Assignment 6

Exercise $\PageIndex{1}$

1. Find subgroups $H$ and $K$ of $D_4$ such that $K$ is normal to $H$ and $H$ is normal to $D_4$ but  $K$ is not normal to $D_4$. 
2. Find subgroups $H$ and $K$ of $S_4$ such that $K$ is normal to $H$ and $H$ is normal to $S_4$ but  $K$ is not normal to $S_4.$
3. Prove or disprove: If all the subgroups of a group $G$ are normal subgroups of $G$, then $G$ is abelian.

Answer

1.  Let $G= D_4 = \{e, r, r^2, r^3, s, sr, sr^2, sr^3\}$ with $r^4=4, s^2=e$ and $srs=r^{-1}$. Since $|D_4|=8$, possible order of subgroups are $1,2,4$ and $8$. Let $H=\{1,r^2,s, sr^2\}.$ We shall show that $H$ is normal to $D_4.$ Consider the following Cayley table for $H$,

Table $\PageIndex{1}$: Cayley Table

	$e$	$r^2$	$s$	$sr^2$
$e$	$e$	$r^2$	$s$	$sr^2$
$r^2$	$r^2$	$e$	$sr^2$	$s$
$s$	$s$	$sr^2$	$e$	$r^2$
$sr^2$	$sr^2$	$s$	$r^2$	$e$

Hence, $H \leq G.$  By Lagranges Theorem, $[G:H] = \frac{|G|}{|H|} = 2, gH = Hg, \forall\ g\in G$. Thus, $H \trianglelefteq G$.

Choose $K= \{e,s\}.$ We shall show that $K$ is normal to $H$ and $K$ is not normal to $D_4.$  Consider the following Cayley table for $H$,

Table $\PageIndex{2}$: Cayley Table

	$e$	$s$
$e$	$e$	$s$
$s$	$s$	$e$

Clearly  $K\leq G$ and $K\le H.$ $\text{Since }[H:K] = 2 \text{, then } hK = Kh, \forall\ h\in H$. Therefore $K \trianglelefteq H$. Let  $k = s \in K$ and $g = r  \in G$. Consider $gkg^{-1} = rsr^{-1} = sr^{-1}r^{-1} = sr^{-2} = sr^2 \notin K$. Thus $K\ntrianglelefteq G.$

2. Let $G = S_4, H= K_4 = \{e,(12)(34), (13)(24), (14)(23)\}$ and $K=\{e, (12)(34)\}.$ Consider the Cayley table for $H$ and $K$,

 

Then clearly, $H\leq G \text{, } K\leq G.$ We shall show that $H \trianglelefteq G$ and $K \trianglelefteq H$.

By Lagrange's Theorem, $[H:K] = \frac{|H|}{|K|} = 2, gK = Kg, \forall\ g\in H$. Thus, $K \trianglelefteq H$.   

Now, by Lagrange's Theorem, $[G:H] = \frac{|G|}{|H|} = \dfrac{24}{4}=6$, Hence $H$ has $6$ cosets in $G$.

 Now, consider the left cosets of $H$ in $G$: $$eH = \{ ee, e(12)(34), e(13)(24), e(14)(23) \}  = \{ e, (12)(34), (13)(24), (14)(23) \}$$ $$= ((1, 2)(3, 4))H = ((1, 3)(2, 4))H  = ((1, 4)(2, 3))H.$$ $$(1, 2, 3)H = \{ (1, 2, 3)e, (1, 2, 3)(12)(34), (1, 2, 3)(13)(24), (1, 2, 3)(14)(23) \} = \{ (1, 2, 3), (1, 3, 4), (2, 4, 3), (4, 2, 1) \} = (1, 3, 4)H  = (2, 4, 3)H  = (4, 2, 1)H .$$ $$(3, 2, 1)H= \{ (3, 2, 1)e, (3, 2, 1)(12)(34), (3, 2, 1)(13)(24), (3, 2, 1)(14)(23) \}  = \{ (3, 2, 1), (2, 3, 4), (1, 2, 4), (1, 4, 3) \} = (2, 3, 4)H = (1, 2, 4)H = (1, 4, 3)H.$$   $$(12)H= \{ (12), (34), (1324), (1243) \}.$$   $$(13)H = \{ (13), (24), (1423), (1234) \}.$$ $$(14)H = \{ (14), (23), (1342), (1432) \}.$$

