Assignment 4

Some of these solutions are written by Fall 2024 MATH 2101 students. 

Exercise $\PageIndex{1}$

Consider the group $\mathbb{Z}_{12}$ with addition modulo $12.$ 
1. Is $(\mathbb{Z}_{12} ,+ mod (12) )$ cyclic group?  If so, what are the possible generators? How many distinct generators are there? What can you say about the number of generators (hint: Euler totient number) and the possible generators (Hint: Divisors)?

2. Let $G=\langle g \rangle$ be a cyclic group with $|g|=n$. Then  $G=\langle g^k \rangle$ if and only if $gcd(k,n)=1.$

3. Find all subgroups of $\mathbb{Z}_{12}$ generated by each element. What are the orders of them?  What can you say about these orders?

4. Let $G=\langle a \rangle$ be a cyclic group with $|G|=n$. Prove the following statements:

a) If $H\leq G$ then $H=\langle a^k \rangle$ for some integer $k$ that divides $n$.

b) If $k|n$ then $\langle a^{\frac{n}{k}} \rangle$ is the unique subgroup of $G$ of order $k$.

 

Answer

Consider  $\mathbb{Z}_{12}=\{0, 1, 2, 3, 4, 5, 6, 7,  8, 9, 10, 11\}$ with addition modulo $12.$  Note that $\{0\}$ is the identity and its order is $1.$ 

Element	Order	Calculations
$0$	$|\langle 0 \rangle |=1$
$1$	$|\langle 1 \rangle |=12$	$1^{12}\equiv 0 \pmod{12}$
$2$	$|\langle 2 \rangle |=6$	$2^{1}\equiv 2\pmod{12}$, $2^2 \equiv 4 \pmod{12}$, $2^{3}\equiv 6 \pmod{12}$, $2^{4} \equiv 8 \pmod{12}$, $2^{5}\equiv 10 \pmod{12}$ and $2^{6}\equiv 0 \pmod{12}$.
$3$	$|\langle 3 \rangle |=4$	$3^1\equiv 3 \pmod{12}$, $3^2\equiv 6 \pmod{12}$ , $3^3\equiv 9 \pmod{12}$, and $3^4\equiv 0 \pmod{12}$.
$4$	$|\langle 4 \rangle |=3$	$4^1 \equiv 4 \pmod{12}$, $4^2 \equiv 8 \pmod{12}$,  and $4^3 \equiv 0 \pmod{12}$.
$5$	$|\langle 5 \rangle |=12$	$5^1 \equiv 5 \pmod{12}$,  $5^2 \equiv 10 \pmod{12}$, $5^3 \equiv 3 \pmod{12}$,  $5^4\equiv 8 \pmod{12}$, $5^5 \equiv 1 \pmod{12}$,  $5^6 \equiv 6 \pmod{12}$, $5^7 \equiv 11 \pmod{12}$,  $5^8 \equiv 4 \pmod{12}$,$5^9 \equiv 9 \pmod{12}$,  $5^{10} \equiv 2 \pmod{12}$, $5^{11} \equiv 7 \pmod{12}$, and  $5^{12} \equiv 0 \pmod{12}$.
$6$	$|\langle 6 \rangle |=2$	$6^1 \equiv 6 \pmod{12}$, and $6^2 \equiv 2 \pmod{12}$.
$7$	$|\langle 7 \rangle |=12$	$7^1 \equiv 7 \pmod{12}$, $7^2 \equiv 2 \pmod{12}$, $7^3 \equiv 9 \pmod{12}$, $7^4 \equiv 4 \pmod{12}$, $7^5 \equiv 11 \pmod{12}$, $7^6 \equiv 6 \pmod{12}$, $7^7 \equiv 1 \pmod{12}$, $7^8 \equiv 8 \pmod{12}$, $7^9 \equiv 3 \pmod{12}$, $7^{10} \equiv 10 \pmod{12}$, $7^{11} \equiv 5 \pmod{12}$,  and  $7^{12} \equiv 0 \pmod{12}$.
$8$	$|\langle 8 \rangle |=3$	$8^1 \equiv 8 \pmod{12}$, $8^2 \equiv 4 \pmod{12}$ and $8^3 \equiv 0 \pmod{12}$.
$9$	$|\langle 9 \rangle |=4$	$9^1 \equiv 9 \pmod{12}$, $9^2 \equiv 6  \pmod{12}$, $9^3 \equiv 3 \pmod{12}$ and $9^4 \equiv 0 \pmod{12}$. $9^5 \equiv 5 \pmod{10}$, $9^6 \equiv 4 \pmod{10}$, $9^7 \equiv 3 \pmod{10}$, $9^8 \equiv 2 \pmod{10}$, $9^9 \equiv 1 \pmod{10}$,  and $9^{10} \equiv 0 \pmod{10}$.
$10$	$|\langle 10 \rangle |=6$	$10^1 \equiv 10 \pmod{12}$, $10^2 \equiv 8 \pmod{12}$, $10^3 \equiv 6 \pmod{12}$, $10^4 \equiv 4 \pmod{12}$, $10^5 \equiv 2 \pmod{12}$ and $10^6 \equiv 0 \pmod{12}$.
$11$	$|\langle 11 \rangle |=12$	$11^1 \equiv 11 \pmod{12}$, $11^2 \equiv 10 \pmod{12}$, $11^3 \equiv 9 \pmod{12}$, $11^4 \equiv 8 \pmod{12}$, $11^5 \equiv 7 \pmod{12}$, $11^6 \equiv 6 \pmod{12}$, $11^7 \equiv 5 \pmod{12}$, $11^8 \equiv 4 \pmod{12}$, $11^9 \equiv 3 \pmod{12}$, $11^{10} \equiv 2 \pmod{12}$, $11^{11} \equiv 1 \pmod{12}$,  and  $11^{12} \equiv 0 \pmod{12}$.

