# For $G$ an infinite cyclic group, and subgroups $H$ and $K$, describe $H\cap K$ and $\langle H,K\rangle$

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?

• 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.
– Gary
May 29, 2021 at 12:05
• Are you familiar with the Euclidean algorithm? May 29, 2021 at 12:17
• Yes, but how can I use Euclidean algorithm to prove this? May 29, 2021 at 12:21
• @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. May 29, 2021 at 14:12
• @user1729 thank you so much. I got it May 29, 2021 at 15:01

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.