And the right cosets of $H$ are: $$He = \{ ee, (12)(34)e, (13)(24)e, (14)(23)e \}  = \{ e, (12)(34), (13)(24), (14)(23) \}  = H((1, 2)(3, 4))  = H((1, 3)(2, 4)) = H((1, 4)(2, 3))$$ $$H(1, 2, 3) = \{ e(1, 2, 3), (12)(34)(1, 2, 3), (13)(24)(1, 2, 3), (14)(23)(1, 2, 3) \}  = \{ (1, 2, 3), (2, 4, 3), (1, 4, 2), (1, 3, 4) \} = H(1, 3, 4)= H(2, 4, 3)= H(1, 4, 2)$$ $$H(3, 2, 1) = \{ e(3, 2, 1), (12)(34)(3, 2, 1), (13)(24)(3, 2, 1), (14)(23)(3, 2, 1) \}  = \{ (3, 2, 1), (1, 4, 3), (2, 3, 4), (1, 2, 4) \} = H(2, 3, 4)  = H(1, 2, 4)  = H(1, 4, 3)$$ $$H(12)= \{ (12), (34), (1324), (1243) \}.$$   $$H (13)= \{ (13), (24), (1423), (1234) \}.$$ $$H (14)= \{ (14), (23), (1342), (1432) \}.$$

Since the right cosets are the same as the left cosets, we know that $H$ is normal in $G$.

We shall show that $K$ is not a normal subgroup of $S_4.$ Since $(1, 2, 3)(12)(34)(3,2,1)=(1,3)(2,4)\notin K$, $K$ is not a normal subgroup of $S_4.$

3. Consider $G=Q_8=\{1, -1, i, -i, j, -j, k, -k\}$. Then, by Lagrange's theorem, possible orders of subgroups of $Q_8$ are $1, 2, 4$ and $8$. The trivial subgroups  $Q_8$ and $\{e\}$ of order $8$ and $1$ are normal subgroups of $G$. Let $H$ be a subgroup of order $4$. Then, by Lagrange's Theorem, $[G:H] = \frac{|G|}{|H|} = \dfrac{8}{4}=2.$ Hence the subgroups of order $4$ (namely $H_1 =\{1, -1, i, -i\}, H_2=\{1, -1, j, -j\}, H_3 =\{1, -1, k, -k\}$) are normal subgroups of $G$. Consider the subgroups of order $2$, $\langle -1\rangle= Z(Q_8)$. Hence, the subgroup of order $4$ is a normal subgroup of $G$. Therefore, all subgroups of $G$ are abelian. However, since $ij =k \ne -k= ji$,  $G$ is not abelian.

Exercise $\PageIndex{2}$

1.  Let $G=<a>$ where   $|a|=12,$ and let $H=<a^4>.$ Find all cosets in $G/H$ and  write down the Cayley table. Is $G/H$ cyclic? Why or why not?
     Find the cosets in $D_6/Z(D_6),$ write down the Cayley table of  $D_6/Z(D_6).$  Is $D_6/Z(D_6)$ cyclic, why or why not?

Answer

1. Let $G = \langle a\rangle$ and let $H = \langle a^4\rangle$. Then, $|H|=\dfrac{12}{gcd(12,4)}=\dfrac{12}{4}=3.$ Hence, by Lagrange's Theorem. $[G:H] = \frac{|G|}{|H|} = \dfrac{12}{3}=4$, Hence $H$ has $4$ cosets in $G$.