 From the table,   $(\mathbb{Z}_{12},+ mod (12) )$ is a cyclic group,   the set of all possible generators are $\{1, 5, 7, 11\}=\{  a \in \mathbb{Z}| gcd(a,12)=1 ,\; 1 \leq a < 12 \}$. Hence, the number of generators is the same as the number of integers relatively prime to $12=2^2\,3$.  That is, the number of generators is the Euler totient number $\phi(12) =(2^2-2)(3^1-1)= 4 \;\; .$

Answer

Let $G=\langle g \rangle$ be a cyclic group with $|g|=n$. Then
$g \in G$  and $g^n=e$.

Assume that  $G = \langle g^k \rangle$. Then $g = (g^k)^ x$ for some $x \in \mathbb{Z}$. 

Thus,
$g^{1 - kx} = e.$
By Theorem 2.4.5 from the text,  $n \mid (1 - kx)$. Therefore, 
$1 - kx = ny,$ for some $y \in \mathbb{Z}.$
Which implies,
$1 = kx + ny,$
hence,
$\gcd(k, n) = 1.$

Conversly, we  shall show that if $\gcd(k, n) = 1$, then $G = \langle g^k \rangle$. 
Suppose that  $\gcd(k, n) = 1$. Then 
$kx + ny = 1$, for some $x, y \in \mathbb{Z}.$
By using the properties of powers in a group and $g^n=e$, we get,
$$g^1 = g^{kx + ny}=(g^k)^x  (g^n)^y=(g^k)^x.$$

Hence $g \in \langle g^k \rangle$.

Therefore, we have $G \subseteq \langle g^k \rangle$. Since $\langle g^k \rangle \ \subseteq G$, 
$G = \langle g^k \rangle.$

Answer

From the above table, we see that the orders of the subgroups are  $1, 2, 3, 4, 6$ and 12, which are all divisors of the order of  $\mathbb{Z}_{12}$. Specifically, the order of the subgroup generated by $g \in \mathbb{Z}_{12}$  is $\dfrac{n}{ gcd(g, n)}.$

Answer

a)  Let $G = \langle a \rangle$ with $|G| = n.$ Assume that $H\leq G$. By Theorem 2.4.3, $H$ is cyclic. Hence  $H=\langle a^k \rangle$ for some integer $k \in \textbf{Z}.$  Let $d=gcd(k,n)$.  We shall show that $H=\langle a^d \rangle$. Since $d=gcd(k,n)$, $kx + ny = d$, for some $x, y \in \mathbb{Z}.$   Consider, $$a^d = a^{kx + ny}=(a^k)^x  (a^n)^y=(a^k)^x.$$ Thus $a^d \in H$. Therefore $\langle a^d \rangle \subseteq H.$
Since $d|k, k=md,$ for some $m\in \mathbb{Z}.$ Thus, $a^k=a^{md}=\left(a^d\right)^m.$ Hence $H \subseteq \langle a^d \rangle$. Thus $H =\langle a^d \rangle$ and $d|n.$

