ \Rightarrow\ a(xaxax)=a(axa)[/tex]
Cancelling [tex]a[/tex] on the left gives [tex]xaxax=axa=x^3[/tex]. On the other hand, [tex]xaxax=x(axa)x=x\,x^3x=x^5[/tex], so [tex]x^5=x^3[/tex], i.e. [tex]x^2x^3=x^2x[/tex], and cancelling [tex]x^2[/tex] on the left gives
[tex]x^3=x[/tex]
Consequently [tex]axa=x^3=x[/tex] for every [tex]x[/tex], and in particular [tex]a^3=a[/tex]. Now [tex]e=a^2[/tex] is an identity element: for every [tex]x[/tex]
[tex]a^2x=a^2(axa)=a^3xa=axa=x,\qquad xa^2=(axa)a^2=axa^3=axa=x[/tex]
Every element is its own inverse: from [tex]x\,x^2=x^3=x=xe[/tex], cancelling [tex]x[/tex] on the left, we get [tex]x^2=e[/tex], i.e. [tex]x^{-1}=x[/tex]. Thus [tex]G[/tex] is a group (the operation is associative, and there is an identity element and inverses). Finally, for [tex]x,y\in G[/tex]
[tex]xy=(xy)^{-1}=y^{-1}x^{-1}=yx[/tex]
so the group [tex]G[/tex] is abelian.

Solution:
First, [tex]\varphi(1)=1[/tex]. Indeed, [tex]\varphi(1)=\varphi(1^2)=\varphi(1)^2[/tex], so [tex]\varphi(1)\in\{0, 1\}[/tex]. If [tex]\varphi(1)=0[/tex], then [tex]\varphi(x)=\varphi(x)\varphi(1)=0[/tex] for all [tex]x[/tex], which contradicts [tex]\varphi\ne 0[/tex].
Next, [tex]\varphi(q)=q[/tex] for every rational [tex]q[/tex]. Additivity gives [tex]\varphi(0)=0[/tex], [tex]\varphi(-x)=-\varphi(x)[/tex] and [tex]\varphi(m)=m\varphi(1)=m[/tex] for every integer [tex]m[/tex]. For [tex]q=\frac{m}{k}[/tex] with [tex]m\in\mathbb{Z}[/tex], [tex]k\in\mathbb{N}[/tex]:
[tex]k\varphi(q)=\varphi(kq)=\varphi(m)=m\ \Rightarrow\ \varphi(q)=\frac{m}{k}=q[/tex]
The map [tex]\varphi[/tex] is non-decreasing. If [tex]t\ge 0[/tex], then [tex]t=s^2[/tex] with [tex]s=\sqrt{t}[/tex], so [tex]\varphi(t)=\varphi(s)^2\ge 0[/tex]. Hence, for [tex]x\le y[/tex],
[tex]\varphi(y)-\varphi(x)=\varphi(y-x)\ge 0[/tex]
Finally, suppose that [tex]\varphi(x)\lt x[/tex] for some [tex]x[/tex]. Since [tex]\mathbb{Q}[/tex] is dense in [tex]\mathbb{R}[/tex], there is a rational [tex]q[/tex] with [tex]\varphi(x)\lt q\lt x[/tex]. As [tex]q\lt x[/tex], monotonicity gives [tex]q=\varphi(q)\le\varphi(x)[/tex], a contradiction. If [tex]\varphi(x)\gt x[/tex], choose a rational [tex]q[/tex] with [tex]x\lt q\lt\varphi(x)[/tex]; then [tex]\varphi(x)\le\varphi(q)=q[/tex], again a contradiction. Therefore
[tex]\varphi(x)=x\quad\forall x\in\mathbb{R}[/tex]
(So the only ring homomorphisms [tex]\mathbb{R}\to\mathbb{R}[/tex] are the zero map and the identity.)

