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 ?

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.