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Problem 59 (The Intersection Of Two Subgroups Of $\mathbb{Z}$)

Let $G=\mathbb{Z}$. Because the group operation is addition, remember that $a^2$ actually means $a+a$, and $a^5=a+a+a+a+a=5a$. Beware of this issue, as $a^n$ actually means $na$ because the group operation is addition when working in $\mathbb{Z}$.

  1. What is $\langle 2\rangle$? What is $\langle 3\rangle$? Convince yourself that $\langle 2\rangle\cap \langle 3\rangle = \langle 6\rangle$.
  2. What is $\langle 4\rangle$? What is $\langle 6\rangle$? Find an integer $c$ so that $\langle c\rangle=\langle 4\rangle\cap \langle 6\rangle$? Prove that your result is true.
  3. If $a,b\in \mathbb{Z}$, then conjecture what $c$ should equal so that $\langle c\rangle=\langle a\rangle\cap \langle b\rangle$. You don't have to prove this result (unless you'd rather prove this result than proving part 2).

Solution

  1. From the definition of the subgroup generated by an element, we know that $\langle 2\rangle = \{ 2^{n}|n\in \mathbb{Z} \}$. But because the group operation with which we are dealing is addition, we know that $\langle 2\rangle = \{ 2^{n}|n\in \mathbb{Z} \} = \{ 2n|n\in \mathbb{Z} \}$. This means $$\langle 2\rangle = \{ 0,\pm 2,\pm 4,\pm 6,\pm 8,\pm 10,\pm 12, ... \}.$$ Similarly, we know then that $$\langle 3\rangle = \{ 3n|n\in \mathbb{Z} \} = \{ 0,\pm 3,\pm 6,\pm 9,\pm 12,\pm 15,\pm 18, ... \}.$$ This means that by examining the elements in the two subgroups generated by $2$ and $3$ we see that $$\langle 2\rangle \cap \langle 3\rangle = \{ 0,\pm 6,\pm 12,\pm 18, ... \} =\{ 6n|n\in \mathbb{Z} \} =\langle 6\rangle.$$
  2. From part 1 we know that $$\langle 6\rangle= \{ 0,\pm 6,\pm 12,\pm 18,\pm 24,\pm 30,\pm 36 ... \}.$$ And we can easily see that $$\langle 4\rangle= \{ 0,\pm 4,\pm 8,\pm 12,\pm 16,\pm 20,\pm 24 ... \}.$$ Also from part 1 we see that perhaps we can find some number, $c\in \mathbb{Z}$, such that $\langle c\rangle=\langle 4\rangle \cap \langle 6 \rangle$. I claim that this is true. Pick $c=12$. We will first show that $\langle c\rangle \subseteq \langle 4\rangle \cap \langle 6 \rangle$. Let $x\in \langle c\rangle$. This means $x=cp=12p$ for some $p\in \mathbb{Z}$. Pick $p\in \mathbb{Z}$ such that $x=12p$. Then we see $$x=12p=(4\cdot 3)p=4(3p)\in \langle 4\rangle$$ and also we see that $$x=12p=(6\cdot 2)p=6(2p)\in \langle 6\rangle.$$ Therefore, because $x\in \langle 4\rangle$ and $x\in \langle 6\rangle$ we know that $x\in \langle 4\rangle \cap \langle 6 \rangle$. So we know $\langle c\rangle \subseteq \langle 4\rangle \cap \langle 6 \rangle$.
    Now we must show that $\langle 4\rangle \cap \langle 6 \rangle \subseteq \langle c\rangle$. Choose $y\in \langle 4\rangle \cap \langle 6 \rangle$. This means $y=4r$ and $y=6s$ for some $r,s\in \mathbb{Z}$. Pick $r,s\in \mathbb{Z}$ such that $y=4r$ and $y=6s$. Then we know $4r=6s$. Because $3|6s$ we know $3|4r$. Hence we know $3|r$ so $r=3t$ for some $t\in \mathbb{Z}$. Therefore, we know $y=4r=4(3t)=(4\cdot 3)t=12t$. So we know $y\in \langle 12\rangle=\langle c\rangle$. Thus $\langle 4\rangle \cap \langle 6 \rangle \subseteq \langle c\rangle$. We conclude that $\langle c\rangle=\langle 12\rangle=\langle 4\rangle \cap \langle 6 \rangle$. $\square$
  3. In both parts 1 and 2 we saw that the intersection of the subgroups generated by two numbers was the same as the subgroup generated by another number. But how are these three numbers related? Let $d,e\in \mathbb{Z}$. I conjecture that given $g=lcm(d,e)$ we have $\langle g\rangle=\langle d\rangle \cap \langle e\rangle$.
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