Suppose that $|a|=24$. Find a generator for $ <a^{21}> \cap <a^{10}>$. In general, what is a generator for the subgroup $<a^m> \cap <a^n>$

Question

Suppose that $|a|=24$. Find a generator for $ \langle a^{21} \rangle \cap \langle a^{10}\rangle$. In general, what is a generator for the subgroup $\langle a^m\rangle \cap \langle a^n\rangle$

I know that intersection of subgroups is again a subgroup.

$\langle a^{21}\rangle=\{e, a^{21}, a^{18},a^{15}, a^{12},a^{9},a^{6},a^{3}\}$

$\langle a^{10} \rangle=\{e, a^{10}, a^{20}, a^{6}, a^{16}, a^{2}, a^{12}, a^{22}, a^{8}, a^{18}, a^{4}, a^{14}\}$

$ \langle a^{21}\rangle \cap \langle a^{10}\rangle = \{e , a^{18}, a^{12}, a^{6}\}= \langle a^{6}\rangle$

This took me a while to find. Is there a faster way to find this out?

What about the general case ?

Answer 1

Hint: If $|a|=n$, then $\langle a^k \rangle = \langle a^d \rangle$, where $d=\gcd(n,k)$.
If $r$ and $s$ divide $n$, then $\langle a^r \rangle \cap \langle a^s \rangle = \langle a^m \rangle$, where $m=lcm(r,s)$.

Answer 2

Find the minimal exponent $m'$ (or $n'$, respectively) of $a$ that generates $\langle a^m\rangle$ (or $\langle a^n\rangle$). Then realize that all elements in $\langle a^m\rangle$ (or $\langle a^n\rangle$) is $a$ raised to some multiple of $m'$ (or $n'$), and $a$ raised to any multiple of $m'$ (or $n'$) gives an element in $\langle a^m\rangle$ (or $\langle a^n\rangle$).

This means that an element that is in both groups must be $a$ raised to some power which is simultaneously a multiple of $m'$ and $n'$, and $a$ raised to any number which is a multiple of both $m'$ and $n'$ is in both groups. Thus what you're looking for is $\langle a^{\operatorname{lcm}(m',n')}\rangle$