Solution:
For [tex]p=2[/tex] the binomial coefficients [tex]4[/tex], [tex]6[/tex], [tex]4[/tex] are even, so in [tex]\mathbb{Z}_2[x][/tex]
[tex]x^4+1=(x+1)^4[/tex]
Let [tex]p[/tex] be odd. In the multiplicative group [tex]\mathbb{Z}_p^*[/tex] of order [tex]p-1[/tex] the map [tex]x\mapsto x^2[/tex] is a homomorphism with kernel [tex]\{1, -1\}[/tex], so the set [tex]Q[/tex] of non-zero squares is a subgroup of index [tex]2[/tex]. The quotient group [tex]\mathbb{Z}_p^*/Q[/tex] has two elements, hence the product of two non-squares is a square. Applying this to [tex]-1[/tex] and [tex]2[/tex], we see that at least one of the numbers [tex]-1[/tex], [tex]2[/tex], [tex]-2=(-1)\times 2[/tex] is a square in [tex]\mathbb{Z}_p[/tex].
If [tex]-1=a^2[/tex], then
[tex]x^4+1=x^4-a^2=(x^2-a)(x^2+a)[/tex]
If [tex]2=b^2[/tex], then
[tex]x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(bx)^2=(x^2-bx+1)(x^2+bx+1)[/tex]
If [tex]-2=c^2[/tex], then
[tex]x^4+1=(x^2-1)^2+2x^2=(x^2-1)^2-(cx)^2=(x^2-cx-1)(x^2+cx-1)[/tex]
In every case [tex]x^4+1[/tex] is a product of two polynomials of degree [tex]2[/tex] over [tex]\mathbb{Z}_p[/tex], so it is reducible over [tex]\mathbb{Z}_p[/tex] for every prime [tex]p[/tex]. (Over [tex]\mathbb{Q}[/tex] none of the numbers [tex]-1[/tex], [tex]2[/tex], [tex]-2[/tex] is a square, which is why these factorizations do not work there.)

Solution:
We have [tex]4036=2^2\times 1009[/tex], and [tex]1009[/tex] is prime. By the Chinese remainder theorem [tex]\mathbb{Z}_{4036}\cong\mathbb{Z}_4×\mathbb{Z}_{1009}[/tex] as rings, so for the group of units
[tex]G\cong\mathbb{Z}_4^*×\mathbb{Z}_{1009}^*\cong C_2×C_{1008}[/tex]
Here [tex]\mathbb{Z}_4^*=\{1, 3\}\cong C_2[/tex], and [tex]\mathbb{Z}_{1009}^*\cong C_{1008}[/tex], because the multiplicative group of a finite field is cyclic. Since [tex]1008=2^4\times 3^2\times 7[/tex], we have [tex]C_{1008}\cong C_{16}×C_9×C_7[/tex], so
[tex]G\cong(C_2×C_{16})×C_9×C_7[/tex]
This is the decomposition of [tex]G[/tex] into its Sylow subgroups. In a finite abelian group every subgroup [tex]H[/tex] is the product of its intersections with the Sylow subgroups: each [tex]h\in H[/tex] splits uniquely as [tex]h=h_2h_3h_7[/tex], and each [tex]h_p[/tex] is a power of [tex]h[/tex], so [tex]h_p\in H[/tex]. Hence, denoting by [tex]s(K)[/tex] the number of subgroups of a group [tex]K[/tex],
[tex]s(G)=s(C_2×C_{16})\,s(C_9)\,s(C_7)[/tex]
A cyclic group of order [tex]m[/tex] has exactly one subgroup for each divisor of [tex]m[/tex], so [tex]s(C_9)=3[/tex] and [tex]s(C_7)=2[/tex]. Now let [tex]C_2×C_{16}=\langle a\rangle×\langle b\rangle[/tex]. For [tex]1\le k\le 4[/tex] the number of elements whose order divides [tex]2^k[/tex] is [tex]2\times 2^k[/tex], so there are [tex]3[/tex], [tex]4[/tex], [tex]8[/tex], [tex]16[/tex] elements of order [tex]2[/tex], [tex]4[/tex], [tex]8[/tex], [tex]16[/tex], respectively. A cyclic subgroup of order [tex]2^k[/tex] has [tex]\varphi(2^k)[/tex] generators, so the number of cyclic subgroups is
