2
$\begingroup$

Let $a$ and $b$ be elements of a group. If $|a|=m, \;|b|=n $ and $(m,n)=1$, show that $\langle a \rangle \cap \langle b \rangle = > \{e\}$

Since $(m,n)=1$

$\langle a \rangle=\langle a^{n}$ and $\langle b \rangle= \langle b^{m} \rangle$

I know $\langle a^{n} \rangle \cap \langle b^{m} \rangle = \langle a^{lcm(m,n)} \rangle = \langle a^{mn} \rangle = \langle {a^{m}}^n \rangle =\langle e \rangle = \{e \}$

Is this proof fine ?

$\endgroup$
  • $\begingroup$ @Arthur sorry fixed here. Fixing previous question too. Please comment on my proof $\endgroup$ – So Lo Dec 4 '17 at 14:37
2
$\begingroup$

No, your proof is not fine. The problem comes in the very first equality: $$ \langle a^n\rangle \cap \langle b^m\rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle $$ This is true by happenstance (they both turn out to be $\langle e\rangle$), but it is not something you have proven, and not a result you can use. Where did $b$ disappear to, for instance? You do prove satisfactory that $\langle a^{\operatorname{lcm}(m, n)}\rangle = \langle e \rangle$, but as I said, it doesn't help you.

Instead, I would use Lagrange's theorem. We have that $\langle a\rangle\cap \langle b\rangle$ is a subgroup of $\langle a\rangle$, and thus has order that divides $m$. It is also a subgroup of $\langle b\rangle$, and therefore has order that divides $n$.

This means that $|\langle a\rangle\cap \langle b\rangle|$ divides both $m$ and $n$, so it must divide their greatest common divisor, $1$. Therefore $|\langle a\rangle\cap \langle b\rangle| = 1$, which is what we wanted to prove.

$\endgroup$

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.