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