[tex]1+\frac{3}{1}+\frac{4}{2}+\frac{8}{4}+\frac{16}{8}=1+3+2+2+2=10[/tex]
A non-cyclic subgroup [tex]H[/tex] is (by the structure theorem for finite abelian groups) a product of two non-trivial cyclic [tex]2[/tex]-groups, so it has at least [tex]3[/tex] elements of order [tex]2[/tex]; hence it contains all three elements of order [tex]2[/tex] of [tex]C_2×C_{16}[/tex], in particular [tex]a[/tex]. If [tex]a\in H[/tex], then [tex]a^eb^m\in H[/tex] implies [tex]b^m\in H[/tex], so [tex]H=\langle a\rangle×K[/tex] with [tex]K=H\cap\langle b\rangle[/tex]; such a subgroup is non-cyclic exactly when [tex]K[/tex] is non-trivial. This gives [tex]4[/tex] non-cyclic subgroups (one for each [tex]K[/tex] of order [tex]2[/tex], [tex]4[/tex], [tex]8[/tex], [tex]16[/tex]). Therefore
[tex]s(C_2×C_{16})=10+4=14,\qquad s(G)=14\times 3\times 2=84[/tex]

Solution:
Expanding the binomial, we get
[tex](x+y)^5-x^5-y^5=5xy(x+y)(x^2+xy+y^2)[/tex]
If [tex]\operatorname{char}F=5[/tex], then [tex]5=0[/tex] in [tex]F[/tex], so the identity holds in every field of characteristic [tex]5[/tex]. Let [tex]\operatorname{char}F\ne 5[/tex]. Then [tex]5\ne 0[/tex] and, since a field has no zero divisors, we need [tex]xy(x+y)(x^2+xy+y^2)=0[/tex] for all [tex]x,y[/tex]. For [tex]x=y=1[/tex] this gives [tex]1\times 1\times 2\times 3=6=0[/tex], so [tex]\operatorname{char}F\in\{2, 3\}[/tex]; in particular, no field of characteristic [tex]0[/tex] works.
Characteristic [tex]3[/tex]: here [tex]x^2+xy+y^2=x^2-2xy+y^2=(x-y)^2[/tex]. Taking [tex]x=1[/tex], we need [tex]y(1+y)(1-y)^2=0[/tex] for all [tex]y\in F[/tex], so every element of [tex]F[/tex] is [tex]0[/tex], [tex]1[/tex] or [tex]-1[/tex], i.e. [tex]F=\mathbb{F}_3[/tex]. Conversely, in [tex]\mathbb{F}_3[/tex] any two non-zero elements satisfy [tex]y=\pm x[/tex], so [tex]x+y=0[/tex] or [tex]x-y=0[/tex], and the identity holds.
Characteristic [tex]2[/tex]: taking [tex]x=1[/tex], every [tex]y\in F[/tex] must be a root of the polynomial [tex]y(y+1)(y^2+y+1)[/tex] of degree [tex]4[/tex], so [tex]|F|\le 4[/tex], i.e. [tex]F=\mathbb{F}_2[/tex] or [tex]F=\mathbb{F}_4[/tex]. Both work: if [tex]x,y\ne 0[/tex] and [tex]x=y[/tex], then [tex]x+y=2x=0[/tex]; in [tex]\mathbb{F}_4[/tex], if [tex]x\ne y[/tex] are non-zero, then [tex]x^3=y^3=1[/tex] (the group [tex]\mathbb{F}_4^*[/tex] has order [tex]3[/tex]), so
[tex](x-y)(x^2+xy+y^2)=x^3-y^3=0\ \Rightarrow\ x^2+xy+y^2=0[/tex]
Answer: all fields of characteristic [tex]5[/tex], and also [tex]\mathbb{F}_2[/tex], [tex]\mathbb{F}_3[/tex] and [tex]\mathbb{F}_4[/tex].