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Let $G$ be the additive group $\mathbb{Z} \oplus \mathbb{Z}$. Let $H$ be the subgroup generated by $(1,3)$ and $(3,1)$.

a) Determine the order of $(0,1) + H$ in $G\:/\:H$.

b) Express $G\:/\:H$ as a direct sum of cyclic groups.

Here are my thoughts so far:

a) In general, let $gH \in G\:/\:H$. Then $|gH| = n$ if and only if $n$ is the smallest positive integer for which $g^n\in H$.

Here, $H$ consists of all elements in $\mathbb Z\oplus\mathbb Z$ of the form $m(1,3) + n(3,1)$, where $m,n \in \mathbb Z$. On the other hand, for the element $gH = (0,1) + H \in G\:/\:H$, we have $(gH)^k = (0,k)$, where $k$ is a positive integer. We need to find the smallest $k$ such that $(0,k) = m(1,3) + n(3,1)$ for integers $m,n \in \mathbb Z$. It's not hard to see that $k$ must be equal to $8$.

Thus, the order of $(0,1) + H$ in $G\:/\:H$ is 8. Am I correct ?

b) I'm not sure how to begin this part. I've never encountered a quotient group being written as a direct sum of cyclic groups. How can I gain some insight into this?

Thanks!

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    $\begingroup$ Shouldn't it be $k(gH) = (0,k)$, not $(k,1)$? $\endgroup$ Dec 7, 2019 at 2:19
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    $\begingroup$ Do you have tools available to tell you that $\#(G/H)=8$ (without looking that the specific point $(0,1)$)? $\endgroup$ Dec 7, 2019 at 2:20
  • $\begingroup$ @GregMartin To observe the index of $H$ in $G$, correct ? $\endgroup$
    – testguy807
    Dec 7, 2019 at 3:27

2 Answers 2

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We first define an isomorphism $\phi: G\rightarrow G$, sending $(a, b)$ to $(a, b - 3a)$. This is an isomorphism, because it has an inverse $\psi$ sending $(a, b)$ to $(a, b + 3a)$.

It suffices to solve the problem after applying $\phi$.

The group $\phi(H)$ is generated by $\phi(1, 3)$ and $\phi(3, 1)$, namely $(1, 0)$ and $(3, -8)$. I claim that it is also generated by $(1, 0)$ and $(0, 8)$. In fact, let $H'$ be the subgroup generated by $(1, 0)$ and $(0, 8)$, we want to show that $\phi(H) = H'$.

Since $(0, 8) = 3(1, 0) - (3, -8)\in H$, we have $H'\subseteq \phi(H)$

Similarly, $(3, -8) = 3(1, 0) - (0, 8)\in H'$ shows that $\phi(H) \subseteq H'$.

Therefore $\phi(H) = H'$.


For a), we want to determine the smallest positive integer $k$ such that $k(0, 1) \in H$. This is of course the same as the smallest $k$ such that $k\phi(0, 1) = (0, k)\in \phi(H)$.

Since $\phi(H)$ is generated by $(1, 0)$ and $(0, 8)$, it's obvious that $k = 8$ is what we need.


For b), we know that the quotient $G/H$ is isomorphic, via $\phi$, to the quotient $G/\phi(H) = (\mathbb Z \oplus \mathbb Z)/(\mathbb Z \oplus 8\mathbb Z) \simeq \mathbb Z/8\mathbb Z$.

More precisely, we define a group homomorphism $\tau: G\rightarrow \mathbb Z/8\mathbb Z$, sending an element $(a, b)$ to the image of $b$ in $\mathbb Z/8\mathbb Z$. It is obviously surjective, because for every $\beta \in \mathbb Z/8\mathbb Z$, there is an integer $b$ whose image in $\mathbb Z/8\mathbb Z$ is $\beta$, and we have $\tau(0, b) = \beta$ by definition.

What is the kernel of $\tau$? It's all the elements $(a, b)\in G$ such that $b$ is a multiple of $8$. But these are exactly those elements that can be written as $x(1, 0) + y(0, 8)$ for integers $x, y$, i.e. they form exactly the subgroup $\phi(H)$.

Hence the homomorphism $\tau$ induces an isomorphism $G/\phi(H)\simeq \mathbb Z/8\mathbb Z$.

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Since $G$ is generated by $x=(1,0)$ and $y=(0,1)$, then $G/H$ is generated by their images $\bar{x}$ and $\bar{y}$. Now you can use the following observations:

  • in $G/H$, $3\bar{x}+\bar{y}=0$, so $\bar{y}=-3\bar{x}$ is in the group generated by $\bar{x}$, which means that $G/H$ is actually generated by $\bar{x}$.

  • we also have $\bar{x}+3\bar{y}=0$, so $9\bar{x}=3(3\bar{x})=3(-\bar{y})=-3\bar{y}=\bar{x}$, which tells you that $8\bar{x}=0$.

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