Algebra - Group Theory

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: nstejas@gmail.com Mobile: +91 9845410469

4. Order of an element


  • Definition 4.1: The order of an element $a$ of a group $G$ is a least positive integer $n$ such that $a^n=e$, denoted by $O(a)=n$. If there exists no such integer, then we say that $a$ is of infinite order.

  • If the binary operation defined on $G$ is addition, then $O(a)=n$ $\Rightarrow$ $na=e$ where $n$ is the least positive integer.

  • Example: Consider the multiplicative group $G=\{1,-1,i,-i\}$. Identity element of $G$ is 1. then,
    $\begin{aligned} 1^1=1 & \Rightarrow &O(1)=1 \\(-1)^2=1 & \Rightarrow &O(-1)=2 \\i^4=1 & \Rightarrow &O(i)=4 \\(-i)^4=1 & \Rightarrow &O(-i)=4 \end{aligned}$

  • Theorems on order of an element of a group

  • Theorem 4.1: The order of every element of a finite group is finite.

  • Theorem 4.2: If the element $a$ of a group $G$ is of order $n$, then $a^m=e$ if and only if $n$ is a divisor of $m$.
    Proof: Let $O(a)=n$ $\Rightarrow$ $a^n=e$. To prove: $a^m=e \Leftrightarrow n|m$.
    Let $a^m=e$ Then $O(a)=n \Rightarrow a^n=e$ and $a^m=e$ $\Rightarrow$ $n\leq m$
    By division algorithm, $\exists$ $q,r \in \mathbb{Z}$ such that $m=qn+r$ with $0\leq r<n$
    Now, $a^m=e$ $\Rightarrow$ $a^{qn+r}=e$
    $\Rightarrow$ $a^{qn}a^r=e$
    $\Rightarrow$ $(a^n)^qa^r=e$
    $\Rightarrow$ $e^qa^r=e$
    $\Rightarrow$ $a^r=e$
    $\Rightarrow$ $r=0$ [$\because$ $O(a)=n$ $\Rightarrow$ $n$ is the least positive integer such that $a^n=e$]
    $\therefore$ $m=qn$ $\Rightarrow$ $n|m$
    Conversely: Suppose if $n|m$. To prove: $a^m=e$.
    Now, $n|m \Rightarrow m=qn$ where $q$ is any integer
    $\Rightarrow$ $a^m=a^{qn}$
    $\Rightarrow$ $a^m=(a^n)^q$
    $\Rightarrow$ $a^m=e^q$
    $\Rightarrow$ $a^m=e$ $\blacksquare$

  • Theorem 4.3: The order of an element is same as that of its inverse in a group $G$.
    Proof: To prove: $O(a)=O(a^{-1})$ $\forall$ $a \in G$.
    Suppose, let $O(a)=n \Rightarrow a^n=e$ and $O(a^{-1})=m \Rightarrow (a^{-1})^m=e$ where $n,m$ is the least positive integer and $e$ is the identity of group $G$
    Consider $a^n=e$
    $\Rightarrow$ $(a^n)^{-1}=e^{-1}=e$
    $\Rightarrow$ $(a^{-1})^n=e$
    But $(a^{-1})^m=e$ and so $m\leq n$ ......(1)
    Consider $(a^{-1})^m=e$
    $\Rightarrow$ $(a^m)^{-1}=e$
    $\Rightarrow$ $((a^m)^{-1})^{-1}=e^{-1}=e$
    $\Rightarrow$ $a^m=e$ [$\because$ $(a^{-1})^{-1}=a$]
    But $a^n=e$ and so $n\leq m$ ......(2)
    From (1) and (2) we have $n=m$ i.e., $O(a)=O(a^{-1})$ $\blacksquare$

