[Math] Write down all the elements of the quotient group $Z_{18} / \langle 6\rangle$. Is any element of order $5?$ Give reasons for your answer.

abstract-algebragroup-theoryquotient-group

Write down all the elements of the quotient group $Z_{18} / \langle 6\rangle.$ Is any element of order $5?$ Give reasons for your answer.

I just know order of $Z_{18} / \langle 6\rangle$ will be
$18 \div 3= 6.$
Where $\langle 6\rangle = \{ 6 , 12 , 0 \}$

But I can't decide how to write elements of this factor group!
I know there can't be a element of order $5.$

Best Answer

The set (group actually, but this is not relevant for this part of the question) $\Bbb Z_{18}/\langle 6\rangle$ is by definition made of the $6$ elements:

  • $\langle 6\rangle=\{0,6,12\}$
  • $1+\langle 6\rangle=\{1,7,13\}$
  • $2+\langle 6\rangle=\{2,8,14\}$
  • $3+\langle 6\rangle=\{3,9,15\}$
  • $4+\langle 6\rangle=\{4,10,16\}$
  • $5+\langle 6\rangle=\{5,11,17\}$

As for the other part, note that the order of an element $aH$ (multiplicative notation) in the quotient group $G/H$ is equal to the least positive integer $n$ such that $a^n\in H$. In fact, by the normality of $H$ in $G$: \begin{alignat}{1} &(aH)^n=H \iff \\ &a^nH=H \iff \\ &a^n\in H \\ \end{alignat} Now back to your case (and to the additive notation): $a+\langle 6\rangle$ has order $5$ in $\Bbb Z_{18}/\langle 6\rangle$ if and only if $5a\in \langle 6\rangle$, which fails for every $a=1,2,3,4,5$. In fact: \begin{alignat}{1} &5\cdot1=5\equiv 5 \pmod {18} \\ &5\cdot2=10\equiv 10 \pmod {18} \\ &5\cdot3=15\equiv 15 \pmod {18} \\ &5\cdot4=20\equiv 2 \pmod {18} \\ &5\cdot5=25\equiv 7 \pmod {18} \\ \end{alignat} Therefore, there isn't any element of order $5$ in $\Bbb Z_{18}/\langle 6\rangle$.

Related Question