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.

Question

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 $5.$

Answer 1

In general, an abelian factor group has the following structure:

You give me some group $G$ and a subgroup $H$. Then the set

$$\{H + g: g\in G\}$$

is a group, where the group operation is $(H + g) + (H + k) = H + (g+k)$. Note that $H + g$ means the set $\{ h + g: h\in H\}$.

So, let’s apply this to $G = \mathbb{Z}/(18)$ and $H = \langle (18) + 6\rangle$. The set of elements is

$$\{ [(18) + 6] + k: k\in \mathbb{Z}/(18)\}.$$

Definitely $[(18) + 6] + 0$ is in this set. And the element $[(18) + 6] + 6$ is actually equal to $[(18) + 6] + 0$. Can you show why they’re equal, and fill out the rest of the elements from here?

Answer 2

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 $\Bbb Z_{18}/\langle 6\rangle$.