First, let's find all elements in $G/H$.

$$\begin{align*}G &= \{a, a^2, \dots\ a^{10}, a^{11}, e\}\\H &= \{a^4, a^8, e\}\\&eH = \{a^4, a^8, e\} = a^4H = a^8H\\&aH = \{a, a^5, a^9\} = a^5H = a^9H\\&a^2H = \{a^2, a^6, a^{10}\} = a^6H = a^{10}H\\&a^3H = \{a^3, a^7, a^{11}\} = a^7H = a^{11}H\\G/H &= \{H, aH, a^2H, a^3H\}\end{align*}$$

From the table, $G/H = \langle aH \rangle$. Hence $G/H$ is a cyclic group.

2. Find the cosets in $D_6/Z(D_6),$ write down the Cayley table of $D_6/Z(D_6).$ Is $D_6/Z(D_6)$ cyclic, why or why not?

Answer

Find all the cosets in $D_6/Z(D_6):$ 

Let $G = D_6 = \{e, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5\}$ and let $H = Z(D_6) = \{e, r^3\}.$ Then $D_6 / Z(D_6) = \{gZ(D_6)| g\in G\}$.

Index of $Z(D_6)$ in $D_6$} The index of $Z(D_6)$ in $D_6$ is calculated as follows: $$[D_6 : Z(D_6)] = \frac{|D_6|}{|Z(D_6)|} = \frac{12}{2} = 6.$$ Therefore, the factor group $D_6 / Z(D_6)$ has 6 elements. 

$$\begin{align*}&er^3 = r^3, && r^3r^3 = r^6 = e, &&\{e, r^3\} = eH = r^3H\\&rr^3 = r^4, && r^4r^3 = r^7 = r, &&\{r, r^4\} = rH = r^4H\\&r^2r^3 = r^5, && r^5r^3 = r^8 = r^2, &&\{r^2, r^5\} = r^2H = r^5H\\&sr^3 = sr^3, && sr^3r^3 = sr^6 = s, &&\{s, sr^3\} = sH = sr^3H\\&srr^3 = sr^4, && sr^4r^3 = sr^7 = sr, &&\{sr, sr^4\} = srH = sr^4H\\&sr^2r^3 = sr^5,&& sr^5r^3 = sr^8 = sr^2, &&\{sr^2, sr^5\} = sr^2H = sr^5H\\ \end{align*}$$

Is $D_6 / Z(D_6)$ Cyclic?

 We analyze the orders of the elements of $D_6 / Z(D_6)$: $$| \langle eZ(D_6) \rangle | = 1, \quad | \langle rZ(D_6) \rangle | = 3, \quad | \langle r^2Z(D_6) \rangle | = 3,$$ $$| \langle sZ(D_6) \rangle | = 2, \quad | \langle rsZ(D_6) \rangle | = 2, \quad | \langle r^2sZ(D_6) \rangle | = 2.$$ Thus, $D_6 / Z(D_6)$ has no elements of order 6, and therefore, it has no possible generators. Hence, $D_6 / Z(D_6)$ is not a cyclic group.

 

3. Prove or disprove the following statements:  If $H$ is a normal subgroup of $G$ such that $H$ and $G/H$ are abelian, then 
$G$ is abelian.

Answer

Counterexample:  Let $G = S_3$ and $H = A_3$. Then  $$|G / H| = [G : H] = 2.$$ Thus, $H$ is a normal subgroup of $G$, and $G / H$ is cyclic. Clearly, $A_3$ is also cyclic. Therefore, both $G / H$ and $H$ are abelian. However, $S_3$ is not abelian.

Exercise $\PageIndex{3}$

1. Let $G$ be a group. Show that the center of the group $Z(G)$ is a normal subgroup of $G.$
2. If $G/Z(G)$  is cyclic , then $G$ is abelian.
3. Let $G$ be a non-ablian group of order $pq$, where $p$ and $q$ are prime. Then the center of $G$, $Z(G)=\{e\}.$

Answer

1. First we shall show that $Z(G)$ is a subgroup of $G.$ Since $e\in G$ and $eg=ge$, $\forall g \in G.$ Thus $e \in Z(G).$ Let $g_1,g_2\in Z(G)$. Then $g_1g=gg_1, g_2g=gg_2$ for all $g\in G.$

