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

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 ?

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

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.