b) Let $G = \langle a \rangle$ with $|G| = n.$ Assume that $k|n$. Suppose $K$ be subgroup of $G$ of order $k$. We shall show that  $K=\langle a^{\frac{n}{k}} \rangle$. Since $K$ is a subgroup of $G$, $K=\langle a^m  \rangle$, for some, $m \in \mathbb{Z}.$ By part a, $m|n.$ Then $\left(a^m\right)^{n/m}=e$. We shall show that $\dfrac{n}{m}$ is the smallest positive integer with this property.

 

Now, $|K|= \dfrac{n}{m}=k$. Thus, $m=\dfrac{n}{k}$. Thus $K= \langle a^\frac{n}{k} \rangle$ which is unique.

 

 

Exercise $\PageIndex{2}$

Let $G$ be a group.

1. Show that $|g|=|g^{-1}|,$  for all $g\in G$. 

2. Show that $|h|=|ghg^{-1}|,$  for all $g,h\in G$. 

3. Show that $|hg|=|gh|$, for all $g,h\in G$.

Answer

Let $G$ be a group.

1. Suppose $g \in G$, then let $n \in \mathbb{Z}$ such that $|g| = n$. Now consider: 
        $$\begin{eqnarray}             g^n &=& e, \nonumber \\             \iff \left(g^n\right)^{-1} &=& e^{-1}, \nonumber \\             \iff (g^{-1})^n &=& e. \nonumber         \end{eqnarray}$$
        Now suppose $k \in \mathbb{Z}$ such that $|g^{-1}| = k$. Since $(g^{-1})^n = e$, it must be true that $k \leq n$.
        Now consider:
        $$\begin{eqnarray}             (g^{-1})^k &=& e, \nonumber \\             \iff (g^{-1})^{-k} &=& e^{-1}, \nonumber \\             \iff g^k &=& e. \nonumber         \end{eqnarray}$$
        Since $g^k = e$ and $|g| = n$, it must be the case that $n \leq k$. 
        Therefore $n \leq k \leq n \implies n = k$. 
        Therefore $|g| = |g^{-1}|$.

2. Suppose $h \in G$, the let $n \in \mathbb{Z}$ such that $|h| = n$. Now consider: 
        $$\begin{eqnarray}             (ghg^{-1})^n &=& (ghg^{-1})(ghg^{-1}) \dots (ghg^{-1}) \nonumber \\             &=& gh(g^{-1}g)h(g^{-1}g)h \dots (g^{-1}g)hg^{-1} \nonumber \\             &=& gh^ng^{-1} \nonumber \\             &=& geg^{-1} \nonumber \\             &=& gg^{-1} \nonumber \\             &=& e. \nonumber         \end{eqnarray}$$
        Now let $k \in \mathbb{Z}$ such that $|ghg^{-1}| = k$. Since $(ghg^{-1})^n = e$, it must be the case that $k \leq n$. Now consider:
        $$\begin{eqnarray}             (ghg^{-1})^k &=& e, \nonumber \\             \iff (ghg^{-1})(ghg^{-1})\dots(ghg^{-1}) &=& e, \nonumber \\             \iff gh(g^{-1}g)h(g^{-1}g)\dots (g^{-1}g)hg^{-1} &=& e, \nonumber \\             \iff gh^kg^{-1} &=& e, \nonumber \\             \iff g^{-1}g h^k g^{-1}g &=& g^{-1}eg, \nonumber \\             \iff e h^k e &=& g^{-1}g, \nonumber \\             \iff h^k &=& e. \nonumber         \end{eqnarray}$$
        Since $h^k = e$ and $|h| = n$, then $n \leq k$. 
        Therefore $k \leq n \leq k \implies n = k$. 
        Therefore $|h| = |ghg^{-1}|$.

