Let $a,b$ be elements of a group $G$ of finite order. Prove if $|a|, |b|$ are relatively prime then,

$\langle a \rangle \cap \langle b \rangle = \{e\}$

proof (what I have so far):

let $a,b$ be elements of a group $G$ of finite order. if $a=e$, then $a^{-1} = a$ and if $b=e$, then $b^{-1}=b$, and we are done.

If a does not equal e, consider the following sequences

$a, a^2, ....$ and

$b, b^2, ....$ and by closure these elements belong to $\langle a \rangle$ and $\langle b \rangle$ respectively. Since $|a|$ and $|b|$ are relatively prime, their $\gcd(|a|,|b|)=1$ which is equivalent to the element since this intersection is finite.

not really sure if I'm even on the right track. any help is appreciated


Well, what might help is a slightly different approach- a proof by contrapositive. Suppose that $\langle a \rangle \cap \langle b \rangle - \{e\} \neq \{\}$; that is, that their intersection contains more than just the identity element. We know, then, that this element... we can call it $c$, generates a subgroup of $G$. More importantly, since it is an element of both $\langle a \rangle$ and $\langle b \rangle$, that $\langle c \rangle$ forms a (sub)group of both.

What theorem do you know that describes a relationship between the order of a subgroup and the order of the group that it is a subgroup of? With that theorem, what can we say about $|c|$ relative to $|a|$ and $|b|$?


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