  • Theorem 4.4: The order of any element $a$ cannot exceed the order of $G$.
    Proof: To prove: $O(a)\leq O(G)$
    Let $G$ be a group and $a \in G$.
    Let $O(a)=n$ and let $H$ be a subgroup of $G$ generated by $a$. Thus $H=\{e,a,a^2,....,a^{n-1} \}$.
    $H$ contains $n$ elements and so $O(H)=n$ (How!)
    Case(i): Suppose if $H=G$
    Then $O(H)=O(G)=n$
    $\Rightarrow$ $O(a)=O(G)$
    Case(ii): Suppose if $H\neq G$
    Then $O(H)<o(G)$
    $\Rightarrow$ $n<O(G)$
    $\Rightarrow$ $O(a)<O(G)$
    From both the cases, we get $O(a)\leq O(G)$ $\blacksquare$
    Remark: Study the cyclic groups to get the clarity of the proof of the above theorem.

  • Theorem 4.5: The order of any integral power of an element $a$ cannot exceed the order of $a$.
    Proof: Let $G$ be a group and $a \in G$. Let $O(a)=n$ $\Rightarrow$ $a^n=e$
    Let $a^k$ be any integral power of $a$ and $O(a^k)=m$ $\Rightarrow$ $(a^k)^m=e$
    To prove: $O(a^k)\leq O(a)$ i.e., $m\leq n$
    Now $a^n=e$ $\Rightarrow$ $(a^n)^k=e^k$
    $\Rightarrow$ $(a^k)^n=e$
    But $(a^k)^m=e$ $\Rightarrow$ $m \leq n$
    $\Rightarrow$ $O(a^k)\leq O(a)$ $\blacksquare$

  • Lemma 4.1: If $a$ and $x$ are any two elements of a group $G$, then $(xax^{-1})^n=xa^nx^{-1}$ for any positive integer $n$.
    Proof: Let $(xax^{-1})^n=xa^nx^{-1}$......(1)
    We prove (1) by induction on $n$.
    For $n=1$, we have $xax^{-1}=xa^1x^{-1}=xax^{-1}$ which is true.
    Assume (1) is true for $n=k$, i.e., $(xax^{-1})^k=xa^kx^{-1}$
    Consider, $(xax^{-1})^{k+1}$
    $\Rightarrow$ $(xax^{-1})^{k+1}=(xax^{-1})^k(xax^{-1})^1$
    $\Rightarrow$ $(xax^{-1})^{k+1}=(xa^kx^{-1})(xax^{-1})$ [By induction hypothesis]
    $\Rightarrow$ $(xax^{-1})^{k+1}=xa^k(x^{-1}x)ax^{-1}$ [By associative law in $G$]
    $\Rightarrow$ $(xax^{-1})^{k+1}=xa^k(ea)x^{-1}$
    $\Rightarrow$ $(xax^{-1})^{k+1}=x(a^ka)x^{-1}$
    $\Rightarrow$ $(xax^{-1})^{k+1}=xa^{k+1}x^{-1}$
    By induction on $n$,(1) holds for all positive integers.

  • Theorem 4.6: Let $G$ be a group and $a,b \in G$ then $O(a)=O(bab^{-1})$
    Proof: Let $O(a)=n$ $\Rightarrow$ $a^n=e$ and $O(bab^{-1})=m \Rightarrow (bab^{-1})^m=e$
    Since, $(bab^{-1})^n=ba^nb^{-1}$ (By lemma 4.1)
    $\Rightarrow$ $(bab^{-1})^n=beb^{-1}$
    $\Rightarrow$ $(bab^{-1})^n=bb^{-1}$
    $\Rightarrow$ $(bab^{-1})^n=e$
    But $(bab^{-1})^m=e$ and so $m\leq n$......(1)
    Now, $O(bab^{-1})=m$ $\Rightarrow$ $(bab^{-1})^m=e$
    $\Rightarrow$ $ba^mb^{-1}=e$
    $\Rightarrow$ $ba^mb^{-1}=bb^{-1}$
    $\Rightarrow$ $a^m=e$ (By cancellation laws)
    But $a^n=e$ and so $n\leq m$......(2)
    From (1) and (2), we have $n=m$ i.e., $O(a)=O(bab^{-1})$ $\blacksquare$

