Exercises: Groups and Group Homomorphisms

Solution. $S_4.$

Solution. Let $G=\mathbb{Z}$ and the operation be the ordinary addition. Take $H$ be the set of positive integers. Then $H$ is a subset of $G$ and is closed under the ordinary addition. But $H$ is not a group, because any element of $H$ does not have an inverse element in $H.$

Solution. If $G$ is commutative, then $$(st{)}^{-1}=(ts{)}^{-1}=s^{-1}{t}^{-1}.$$ Now we prove the converse. Suppose (*) holds. Let $x,$ $y$ be elements of $G.$ Since $G$ is a group, the inverse elements $x^{-1}$ and $y^{-1}$ exist in $G.$ Therefore, taking $s=x^{-1}$ and $t=y^{-1},$ we have $$xy=(x^{-1}{)}^{-1}(y^{-1}{)}^{-1}=(x^{-1}{y}^{-1}{)}^{-1}=((yx{)}^{-1}{)}^{-1}=yx$$ by (*), that is, $G$ is a commutative group.

Solution.

Solution.

• It is trivial that $G_0\times G_1$ is closed under the operation.
• Let $e_0$ and $e_1$ be the identity elements of $G_0$ and $G_1$ respectively. Then $(e_0,e_1)$ is the identity element of $G_0\times G_1.$
• Associativity directly comes from that of $G_0$ and $G_1.$
• Let $(s,t)$ be in $G_0\times G_1,$ then $(s^{-1},t^{-1})$ is the inverse element of $(s,t)$, where $s^{-1}\in G_0$ and $t^{-1}\in G_1.$

Solution. Take $\varphi ((s,t))=(t,s)$ then $\varphi$ is an isomorphism from $G_0\times G_1$ to $G_1\times G_0.$

Solution. WLOG, we only have to prove for ${\rho }_0.$

• Let $e_1$ be the identity element of $G_1.$ For any $s\in G_0,$ ${\rho }_0(s,e_1)=s,$ that is, ${\rho }_0$ is surjective.
• Let $(s_0,s_1)$ and $(t_0,t_1)$ be elements of $G_0\times G_1.$ Then $${\rho }_0((s_0,s_1)(t_0,t_1))={\rho }_0((s_0{t}_0,s_1{t}_1))=s_0{t}_0={\rho }_0((s_0,s_1)){\rho }_0((t_0,t_1)),$$ that is, ${\rho }_0$ is a homomorphism.
• Let $e_0$ be the identity element of $G_0.$ Then the kernel of ${\rho }_0$ is $$\{(e_0,s_1)|s_1\in G_1\}.$$

Solution. First, $D$ is closed under the operation. Besides, for any $(s,s)$ and $(t,t)$ in $D,$ we have $$s\in G,t\in G,t^{-1}\in G,s{t}^{-1}\in G$$ and $$(s,s)(t,t{)}^{-1}=(s,s)(t^{-1},t^{-1})=(s{t}^{-1},s{t}^{-1})\in D.$$ Therefore, by one-step test, $D$ is a subgroup of $G.$

Solution. Let $(x_1,y_1)$ and $(x_2,y_2)$ be elements of $\mathbb{R}\times \mathbb{R}.$ Then $$\begin{array}{rl}f((x_1,y_1)+(x_2,y_2))& =f((x_1+x_2,y_1+y_2))\\ & =2(x_1+x_2)-(y_1+y_2)\\ & =(2x_1-y_1)+(2x_2-y_2)\\ & =f((x_1,y_1))+f((x_2,y_2)),\end{array}$$ that is, $f$ is a homomorphism. Its kernel is $$\{(x,y)|2x-y=0\}$$ and its image is $\mathbb{R}.$

Solution. $f(0)\ne 0.$

Solution. For any integers $m$ and $n,$ $$\phi (m+n)=s^{m+n}=s^m{s}^n=\phi (m)\phi (n),$$ which shows that $\phi$ is a homomorphism.

If $s=e,$ then the kernel of $\phi$ is $\mathbb{Z}$ and $s^1=e.$ If $s\ne e,$ then the set $$S=\{s^n|n\in \mathbb{Z}\}$$ must contain $e=s^0.$ Now consider the enumeration of elements: $$s^1,s^2,s^3,\cdots .$$ Since $S$ has only finite number of elements, there must be two different positive integers $i$ and $j$ such that $s^i=s^j.$ Taking $m=|i-j|>0,$ we have $s^m=e.$

Solution. Let's check the conditions for $⟨s⟩$ to be a group.

• $e=s^0\in ⟨s⟩.$
• If $s^m$ and $s^n$ belong to $⟨s⟩,$ then $s^m{s}^n=s^{m+n}\in ⟨s⟩.$
• If $s^n$ belongs to $⟨s⟩,$ then $s^{-n}$ is the inverse element in $⟨s⟩.$

Solution. First, $H$ must contain $e=s^0.$ Next, since $H$ is closed under the group operation, $s^n\in H$ for all positive integers $n.$ Finally, since $s^{-n}$ is the inverse element of $s^n,$ $s^{-n}\in H$ for all positive integers $n.$ Therefore $H$ contains all the elements of $⟨s⟩.$

Solution. In ${\mathbb{Z}}_4,$ we have $$1+1=2,2+1=3,3+1=0$$ and $$3+3=2,2+3=1,1+3=0.$$ Therefore each of $1$ and $3$ generates whole ${\mathbb{Z}}_4.$

Solution. Each nonzero element of ${\mathbb{Z}}_5$ generates ${\mathbb{Z}}_5.$ Therefore $\left\{0\right\}$ and ${\mathbb{Z}}_5$ are all the subgroups.

