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

Solution:
A polynomial of degree [tex]2[/tex] over a field is reducible exactly when it has a root in that field, because a nontrivial factorization consists of two linear factors. So we look for the roots of [tex]x^2+1[/tex] in [tex]\mathbb{Z}_5=\{0, 1, 2, 3, 4\}[/tex]:
[tex]0^2+1=1,\quad 1^2+1=2,\quad 2^2+1=5\equiv 0,\quad 3^2+1=10\equiv 0,\quad 4^2+1=17\equiv 2 \pmod 5[/tex]
The roots are [tex]2[/tex] and [tex]3[/tex]. Since [tex]-2\equiv 3[/tex] and [tex]-3\equiv 2 \pmod 5[/tex], we get
[tex]x^2+1=(x-2)(x-3)=(x+3)(x+2)[/tex]
Check:
[tex](x+2)(x+3)=x^2+5x+6\equiv x^2+1 \pmod 5[/tex]
Linear factors are always irreducible, so this is the required factorization.

Solution:
All seven maps take nonzero real values, so they map [tex]\mathbb{C}^*[/tex] to [tex]\mathbb{R}^*[/tex]. A map [tex]f[/tex] is a homomorphism exactly when [tex]f(zw)=f(z)f(w)[/tex] for all [tex]z,w\in\mathbb{C}^*[/tex]. Since [tex]|zw|=|z|\,|w|[/tex], the maps [tex]f(z)=|z|[/tex], [tex]f(z)=\frac{1}{|z|}[/tex], [tex]f(z)=|z|^2[/tex] and the constant [tex]f(z)=1[/tex] (the trivial homomorphism) are homomorphisms:
[tex]|zw|=|z|\,|w|,\qquad \frac{1}{|zw|}=\frac{1}{|z|}\times \frac{1}{|w|},\qquad |zw|^2=|z|^2\,|w|^2,\qquad 1=1\times 1[/tex]
Every homomorphism maps the identity element to the identity element: [tex]f(1)=f(1\times 1)=f(1)^2[/tex] and [tex]f(1)\ne 0[/tex], hence [tex]f(1)=1[/tex]. The other three maps do not satisfy this: for [tex]f(z)=2|z|[/tex], [tex]f(z)=1+|z|[/tex] and [tex]f(z)=2[/tex] we get [tex]f(1)=2\ne 1[/tex], so they are not homomorphisms.
Hence exactly [tex]4[/tex] of the maps are homomorphisms.

Solution:
We decompose [tex]\sigma[/tex] into disjoint cycles by following the images of each element:
[tex]0\to 1\to 0,\quad 2\to 4\to 7\to 2,\quad 3\to 8\to 3,\quad 5\to 5,\quad 6\to 9\to 6[/tex]
Hence ([tex]5[/tex] is a fixed point)
[tex]\sigma=(0\,1)(2\,4\,7)(3\,8)(6\,9)[/tex]
Disjoint cycles commute, so
[tex]\sigma^n=(0\,1)^n(2\,4\,7)^n(3\,8)^n(6\,9)^n[/tex]
This is the identity exactly when each cycle raised to the power [tex]n[/tex] is the identity. A cycle of length [tex]k[/tex] has order [tex]k[/tex], so [tex]n[/tex] must be divisible by [tex]2[/tex] and by [tex]3[/tex]. Hence the order of [tex]\sigma[/tex] is
[tex]\operatorname{lcm}(2, 3, 2, 2)=6[/tex]

Solution:
An isomorphism [tex]f\colon G\to H[/tex] of additive groups preserves the order of every element. Indeed, [tex]f(ng)=n\,f(g)[/tex] because [tex]f[/tex] is a homomorphism, and [tex]f(x)=0[/tex] only for [tex]x=0[/tex] because [tex]f[/tex] is injective; hence for every natural [tex]n[/tex]
[tex]ng=0\iff f(ng)=0\iff n\,f(g)=0[/tex]
i.e. [tex]g[/tex] and [tex]f(g)[/tex] have the same order.
In [tex]\mathbb{Z}_4[/tex] the element [tex]1[/tex] has order [tex]4[/tex], because [tex]k\times 1=k\ne 0[/tex] for [tex]k=1, 2, 3[/tex] and [tex]4\times 1=0[/tex]. In [tex]\mathbb{Z}_2×\mathbb{Z}_2[/tex] every element [tex]\left(a, b\right)[/tex] satisfies
[tex]2\left(a, b\right)=\left(2a, 2b\right)=\left(0, 0\right)[/tex]
so every element has order [tex]1[/tex] or [tex]2[/tex], and there is no element of order [tex]4[/tex]. If there were an isomorphism [tex]f\colon\mathbb{Z}_4\to\mathbb{Z}_2×\mathbb{Z}_2[/tex], then [tex]f(1)[/tex] would be an element of order [tex]4[/tex] in [tex]\mathbb{Z}_2×\mathbb{Z}_2[/tex], which is impossible. Hence the groups are not isomorphic ([tex]\mathbb{Z}_4[/tex] is cyclic, while [tex]\mathbb{Z}_2×\mathbb{Z}_2[/tex] is not).

Solution:
Let [tex]e[/tex] be the identity element of [tex]G[/tex]. The element [tex]ba[/tex] is a conjugate of [tex]ab[/tex]:
[tex]a^{-1}(ab)a=(a^{-1}a)ba=ba[/tex]
Hence, for every natural number [tex]n[/tex], all inner products [tex]a\,a^{-1}[/tex] cancel and
[tex](ba)^n=\left(a^{-1}(ab)a\right)\left(a^{-1}(ab)a\right)\ldots\left(a^{-1}(ab)a\right)=a^{-1}(ab)^n a[/tex]
Multiplying on the left by [tex]a[/tex] and on the right by [tex]a^{-1}[/tex], we get
[tex](ba)^n=e\iff a^{-1}(ab)^n a=e\iff (ab)^n=e[/tex]
So the sets [tex]\{n\in\mathbb{N}\mid (ab)^n=e\}[/tex] and [tex]\{n\in\mathbb{N}\mid (ba)^n=e\}[/tex] coincide. If they are empty, both elements have infinite order; otherwise their smallest elements coincide, i.e. [tex]ab[/tex] and [tex]ba[/tex] have the same finite order.