    Corollary 4.6.1: Let $G$ be a group and $a,b \in G$ then $O(ab)=O(ba)$
    Proof: From the above, we have $O(a)=O(bab^{-1})$
    $\Rightarrow$ $O(ab)=O[b(ab)b^{-1}]$
    $\Rightarrow$ $O(ab)=O[(ba)(bb^{-1})]$
    $\Rightarrow$ $O(ab)=O(ba)$ $\blacksquare$

  • Theorem 4.7: In a group $G$ if $O(a)=n$ and $d=(m,n)$ then prove that $O(a^m)=\frac{n}{d}$
    Proof: Given $O(a)=n$ $\Rightarrow$ $a^n=e$.
    Let $O(a^m)=k$ where $k$ is a least positive integer such that $(a^m)^k=e$
    Since $d=(m,n)$ $\Rightarrow$ $d|m$ and $d|n$
    $\Rightarrow$ $m=dp$ and $n=dq$ for some $p,q \in \mathbb{Z}$
    Consider $(a^m)^k=e$ $\Rightarrow$ $a^{mk}=e$
    Since $a^n=e$ we have $n|mk$ [by theorem 4.2]
    $\Rightarrow$ $dq|mk$
    $\Rightarrow$ $dq|(dp)k$
    $\Rightarrow$ $q|pk$
    $\Rightarrow$ $q|k$ [$\because$ $(p,q)=1$]
    Now consider $(a^m)^q=a^{mq}$
    $\Rightarrow$ $(a^m)^q=a^{dpq}$
    $\Rightarrow$ $(a^m)^q=a^{np}$
    $\Rightarrow$ $(a^m)^q=(a^n)^{p}$
    $\Rightarrow$ $(a^m)^q=e^{p}=e$
    But $(a^m)^k=e$ $\Rightarrow$ $k|q$
    Hence $q|k$ and $k|q$ $\Rightarrow$ $k=q$ and so $n=dk$ $\Rightarrow$ $k=\frac{n}{d}$ $\blacksquare$

Problems 4

  1. Find the order of 8 in the group $\{1,5,8,12\}$ under multiplication mod13.
    Solution: Let $G=\{1,5,8,12\}$

    Then $G$ is a group with identity 1.
    $8\otimes_{13}8=64\equiv 12\neq 1$
    $8\otimes_{13}8\otimes_{13}8=512\equiv 5\neq 1$
    $8\otimes_{13}8\otimes_{13}8\otimes_{13}8=4096\equiv 1=1$ and so $8^4=1$ $\Rightarrow$ $O(8)=4$ $\spadesuit$

  2. Show that $0$ is the only element of finite order in $\mathbb{Z}$.
    Solution: 0 is the identity element of an abelian group $(\mathbb{Z},+)$ . For any $a \in \mathbb{Z}$ and $n \in \mathbb{Z}^+$ Consider, $na=0$
    $\Rightarrow$ $a+a+....+a=0$ For any $a \in \mathbb{Z}$ other than 0, $na=0$ does not hold good. So no such $n$ exists for any element other than 0 in $\mathbb{Z}$.
    $\therefore$ $0$ is the only element of finite order in $\mathbb{Z}$. $\spadesuit$

  3. Give an example to show that $O(ab)$ may not be equal to $O(a)O(b)$.
    Solution: Consider the multiplicative group, $G=\{1,-1,i,-i\}$.
    Let $a=-1$ we have $O(a)=O(-1)=2$
    Let $b=i$ we have $O(b)=O(i)=4$
    So, $ab=-i$ we have $O(ab)=O(-i)=4$
    Hence, $O(a)O(b)\neq O(ab)$. $\spadesuit$

  4. Find the order of the elements of the multiplicative group $\{1,5,7,11\}$mod12.
    Solution: Let $G=\{1,5,7,11\}$

    Clearly $G$ is a group with 1 as the identity.
    From the table, we have $5^2=1$ $\Rightarrow$ $O(5)=2$, $7^2=1$ $\Rightarrow$ $O(7)=2$, $11^2=1 \Rightarrow O(11)=2$ and $O(1)=1$ $\spadesuit$