Solution. First, it is trivial that ${\phi }_1(0)=e={\phi }_2(0).$ Next, use mathematical induction for $n$ to prove that ${\phi }_1(n)={\phi }_2(n),$ where $n$ is any natural number. Finally, we have $${\phi }_1(-n)=-{\phi }_1(n)=-{\phi }_2(n)={\phi }_2(-n)$$for a natural number $n.$

Solution. Since the sign map is a homomorphism, we have $$\sigma ({\pi }^{-1})=(\sigma (\pi ){)}^{-1}$$for a permutation $\pi .$ But $\sigma$ assumes only $\pm 1$ and it completes the proof.

Solution. The sign map $\sigma :S_n\to \{-1,1\}$ is a homomorphism and $A_n$ is the kernel of the sign map. Therefore $A_n$ is a subgroup of $S_n.$

Solution.

1. For any $x\in G,$ $$L_s(s^{-1}x)=s(s^{-1})x=x,$$ which shows that $L_s$ is surjective. If $L_c(x)=L_c(y)$ for $x,$ $y\in G,$ then $cx=cy$ and $x=y,$ which shows that $L_s$ is injective. Therefore $L_s$ is bijective and $L_s$ belongs to $\mathrm{Sym}(G).$
2. For any $x\in G,$ $$L_{st}(x)=stx=s{L}_t(x)=L_s(L_t(x))=(L_s\circ L_t)(x),$$ that is, $L_{st}=L_s\circ L_t.$ Therefore $$\mathrm{\Lambda }(st)=L_{st}=L_s\circ L_t=\mathrm{\Lambda }(s)\mathrm{\Lambda }(t),$$ which shows that $\mathrm{\Lambda }$ is a group-homomorphism.
3. Suppose $\mathrm{\Lambda }(x)=\mathrm{\Lambda }(y)$ then $L_x=L_y$ and $$x=L_x(e)=L_y(e)=y,$$which shows that $\mathrm{\Lambda }$ is injective.

Solution. If $A$ is commutative, then (*) holds. Now we prove the converse. Let (*) holds for all $a$ and $b.$ Expanding LHS of (*) and substitute $ab+ab$ into $2ab,$ we have $$a^2+ab+ba+b^2=a^2+ab+ab+b^2.$$ Cancellation law gives that $ba=ab,$ which shows the commutativity.

Solution. Take $A={\mathbb{Z}}_6$ and $a=2,$ $b=3.$

Solution. Suppose $ab=0$ but $a\ne 0.$ Then $b=a^{-1}(ab)=a^{-1}0=0$ and we have $b=0.$

Solution.

Solution. This ring has zero divisors $2$ and $3,$ therefore this ring is not a field.

Solution. Since $n$ is not a prime number, there exist positive integers $p$ and $q$ such that $n=pq$ and $p<n,$ $q<n.$ Then $p$ and $q$ are zero divisors of ${\mathbb{Z}}_n,$ which shows that ${\mathbb{Z}}_n$ is not a field.

(In fact, ${\mathbb{Z}}_n$ is a field if and only if $n$ is a prime number.)

Solution. Suppose (*) holds. If $a$ and $b$ were zero divisors, from $ab=0=a0$ we have $b=0,$ which is a contradiction. Thus $A$ has no zero divisors. Conversely, suppose that $A$ is an integral domain. If $a\ne 0,$ then $a^{-1}$ must be in $A$, and from $ab=ac$ we have $$b=a^{-1}(ab)=a^{-1}(ac)=c,$$ which proves the cancellation law.

Now we prove that every finite integral domain is a field. Suppose $D$ be a finite integral domain, consisting of the elements $$0,1,a_1,a_2,\cdots ,a_n.$$ If $D$ consists of $0$ and $1$ only, the proof is done. Therefore, suppose that $D$ has more than $2$ elements, and take any $a\in D$ such that $a\ne 0$ nor $a\ne 1.$ Consider the elements $$\begin{array}{}\text{(**)}& a1,a{a}_1,a{a}_2,\cdots ,a{a}_n.\end{array}$$ Note that none of these elements equals $0,$ since if $a{a}_i=0,$ then $a_i=0.$ Furthermore, these elements are all distinct, since if $a{a}_i=a{a}_j,$ then $a_i=a_j.$ Thus these elements are just an rearrangement of the elements $$1,a_1,a_2,\cdots ,a_n.$$ Therefore one of the enumeration (**) is equals to $1,$ say $a1=1$ or $a{a}_k=1,$ which implies that $a$ has the multiplicative inverse in $D,$ that is, $D$ is a field.

Solution.

Solution. Suppose that $d=gcd(m,n)\ne 1.$ Since $$m\left(\frac{n}{d}\right)=\left(\frac{m}{d}\right)n=0,$$ we have $m(n/d)=0.$ For $m\ne 0$ nor $n/d\ne 0$ in ${\mathbb{Z}}_n,$ $m$ is a zero divisor of ${\mathbb{Z}}_n.$

Conversely, suppose that $m$ is relative prime to $n.$ If for $s\in {\mathbb{Z}}_n$ we have $ms=0,$ then $n$ divides $ms.$ Since $n$ is relatively prime to $m,$ $n$ divides $s,$ so $s=0$ in ${\mathbb{Z}}_n.$

Solution. Since each nonzero element in ${\mathbb{Z}}_p$ is not a zero divisor, we have by Problem 2.28 that ${\mathbb{Z}}_p$ is an integral domain. Furthermore ${\mathbb{Z}}_p$ consists of finite number of elements, we conclude by Problem 2.26 that ${\mathbb{Z}}_p$ is a field.