3. Suppose $g,h \in G$. Let $n \in \mathbb{Z}$ such that $|hg| = n$. Now consider:
        $$\begin{eqnarray}             (hg)^n &=& e, \nonumber \\             \iff (hg)(hg)\dots(hg)(hg) &=& e, \nonumber \\             \iff h(gh)(gh) \dots (gh)g &=& e, \nonumber \\             \iff h(gh)^{n-1}g &=& e, \nonumber \\             \iff h^{-1}h(gh)^{n-1}gh &=& h^{-1}eh, \nonumber \\             \iff e (gh)^n &=& h^{-1}h, \nonumber \\             \iff (gh)^n &=& e. \nonumber         \end{eqnarray}$$
        Now let $k \in \mathbb{Z}$ such that $|gh| = k$. Since $(gh)^n = e$, it must be the case that $k \leq n$. Now consider:
        $$\begin{eqnarray}             (gh)^k &=& e, \nonumber \\             \iff (gh)(gh)\dots(gh)(gh) &=& e, \nonumber \\             \iff g(hg)(hg) \dots (hg)h &=& e, \nonumber \\             \iff g(hg)^{k-1}h &=& e, \nonumber \\             \iff g^{-1}g(hg)^{k-1}hg &=& g^{-1}eg, \nonumber \\             \iff e (hg)^k &=& g^{-1}g, \nonumber \\             \iff (hg)^k &=& e. \nonumber         \end{eqnarray}$$
        Since $(hg)^k = e$ and $|hg| = n$, then it must be the case that $n \leq k$. 
        Therefore $n \leq k \leq n \implies n = k$. 
        Therefore $|hg| = |gh|$.

Exercise $\PageIndex{3}$

Let $G$ be a group.

1. If $G$ is  an abelian group then show that $H=\{g^2|g\in G\}$ is a subgroup of $G$.  Show that $H$ is not necessarily a subgroup if $G$ is not abelian by taking $G=A_4.$

2. If $a^2=e$, for all $a \in G$, show that G is abelian. Give an example to show that the converse of the above statement is not true.

Answer

1.To show that $H$ is a subgroup of $G$ when $G$ is abelian, we use the subgroup criterion:

Since $e \in G$ and $e^2 = e$ , $e \in H$ . Let $a,b \in H$. Then $a=g_1^2, b=g_2^2$ for some $g_1,g_2 \in G$ .

Consider, $ab^{-1}=g_1^2 \left(g_2^2 \right)^{-1} = g_1^2 \left(g_2^{-1} \right)^{2} = g_1(g_1g_2^{-1})g_2^{-1}= g_1(g_2^{-1}g_1)g_2^{-1} \quad \text{(since \(G \) is abelian)}$. Thus $ab^{-1} =\left( g_1g_2^{-1}\right)^2.$ Since $g_1g_2\in G, \left( g_1g_2^{-1}\right)^2 \in H.$

Therefore, $H$ is a subgroup of $G$ when $G$ is abelian.