  5. Let $G$ be a group and $a,b \in G$ then $G$ is abelian if and only if $(ab)^2=a^2b^2$.
    Solution: Suppose $G$ is abelian then $ab=ba$ $\forall$ $a,b \in G$.
    Consider, $(ab)^2=(ab)(ab)$
    $\Rightarrow$ $(ab)^2=a(ba)b$
    $\Rightarrow$ $(ab)^2=a(ab)b$ [$G$ is abelian]
    $\Rightarrow$ $(ab)^2=(aa)(bb)$
    $\Rightarrow$ $(ab)^2=a^2b^2$
    Conversely: suppose $(ab)^2=a^2b^2$
    $\Rightarrow$ $(ab)(ab)=(aa)(bb)$
    $\Rightarrow$ $a(ba)b=a(ab)b$
    $\Rightarrow$ $ba=ab$ [By cancellation law in $G$]
    $\Rightarrow$ $G$ is abelian. $\spadesuit$

  6. Show that if $G$ is abelian then $(ab)^n=a^nb^n$ $\forall$ $a,b \in G$ and $n \in \mathbb{Z}$.
    Solution: Suppose $G$ is abelian.
    Case i) if $n=0$ then clearly $(ab)^n=a^nb^n$
    Case ii) if $n>0$
    Let $(ab)^n=a^nb^n$......(1)
    We prove (1) by induction on $n$
    For $n=1$, (1) $\Rightarrow$ $ab=ab$ which is true.
    Assume that (1) holds for $k$ i.e., $(ab)^k=a^kb^k$
    Consider, $(ab)^{k+1}=(ab)^k(ab)$
    $\Rightarrow$ $(ab)^{k+1}=a^kb^k(ab)$
    $\Rightarrow$ $(ab)^{k+1}=a^kb^k(ba)$ [$G$ is abelian]
    $\Rightarrow$ $(ab)^{k+1}=a^k(b^kb)a$ [By associative law in $G$.]
    $\Rightarrow$ $(ab)^{k+1}=a^k(b^{k+1}a)$
    $\Rightarrow$ $(ab)^{k+1}=a^k(ab^{k+1})$ [$G$ is abelian]
    $\Rightarrow$ $(ab)^{k+1}=(a^ka)b^{k+1}$
    $\Rightarrow$ $(ab)^{k+1}=a^{k+1}b^{k+1}$
    So, by induction $(1)$ is true for all $n >0$.
    Case iii) if $n<0$ then $n=-m$ where $m$ is a positive integer.
    Consider $(ab)^n=(ab)^{-m}$
    $\Rightarrow$ $(ab)^n=((ab)^m)^{-1}$
    $\Rightarrow$ $(ab)^n=(a^mb^m)^{-1}$ [By case ii]
    $\Rightarrow$ $(ab)^n=b^{-m}a^{-m}$ [$\because (ab)^{-1}=b^{-1}a^{-1}$]
    $\Rightarrow$ $(ab)^n=a^{-m}b^{-m}$ [Since $G$ is abelian]
    $\Rightarrow$ $(ab)^n=a^nb^n$
    Hence $(ab)^n=a^nb^n$ is true for all $n \in \mathbb{Z}$. $\spadesuit$

  7. Show that if every element of a group $G$ has its own inverse then $G$ is abelian.
    Solution: Given, Every element of a group $G$ has it own inverse. i.e., if $a \in G$ then $a^{-1}=a$. Similarly, for $b \in G$, we have $b^{-1}=b$.
    Now for $a,b \in G$ $\Rightarrow$ $ab \in G$ $\Rightarrow$ $(ab)^{-1}=ab$ $\Rightarrow$ $b^{-1}a^{-1}=ab$ $\Rightarrow$ $ba=ab$ $\Rightarrow$ $G$ is abelian. $\spadesuit$