Find the order of the intersection of two cyclic groups

Question

Suppose $G=\langle a\rangle $ has order $140$. What is the order of the intersection $\langle a^{100}\rangle$ and $\langle a^{30}\rangle$?

My attempt: I know that the order of $\langle a^{100}\rangle$ is $7$ and the order of $\langle a^{30}\rangle$ is $14.$ Then, the order of the intersection $\langle a^{100}\rangle$ and $\langle a^{30}\rangle$ would be $14$? And a generator is $a^3$?

Answer 1

Hint: $G\cap H$ is a subgroup of both $G$ and $H$. So $|G\cap H|\Large|$$\gcd(|G|,|H|)$. (So, number of elements cannot be 14)

Now, by properties of cyclic subgroup if $|a|=n$, then $\langle a^k\rangle=\langle a^{\gcd(k,n)}\rangle$

So $\langle a^{100}\rangle=\langle a^{20}\rangle$ and $\langle a^{30}\rangle=\langle a^{10}\rangle$

Now can you prove that $\langle a^p\rangle\cap \langle a^q\rangle=\langle a^{lcm(p,q)}\rangle$?

Answer 2

The order of the intersection cannot be $14$ because the intersection is at most the size of the smallest of the two sets being intersected. Better generators for these groups are $a^{20}$ for the first and $a^{10}$ for the second. These work because a cyclic group has at most one subgroup of a given order. From this it is easy to deduce that the intersection is the entire subgroup of order 7.

Answer 3

There are only seven numbers in $\langle a^{100}\rangle$, so the intersection won't have 14 numbers in it.
Can you show that $\langle a^{20}\rangle$ is a power of both $a^{100}$ and $a^{30}$