Consider $G = A_{4}$ , we will show that $H$ is not a subgroup of $G$ .
$$\begin{align*}     ( \ 1 \ 2 \ 3 \ )^{2} = ( \  1 \ 3 \ 2 \ ) &= ( \ 3 \ 2 \ 1 \ ) \\     ( \ 1 \ 2 \ 4 \ )^{2} = ( \  1 \ 4 \ 2 \ ) &= ( \ 4 \ 2 \ 1 \ ) \\     ( \ 1 \ 3 \ 4 \ )^{2} = ( \  1 \ 4 \ 3 \ ) &= ( \ 4 \ 3 \ 1 \ ) \\     ( \ 2 \ 3 \ 4 \ )^{2} = ( \  2 \ 4 \ 3 \ ) &= ( \ 4 \ 3 \ 2 \ ) \\     ( \ 4 \ 3 \ 2 \ )^{2} = ( \  4 \ 2 \ 3 \ ) &= ( \ 2 \ 3 \ 4 \ ) \\     ( \ 4 \ 3 \ 1 \ )^{2} = ( \  4 \ 1 \ 3 \ ) &= ( \ 1 \ 3 \ 4 \ ) \\     ( \ 4 \ 2 \ 1 \ )^{2} = ( \  4 \ 1 \ 2 \ ) &= ( \ 1 \ 2 \ 4 \ ) \\     ( \ 3 \ 2 \ 1 \ )^{2} = ( \  3 \ 1 \ 2 \ ) &= ( \ 1 \ 2 \ 3 \ ) \\      ( \ 1 \ 2 \ )( \ 3 \ 4 \ )( \ 1 \ 2 \ )( \ 3 \ 4 \ ) &= e  \\     ( \ 1 \ 3 \ )( \ 2 \ 4 \ )( \ 1 \ 3 \ )( \ 2 \ 4 \ ) &= e  \\     ( \ 1 \ 4 \ )( \ 2 \ 3 \ )( \ 1 \ 4 \ )( \ 2 \ 3 \ ) &= e  \end{align*}$$
$$\begin{align*}     H &= \Bigl\{ \ e, \ ( \ 1 \ 2 \ 3 \ ), \ ( \ 1 \ 2 \ 4 \ ), \ ( \ 1 \ 3 \ 4 \ ), \ ( \ 2 \ 3 \ 4 \ ), \ ( \ 4 \ 3 \ 2 \ ), \ ( \ 4 \ 3 \ 1 \ ), \ ( \ 4 \ 2 \ 1 \ ), \ ( \ 3 \ 2 \ 1 \ ) \ \Bigr\} \end{align*}$$
We can see that $H \subseteq G$ , we will show that $H$ is not closed. We know that $( \ 1 \ 2 \ 3 \ ) \in H$ and $( \ 1 \ 2 \ 4 \ ) \in H$ . Consider
$$\begin{align*}     ( \ 1 \ 2 \ 3 \ )( \ 1 \ 2 \ 4 \ ) &=      \begin{pmatrix}     1 & 2 & 3 \\     2 & 3 & 1      \end{pmatrix}      \begin{pmatrix}     1 & 2 & 4 \\     2 & 4 & 1      \end{pmatrix}     \\ &=      \begin{pmatrix}     1 & 2 & 4 \\     3 & 4 & 2      \end{pmatrix}      \\ &= ( \ 1 \ 3 \ )( \ 2 \ 4 \ ) \notin H \end{align*}$$
$\therefore \ H \nleq G .$  

2. The statement is false.  Answers may vary.

Consider the quaternion group $Q_8$ , which is non-abelian and defined by:
$$Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$$
with multiplication rules:
$$i^2 = j^2 = k^2 = -1, \qquad ij = k, \qquad ji = -k$$
Consider
$$\begin{align*}     (\pm 1)^2 &= 1, \\     (\pm i)^2 = (\pm j)^2      = (\pm k)^2      &= -1  \end{align*}$$
Then
$$\begin{align*}     H = \bigl\{ 1, -1 \bigr\} \end{align*}$$

Clearly, $H$ is a subgroup of $Q_8$. (You can use a Cayley table to show this!)

Now we will show that $Q_8$ is non-abelian. consider $i, j, k \in Q_8$
$$\begin{align*}     i \cdot j &= k, \\     j \cdot i &= -k \\     k &\neq -k \\     \iff i \cdot j &\neq j \cdot i \end{align*}$$
$\therefore$ the existence of $H = \{ g^2 \mid g \in G \}$ as a subgroup of $G$ does not imply that $G$ is abelian.

2. 

Assume that $a^2 = e$ for all $a \in G$ . Let $a, b \in G$ . Consider the product $a b$ . By using the properties of an abelian group, we have
$$\begin{align*}     (a b)^2 &= (a b) (a b) \\     &= a(b a) b \\     &= a (a b) b \\     &= (a a) (b b) \\     &= a^{2} \cdot b^{2} \\     &= e \cdot e \\     &= e \end{align*}$$
Since $(a b)^2 = e$ , we have:
$$(a b)^2 = e \implies a b a b = e.$$
Multiply both sides on the left by $a$ and on the right by $b$ :
$$a (a b a b) b = a e b \implies (a a) b a b b = a b.$$
Simplify using $a^2 = e$ and $b^2 = e$ :
$$e \cdot b a \cdot e = a b \implies b a = a b.$$
Hence, $a b = b a$ , $\ \therefore \ G$ is abelian.
 

Showing the converse is false:

Consider the cyclic group $\mathbb{Z}_4 = \{ 0, 1, 2, 3 \}$ under $+(\text{mod} 4)$ . $\mathbb{Z}_4$ is abelian, but not all elements satisfy $a^2 = e$ . 
Consider  $1^2 \in \mathbb{Z}_4$. 
$$\begin{align*}     1 + 1 &= 2 \\         &\neq 0 \end{align*}$$

Thus, $\mathbb{Z}_4$ is abelian, but $a^2 = e$ does not hold for all $a \in \mathbb{Z}_4$ . This shows that the converse is not true. 