Let $g\in G.$ Consider $g(g_1g_2)=( gg_1)g_2= (g_1g)g_2= g_1(gg_2)= g_1(g_2g)= (g_1g_2)g.$ Thus $(g_1g_2\in Z(G).$

Since $g_1g=gg_1$, $g_1^{-1}( g_1g)=g_1^{-1}(gg_1) \implies g g_1^{-1}=g_1^{-1}g.$ Thus $g_1^{-1}\in Z(G).$ Thus $Z(G)$ is a subgroup of $G.$

Since $g_1 \in Z(G),$ $g^{-1} g_1g= g_1 \in Z(G).$ Thus $g^{-1} Z(G) g \subset Z(G).$ Hence $Z(G)$ is a normal subgroup of $G.$

2. Suppose $G / Z(G)$ is cyclic. Then $G / Z(G) =\langle (gZ(G))\rangle,$ for some $g \in G.$ Let $g_1,g_2 \in G$ Then $g_1 Z(G)= g^m Z(G), \quad g_2 Z(G)= (gZ(G))^n = g^n Z(G),$ for  some $m, n \in \mathbb{Z}$. Thus $g_1 = g^m f_1, \quad g_2 = g^n f_2,$ for some $f_1, f_2 \in Z(G)$. Now consider: $$g_1 g_2 = g^m f_1 g^n f_2,$$ $$= g^m g^n f_1 f_2,$$ $$= g^{m+n} f_1 f_2,$$ $$= g^{n+m} f_1 f_2,$$ $$= g^n f_2 g^m f_1,$$ $$= g_2 g_1.$$ Thus, $g_1 g_2 = g_2 g_1$, which shows that $G$ is abelian.

3. Let $H := Z(G)$. Since $H \leq G$, by Lagrange's Theorem (Theorem 4.1.4), we know that $|H|\mid |G|$. Hence the potential values of $|H|$ are $1, p, q, pq$. Let's consider each.

Case 1:$|H| = pq$,  If $|H| = pq$ then $|H| = |G|$, and $H = G$. This would mean that $G = Z(G)$, but since $G$ is non-abelian, this is a contradiction. Therefore, $|H| \neq pq$.

Case 2:$|H| = p$ or $|H| = q$. This means that $H$ is of prime order, and by the previous problem, this implies that $G$ is abelian. Again, this is a contradiction. Therefore $|H| \neq p$ and $|H| \neq q$. 

Therefore, if $G$ is a non-abelian group of order $pq$, then $Z(G) = \{ e\}$.

Exercise $\PageIndex{4}$

1.  Let  $h: G\to G_1$ be a group homomorphism. If  $K$ is normal to subgroup of $G,$ then show that $h(K)$ is a normal subgroup of $h(G).$
2.  If $h_1: G\to G_1$ is  a group homomorphism and $G=\langle X\rangle$ is generated by a subset $X$, then  show that $h=h_1$ if and only if   $h(x)=h_1(x), \forall x\in X.$
3.   Show that there are almost six  homomorphisms from $S_3$ to a cyclic group of order $6$, $\mathbb{Z}_6$.

Answer

1. Let $k_1, k_2 \in K$, then since $K$ is a subgroup of $G$,  then $h(k_1k_2) = h(k_1)h(k_2) \in h(K)$. The identity element $e \in K$ maps to $h(e) = e_{G_1}$ (the identity in $G_1$), so $e_{G_1} \in h(K)$. If $k \in K$, then $k^{-1} \in K$ and hence $h(k^{-1}) = h(k)^{-1} \in h(K)$. Thus, $h(K)$ is a subgroup of $h(G)$. To show $h(K) \trianglelefteq h(G)$, we need to show that for every $h(g) \in h(G)$ and $h(k) \in h(K)$, $$h(g) h(k) h(g)^{-1} \in h(K).$$ Given $h(g) \in h(G)$, there exists some $g \in G$ such that $h(g) = h(g)$. Similarly, for $h(k) \in h(K)$, there exists some $k \in K$. Consider the element $gkg^{-1}$ in $G$. Since $K \trianglelefteq G$, we have: $$gkg^{-1} \in K.$$ Consider, $$h(gkg^{-1}) = h(g)h(k)h(g^{-1}) = h(g)h(k)h(g)^{-1}.$$ Since $gkg^{-1} \in K$ and $K$ is mapped into $h(K)$, we get: $$h(gkg^{-1}) \in h(K).$$

