Let $H= \langle n \rangle$ and $K= \langle m \rangle$ be two cyclic groups. Show that their intersection is a cyclic subgroup generated by the lcm of $n$ and $m$.

I took an element, say $a$, belonging to $H \cap K$. Then $a$ can be written as a multiple of $m$ and as a multiple of $n$. Then I want to show that it can be written as a multiple of lcm of $n$ and $m$.

  • $\begingroup$ Just to be sure: these are cyclic groups under addition, not multiplication? $\endgroup$ – 6005 Sep 4 '13 at 2:53
  • $\begingroup$ @Goos: The exercise wouldn't make sense otherwise. $\endgroup$ – anon Sep 4 '13 at 3:04
  • $\begingroup$ yes they are cyclic groups under addition.can you help me. $\endgroup$ – abc Sep 4 '13 at 4:13
  • $\begingroup$ please someone help . $\endgroup$ – abc Sep 4 '13 at 5:58
  • $\begingroup$ @abc I've added an answer :) $\endgroup$ – 6005 Sep 4 '13 at 5:59

Your approach is good. You're trying to show $H \cap K = \langle \text{lcm}(m,n) \rangle$. To show this, first prove $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$; then prove $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$.

To show $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$, you're doing the right thing. Take any element of $H \cap K$ and call it $a$. Since $a$ is a multiple of both $m$ and $n$, it's a multiple of $\text{lcm} (m,n)$. (If you're not allowed to state this without proof, use unique prime factorization.) Hence $a \in \langle \text{lcm}(m,n) \rangle$.

Now you just need to show $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$. Go the other way around - take some element of $\text{lcm}(m,n)$, call it $b$. You know that $b$ is a multiple of $\text{lcm} (m,n)$; you just need to show it's in both $H$ and $K$. I trust you can do this. Then you'll be done - since $H \cap K \subseteq \langle \text{lcm}(m,n) \rangle$ and $H \cap K \supseteq \langle \text{lcm}(m,n) \rangle$, we must have $H \cap K = \langle \text{lcm}(m,n) \rangle$.

  • $\begingroup$ thanks for your hint because of this hint i abled to complete the question. thanks once again. $\endgroup$ – abc Sep 4 '13 at 6:06

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