 

Exercise $\PageIndex{4}$

Consider the symmetric group $S_4$

1.Show that $S_4$ is  not abelian.

2. Find the centre of $S_4$.

3. Given $\alpha, \beta \in S_4$ with $\alpha\beta=(1432)$,$\beta\alpha=(1243)$, and 
$\alpha(1)=4$, determine $\alpha$ and $\beta$.

4. Find the cyclic subgroup of $S_4$ that has an order of 4.

5. Find a non-cyclic subgroup of $S_4$ that has order 4.
 

Answer

1.

Let $a=(1,2)$ and $b=(2,3)$ both elements of $S_4$.

Consider $ab=(1,2)(1,3)=(1,3,2)$ and $ba=(1,3)(1,2)=(1,2,3)$.

Since $(1,3,2) \ne (1,2,3)$, $ab\ne ba \; for some  a,b \in S_4$, $S_4$ is non-abelian.

2. 

$S_4=\{e, (12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23), (123), (124), (134), (234), (432), (431), (421), (321), (1234), (1243), (1324),  (1342), (1423),( 1432)\}.$

Method I:

Let $s ∈ S_4,$ and $N = \{1,2,3,4\}.$
case 1: $s = e. e$  is commutative by definition. Therefore $e ∈ Z(S_4)$
case 2: $s$ is of type $[2,1,1].$
It follows that $s = (i, j)$for some distinct $i, j ∈ N.$  Consider $t = (i, k)$ for some $k ∈ N$
such that $i ̸= k ̸= j.$ Then $st = (i, k, j).$ But $ts = (i, j, k) ̸= st.$ So those of type [2,1,1]
are not in $Z(S_4)$.

case 3: s is of type [2,2].
It follows that s = (i, j)(k,l) for some distinct i, j, k,l ∈ N. Consider t = (i, j, k). Then
st = (j,l, k) ̸= (i, k,l) = ts. So $s ̸∈ Z(S_4).$

case 4: s is of type [3,1].
It follows that s = (i, j, k) for some distinct i, j, k ∈ N. Consider t = (i,l) such that
l ̸∈ {i, j, k}. Then st = (i,l, j, k) ̸= (i, j, k,l) = ts.

Case 5: s is of type [4].
It follows that s = (i, j, k,l) for some distinct i, j, k,l ∈ N. Consider t = (i,l). Then
st = (j, k,l) ̸= (i, j, k) = ts.

Except for e, all elements have some other element they do not commute with.
So $Z(S_4)$ = \{e\}.\)

Method II:

Proof by exhaustion: 

Clearly $e \in Z(S_4)$. \) 
$(12)(13) = (132), \quad (13)(12) = (123), \quad (12), (13) \notin Z(S_4)$
$(14)(12) = (124), \quad (12)(14) = (142), \quad (14) ,(1 2)  \notin Z(S_4)$
$(23)(12) = (132), \quad (12)(23) = (123), \quad (23) ,(1 2) \notin Z(S_4)$
$(24)(12) = (142), \quad (12)(24) = (124), \quad (2 4),(1 2) \notin Z(S_4)$
$(34)(13) = (143), \quad (13)(34) = (341), \quad (34) (13) \notin Z(S_4)$
$(12)(34)(13) = (1432), \quad (13)(12)(34) = (3412), \quad (12)(34) \notin Z(S_4)$
$(13)(24)(12) = (1423), \quad (12)(13)(24) = (3421), \quad (13)(24) \notin Z(S_4)$
$(14)(23)(12) = (1324), \quad (12)(14)(23) = (2314), \quad (14)(23) \notin Z(S_4)$
$(123)(12) = (13), \quad (12)(123) = (23), \quad (123) \notin Z(S_4)$
$(132)(12) = (23), \quad (12)(132) = (13), \quad (132) \notin Z(S_4)$
$(124)(12) = (14), \quad (12)(124) = (24), \quad (124) \notin Z(S_4)$
$(142)(12) = (24), \quad (12)(142) = (14), \quad (142) \notin Z(S_4)$
$(134)(12) = (4123), \quad (12)(134) = (4213), \quad (134) \notin Z(S_4)$
$(143)(12) = (1243), \quad (12)(143) = (1432), \quad (143) \notin Z(S_4)$
$(234)(12) = (4213), \quad (12)(234) = (4123), \quad (234) \notin Z(S_4)$
$(243)(12) = (1432), \quad (12)(243) = (4312), \quad (243) \notin Z(S_4)$
$(1234)(12) = (134), \quad (12)(1234) = (234), \quad (1234) \notin Z(S_4)$
$(1243)(12) = (143), \quad (12)(1243) = (243), \quad (1243) \notin Z(S_4)$
$(1324)(12) = (14)(23), \quad (12)(1324) = (13)(24), \quad (1324) \notin Z(S_4)$
$(1342)(12) = (342), \quad (12)(1342) = (134), \quad (1342) \notin Z(S_4)$
$(1423)(12) = (13)(24), \quad (12)(1423) = (14)(23), \quad (1423) \notin Z(S_4)$
$(1432)(12) = (243), \quad (12)(1432) = (143), \quad (1432) \notin Z(S_4)$
Having tried all $x \in S_4$, the only $x \in S_4$ that satisfies $xg=gx, \; \forall g \in S_4$ is the identity. Hence, $Z(S_4) = \{e\}$.