But $h(gkg^{-1}) = h(g)h(k)h(g)^{-1}$, hence: $$h(g)h(k)h(g)^{-1} \in h(K).$$ This shows that $h(K)$ is normal in $h(G)$. Hence, $$K \trianglelefteq G \implies h(K) \trianglelefteq h(G).$$

2. Discussed in Tutorial.

3.  Discussed in Tutorial.

Exercise $\PageIndex{5}$

 1. Let $G$ be a group and let $a \in G$ then show that $\sigma_a :G \to G$ defined by $\sigma_a(g)=aga^{-1}$is an isormorphism.
2. Prove or disprove the following statements: 

a. $U(5)$ is isormorphic to $\mathbb{Z}_4$ .

b. $\mathbb{Z}$ under addition is isormorphic to $\mathbb{Q}$ under addition.

Answer

a. Let $\phi: \; \mathbb{Z}_4 \rightarrow U(5)$ be defined by:  $\phi(n)=2^n \pmod 5.$ That is 

$\phi(0)=1$,  $\phi(1)=2$,  $\phi(3)=8 \equiv 3 \pmod{5}$, and $\phi(2)=4$.

We will show $\phi(a+b)=\phi(a)\phi(b), \; \forall a,b\in \mathbb{Z}_4$.

We will proceed by proof by exhaustion.  

There are 10 cases to consider since addition and multiplication are both commutative.

$\phi(0+0)=\phi(0)=1=(1)(1)=\phi(0)\times \phi(0)$.

$\phi(0+1)=\phi(1)=2=(1)(2)=\phi(0)\times \phi(1)$.

$\phi(0+2)=\phi(2)=4=(1)(4)=\phi(0)\times \phi(2)$.

$\phi(0+3)=\phi(3)=3=(1)(4)=\phi(0)\times \phi(3)$.

$\phi(1+1)=\phi(2)=4=(2)(2)=\phi(1)\times \phi(1)$.

$\phi(1+2)=\phi(3)=3=(1)(3)=\phi(1)\times \phi(3)$.

$\phi(1+3)= \phi(0)=1\equiv (2)(3)=\phi(1)\times \phi(3)$.

$\phi(2+2) = \phi(0)=1=(1)(1)=\phi(0)\times \phi(0)$.

$\phi(2+3)= \phi(1)=2\equiv (4)(3)=\phi(2)\times \phi(3)$.

$\phi(3+3)= \phi(2)=4 \equiv(3)(3)=\phi(3)\times \phi(3)$.

Having examined all the possible cases, $\phi(ab)=\phi(a)\phi(b), \; \forall a,b\in \mathbb{Z}_4$.

b. Counterexample:

We know that $(\mathbb{Z},+)$ is a cyclic group generated by $\langle -1\rangle$ and $\langle +1 \rangle$.

Assume that $(\mathbb{Q}, +)$ is cyclic. We will show that $(\mathbb{Q}, +)$ is not cyclic.

Since $\mathbb{Q}$ is cyclic, it can be generated by some $\frac{p}{q}, p,q(\ne 0) \in \mathbb{Z}$. 

Thus $\frac{p}{2q} = n \frac{p}{q}$ for some $n \in \mathbb{Z}$. That is $\frac{p}{2q} =  \frac{np}{q} \implies n=\frac{1}{2}.$

This is a contradiction, and therefore $(\mathbb{Q}, +)$ is not cyclic.

 

 

Exercise $\PageIndex{6}$

 List or characterize all of the following rings' units. Justify your answer.

