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The problem says:

"If $g$ and $h$ have orders $15$ and $16$ respectively in a group $G$, what is the order of $\left \langle g \right \rangle\cap \left \langle h \right \rangle$?"

I have to say that I am using the book "Abstract Algebra: Theory and Aplications" and at this moment I cannot use the Lagrange Theorem. This was my try:

"Let $a\in \left \langle g \right \rangle\cap \left \langle h \right \rangle$, then there are $k,q\in \mathbb{Z}$ now suppose there is $n$ such that $e=a^{n}=(g^k)^n=(h^q)^n$. Then, using the theorem 4.6 (page 60) of the book we can deduce that $n=\frac{15}{d_{1}}=\frac{16}{d_{2}}$ where $d_{1}=\gcd(15,k)$ and $d_{2}=\gcd(16,k)$. But this is impossible since $15$ and $16$ are coprimes, therefore there is not $n$ such that $e=a^{n}$."

At this point I am not sure how to continue, I was reading these links but a cannot use the theorems of this answers yet because I am avaible to use only very basic theorems about cyclic groups. Any help will be appreciated because I am completely lost in this.

proof of the intersection of 2 cyclic subgroups of G is the set e

How can I calculate the order of intersections of cyclic subgroups?

Intersection of subgroups of orders 3 and 5 is the identity

Shaun
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HeyHéctor
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    This is definitely the right idea, but your argument doesn't show that there is no such $n$; rather it shows that $n$ must divide both $15$ and $16$ and therefore must divide $\gcd(15,16)=1$. (Remember: coprime doesn't mean there are no divisors in common—it means that $1$ is the only positive divisor in common.) – Greg Martin Feb 28 '22 at 02:51
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    I don't think the theorem says what you think it says. That is, you will not "deduce" that $n$ is exactly equal to $\frac{15}{d_1}$; rather, you can conclude that $n$ must be a multiple of that number, etc. – Arturo Magidin Feb 28 '22 at 03:29

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