Method III:  Proof by contradiction

Let $\gamma (\ne e) \in Z(S_4).$  We shall find an $\alpha \in S_4$ such that $\gamma \alpha \ne \alpha \gamma.$ Since $\gamma (\ne e)$, there exist $l,m \in \{1,2,3,4\}$ such that  $\gamma (l) =m \ne l.$ Now we can choose an element $k \in \{1,2,3,4\}$ such that $k\ne l$ and $k\ne m$. Choose $\alpha=(l,k).$ Then $\gamma \alpha(l)=\gamma(k)$ and $\alpha \gamma (l)= \alpha(m)=m$. Since $\gamma$ is a bijection and  $\gamma (l) =m \ne l$, $\gamma(k) \ne m$.  This is a contradiction. Hence, $Z(S_4) = \{e\}$.

3.  Let  $\alpha, \beta \in S_4$ with $\alpha\beta=(1432)$,$\beta\alpha=(1243)$, and 
$\alpha(1)=4$, $$\begin{align}     \beta\alpha(1) = \beta(\alpha(1)) = \beta(4) = 2 \nonumber\\     \alpha\beta(4) = \alpha(\beta(4)) =\alpha(2) = 3 \nonumber\\     \beta\alpha(2) = \beta(\alpha(2)) =\beta(3) = 4 \nonumber\\     \alpha\beta(3) = \alpha(\beta(3)) =\alpha(4) = 2 \nonumber\\     \beta\alpha(4) = \beta(\alpha(4)) =\beta(2) = 3 \nonumber\\     \alpha\beta(2) = \alpha(\beta(2)) =\alpha(3) = 1 \nonumber\\     \beta\alpha(3) = \beta(\alpha(3)) =\beta(1) = 1 \nonumber\end{align}$$
$\text{from above}, \alpha =(1423)  \beta= (234).$   

4. 

The subgroup of $S_4, C_4 =\{e, (1,2,3,4),(1,3)(2,4),(1,4,3,2) \}$, has an order 4 and is generated by $(1,2,3,4)$. 

$(1,2,3,4)^2=(1,3)(2,4)$, $(1,2,3,4)^3=(1,4,3,2)$, and $(1,2,3,4)^4=e$.

5. 

The 2-2 cycles together will form an order 4 group. 

$\{e, (1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$ is a subgroup of $S_4$ and has order 4.  Note the inverse of $(1,2)(3,4)$ is $(4,3)(2,1)$, which is $(1,2)(3,4)$.  Since a similar argument can be made for the other 2-2 cycles, the inverses of each element exist.  In addition, the set is closed under composition.

 

Exercise $\PageIndex{5}$

Consider the symmetric group $S_7$.

1. Find all possible types of permutations and state how many of each type.

2. Find  the possible order of all  the $7!$ elements of  the symmetric group $S_7.$

3. Factor the permutation $(1 2 3 4 5)(67)(1357)(163)$ into disjoint cycles.

4. Find an element of maximum order in $S_7.$

Answer

3. $(1723456)$

4.The cycle type in $S_7$ with the largest order is $\underline{4} \underline{3}$. An example of an element of this class would be: $(1234)(567)$.

 

Assignment 5