What are the orders of a element $a + H$ of a quotient group $G/H$

Question

I would like somebody to check my answers

1. $G = (\mathbb{Q, +}), H = (\mathbb{Z, +}), a = \frac{2}{3}$

I think the answer is $\infty$, because $a + H = \{\frac{2}{3} + h: h \in \mathbb{Z}\}$

2. $G = (\mathbb{Z_{12}, +_{12}}), H = \{0,3,6,9\}, a = 5$

$a +H = \{5,8,11,2\}$, so the answer is 4.

Answer 1

I think that you are misunderstanding the meaning of “order” here. It is the order of an element $g$ of a group $G$, which is the smallest natural $n$ such that $g^n=e_G$ (and it is $\infty$ if no such $n$ exists).

In your first question the order is $3$, because $3.(a+H)=H$, which is the identity element of $G/H$. And the answer is also $3$ in the other case, since$$5+H\ne H,\ 2.(5+H)=10+H\ne H\text{, and }3.(5+H)=H.$$

Answer 2

By definition, the order of the coset $aH$ in the quotient group $G/H$ (so, $H\unlhd G$) is the least positive integer $n$ such that $(aH)^n=H$, being $H$ the identity of $G/H$. But, by normality of $H$ in $G$, it is $(aH)^n=a^nH$ (induction on $n$), so you are looking for the least $n$ such that $a^nH=H$ or, equivalently, such that $a^n\in H$. In your case (both $G$ are abelian, so I use the additive notation):

1. $na \in H=\Bbb Z\Longrightarrow n=3$
2. $na \in H=\{0,3,6,9\}\Longrightarrow n=3$

Answer 3

The order is simply how many times can we do the operation so that the result yields an element of the subgroup. For example, how many times you add $ =\frac{2}{3}$ to become inside $H = Z$, the integer numbers. $\frac{2}{3}+\frac{2}{3}+\frac{2}{3}=\frac{6}{3}=2$ which is an integer number. So the order of $a$ is 3.