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Let $G=\langle x\rangle$ an infinite cyclic group, put $H=\langle x^i\rangle$ and $K=\langle x^j\rangle$. Prove that $H\cap K =\langle x^l\rangle$ and $\langle H,K\rangle=\langle x^d\rangle$ where $d=\gcd (i, j)$ and $l=\operatorname{lcm} (i, j)$.

It is given that $G$ is an infinite cyclic group with subgroups $H$ and $K$ then obviously $H\cap K$ and $\langle H,K \rangle$ are also cyclic, because they are subgroups of $G$. Since, by the definition of cyclic group, we know that $\langle x\rangle=\{x^n : n \in \mathbb{Z}\}$.

How should I proceed it further?

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    $\begingroup$ Please have a look at math.stackexchange.com/help/notation and re-fromat your question accordingly. Also, please put the question in the main body and give a brief version of it in the title instead. $\endgroup$
    – Gary
    May 29, 2021 at 12:05
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    $\begingroup$ Are you familiar with the Euclidean algorithm? $\endgroup$
    – user1729
    May 29, 2021 at 12:17
  • $\begingroup$ Yes, but how can I use Euclidean algorithm to prove this? $\endgroup$ May 29, 2021 at 12:21
  • $\begingroup$ @Attika Given $p,q\in\mathbb{Z}$ there exist $k_1,k_2\in\mathbb{Z}$ such that $\gcd(p,q)=pk_1+qk_2$ (which we prove via the Euclidean algorithm). From here, we can prove that $\gcd$ is your generator. $\endgroup$
    – user1729
    May 29, 2021 at 14:12
  • $\begingroup$ @user1729 thank you so much. I got it $\endgroup$ May 29, 2021 at 15:01

1 Answer 1

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Suppose $g$ is an element of $H \cap K$, then it is an element of $H$ and $K$. The elements of $H$ are all of the form $x^{im}$ for some $m$ and the elements of $K$ are of the form $jn$ for some $n$. So $g = x^o$ where $o$ is a multiple of $i$ and $j$. That means $o$ is a multiple of the least common multiple. This shows that $H \cap K \subseteq \langle x^l \rangle$ where $l = \text{lcm}(i,j)$. Also $x^l = x^{i\frac{j}{\text{gcd}(i,j)}} = x^{j \frac{i}{\text{gcd}(i,j)}}$. The first shows that $x^l \in H$ and the second shows $x^l \in K$. So $\langle x^l \rangle \subseteq H \cap K$. This gives the first part.

For the second part $\langle H, K \rangle$ contains all of $x^{im} x^{jn} = x^{im + jn}$ for all $m$ and $n$. By the Euclidean algorithm we know that there exists $m_0$ and $n_0$ such that $im_0 + j n_0 = \text{gcd}(i,j)$. Therefore $x^{\text{gcd}(i,j)}$ is in $\langle H , K \rangle$. So $\langle x^d \rangle \subseteq \langle H, K \rangle$. Conversely because we know everything in $\langle H , K \rangle$ is of the form $x^{im + jn}$ we know that it is a power of $x^d$. This is because $x^{im + jn} = x^{d \frac{im}{d} + d \frac{jn}{d}} = (x^d)^{\frac{im}{d} + \frac{jn}{d}}$ where $\frac{im}{d}$ and $\frac{jn}{d}$ are both integers because $d \mid i$ and $d \mid j$ respectively.

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  • $\begingroup$ thank you so much $\endgroup$ May 29, 2021 at 15:02

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