Let $G=\mathbb Z^2=\mathbb Z\times \mathbb Z$,and consider the subgroup $H=\{(6b,12b):b\in \mathbb Z\} \leq G$.
(a) Is $H$ a cyclic group?
(b) Describe the elements of $G/H$ and give two examples of distinct elements.
(c) Is $G/H$ a group? (And if so, under what operation?)
(d) For what integers $a \in \mathbb Z$ are the cosets $(0, a) + H$ and $(−18, 1) + H$ equal?
(e) If $\alpha=(1,0)+H$ and $\beta=(−2,−4)+H$, what are their additive orders?
My attempt:
a) $H=\langle(6,12)\rangle$, so it is cyclic.
b) I tried to write down some elements of $G/H$: $(0,0) + H = \{\ldots,(6,12),(12,24),(18,36),\ldots\}$; $(1,0) + H = \{\ldots,(7,12),(13,24),(19,36),\ldots\}$.
So, I think elements of $G/H$ are: $(i,j)+H$, where $ 0\leq i \leq 5$, $0\leq j \leq 11.$ But then to which coset does $(6,0)$ belong?
c) $G/H$ is a quotient group, which is indeed a group with group operation addition: $(a + H) + (b + H) = (a + b) + H$.
d) $(a + H) = (b + H) \iff (a-b) \in H$, so $(18, a-1) \in H \Rightarrow a = 37 $
e) $(−2,−4)+H +(−2,−4)+H + (−2,−4)+H = (-6,-12)+H=H$, so $\beta$ has order $3$. As for $\alpha$ I can't get $H$ using the same trick, so I don't know…
Can somebody please help me improve my solution?
Best Answer
For (b), if you take any element of $G/H$, it is of the form $(i_0,j_0)+H$. Then, you can add or subtract enough elements of $(6,12)$ to get in the form $(i,j)+H$ where $0\leq i\leq 5$, but there is no guarantee that $0\leq j\leq 11$: if $(i,j)+H$ and $(i',j')+H$ are equal, and both satisfy $0\leq i,i'\leq 5$, then $(i-i',j-j')\in H$ means that $i-i'=0$. But then $j-j'=0$ because the only $(i,j)\in H$ with $i=0$ is the element $(0,0)$. So a list of each distinct element of $G/H$ would be $(i,j)+H$ where $0\leq i\leq 5$ and $j\in \mathbb{Z}$.
For (e), you're exactly right with your calculation for $\beta$. The additive order of an element $\alpha$ of $G/H$ is the smallest number of times you have to add $\alpha$ to itself until it lands back in $H$, or $\infty$ if it will never land back in $H$. If you can show $\alpha+\alpha+\cdots+\alpha \not \in H$ no matter how many terms you add, then that means the order of $\alpha$ is $\infty$.