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 $H \cap K = \langle \text{lcm}(m,n) \rangle$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.