1.   $\mathbb{Z}_{12}$
2. $M_{22}(\mathbb{Z}_2)$
3. $\mathbb{Z}(i)$

Answer

1. Let $a \in \mathbb{Z}_{12}$ be a unit. Then there exists $x \in \mathbb{Z}_{12}$ such that $ax \equiv 1\pmod{12}.$ Thus $12 \mid (ax-1) \implies (ax-1)=12y,$ for some $y \in \mathbb{Z}.$ Therefore, $ax-12y=1$. Hence $gcd(a,12)=1.$ Therefore the units are $1, 5, 7, 11.$

2. Let $A \in M_{22}(\mathbb{Z}_2)$ be a unit. . Then there exists $X \in M_{22}(\mathbb{Z}_2)$ such that $$\begin{align*}AX= \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}\end{align*}$$ , and $det(A) \ne 0.$ The determinant of a $2 \times 2$ matrix $A$ is:
$$A =  \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad \text{and} \quad \det(A) = ad - bc \pmod{2}.$$

We now list all matrices in $M_2(\mathbb{Z}_2)$ with $\det(A) \equiv 1 \pmod{2}$. These matrices are:
$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}.$$

Thus, the units in $M_2(\mathbb{Z}_2)$ are these six matrices. Each of these has an inverse in $M_2(\mathbb{Z}_2)$, confirming that they are units.

 3. Let $z = a + bi \in \mathbb{Z}[i]$ where $a, b \in \mathbb{Z}$ be a unit. Then multiplicative inverse of $z$ is given by $z^{-1}$. Then $z  z^{-1} = 1 \implies |z || z^{-1}| = 1$. Since $|z|$ is a nonnegative integer, $|z|=1.$ Thus $a^2+b^2=1.) Hence \(a=\pm1$ or $b=\pm 1$.

 Therefore the units of $\mathbb{Z}[i]$ are $\{-1, 1, -i, i\}$.

 

Exercise $\PageIndex{7}$

An element $e$ of a ring $R$ is said to be idempotent if $e^2=e.$

1.   If $e$ is an idempotent of a ring $R$, then show that $1-2e$ is a unit.
2.  Find all idempotents in $\mathbb{Z}_{12}$.
3.   Find all the idempotents in an integral domain $R$.

Answer

1. Let $e \in R$ be an idempotent of a ring$R$. We shall show that $1-2e$ is a unit of $R.$

Consider, $$\begin{eqnarray}(1-2e)(1-2e) &=& 1 - 2e - 2e + 4e^2 \nonumber \\&=& 1 - 4e + 4e \nonumber \\&=& 1. \nonumber\end{eqnarray}$$

Since $(1-2e)(1-2e) = 1$, $(1-2e)$ is a unit of $R$.

2. Let $x$ be an idempotent in $\mathbb{Z}_{12}$, Then $x^2 = 1 \pmod{12}$. Let's find these by exhaustion. 

$$\begin{align}\textbf{Idempotents} && \textbf{Non-Idempotents} \nonumber \\0^2 = 0 \equiv 0\pmod{12} && 2^2 = 4 \equiv 4\pmod{12}\nonumber \\1^2 = 1 \equiv 1\pmod{12} && 3^2 = 9 \equiv 9\pmod{12}\nonumber \\4^2 = 16 \equiv 4\pmod{12} && 5^2 = 25 \equiv 1\pmod{12} \nonumber \\9^2 = 81 \equiv 9\pmod{12} && 6^2 = 36 \equiv 0\pmod{12} \nonumber \\&& 7^2 = 49 \equiv 1\pmod{12} \nonumber \\&& 8^2 = 64 \equiv 4\pmod{12} \nonumber \\&& 10^2 = 100 \equiv 4\pmod{12} \nonumber \\&& 11^2 = 121 \equiv 1\pmod{12} \nonumber\end{align}$$

Therefore, the idempotents of $\mathbb{Z}_{12}$ are $\{0, 1, 4, 9\}.$

3. Let $e$ be an  idempotent in an integral domain $R$. Then $e^2 = e \implies e^2 - e = 0 \implies e(e-1) = 0.$

Since the integral domain $R$ contains no zero divisors,  $e=0$ or $e-1 = 0$. Therefore, the idempotents of $R$ are $0,$ and $1$.