Let $a$ and $b$ belong to a group. If $|a|$ and $|b|$ are relatively prime, show that $\langle a \rangle \cap \langle b \rangle = \{e \}$ Let $a$ and $b$ belong to a group.  If $|a|$ and $|b|$ are relatively prime, show that $\langle a \rangle \cap \langle b \rangle = \{e \}$.
The most I can figure to do is to let $a^k = b^j$ for some $a^k$ and $b^j$ in $\langle a \rangle$ and $\langle b \rangle$, respectively.  I'd assume that I'd need to show that the only possible element either could be would be $e$ (the identity element) but I'm not seeing a way.  Would it make sense to say that $aa^k \neq bb^j$ because then otherwise $\langle a \rangle = \langle b \rangle$, therefore the only option is to have $a^k = e$ and $b^j = e?$
 A: How about this approach: Let $x$ and $y$ be the order of $a$ and $b$ respectively. Since they are relatively prime, there exist some integers $p,q$ satisfying $$px+qy=1.$$ Now suppose $g$ is an element of $<a>\cap<b>$. Then the caculation $$g=g^{px+qy}=g^{px}\cdot g^{qy}=(g^{x})^{p}\cdot (g^{y})^{q}=1^{p}\cdot 1^{q}=1$$ shows that $g$ must be 1.
A: By Lagrange, being in $\langle a\rangle$ its order divides $|a|,\,$ and being in $ \langle b\rangle$ its order divides $|b|.\,$ Being a divisor of coprime integers,  the order  $= 1$, so the only common element is the identity.
Remark $\ $ This simple order-form form of coprimality is often handy. For example an additive analog is that a rational representable in two ways with coprime denominators is an integer, which is discussed further here. 
Generally we have the following combination of the universal properties of order and gcd
$$ n = {\rm ord}(g)\,\iff\, {\Large [} \ g^{\large J_1}\! = 1,\ldots,g^{\large J_k}\!=1\!\iff\!  n\mid (J_1,\ldots,J_k)\:{\Large ]}$$
i.e. $\,{\rm ord}(g)\,$ is the (divisibility) minimal element of the set $S$ of integers $\,J\,$ such that $\,g^J = 1,\,$ which is the least positive element of $S$ since $S$ is closed under subtraction (i.e. ideals in Euclidean domains are generated by any element of least value).
The proof in Dilemian's answer is essentially a proof of the case $\,k=2\,$ above using the Bezout identity instead of the universal property of the gcd (the final $\iff$ above).
A: More generally, if $H$ and $K$ are finite subgroups of a group $G$ such that $|H|$ and $|K|$ are relatively prime, then $H$ and $K$ intersect trivially.  A proof consists of observing that $H\cap K$ is a subgroup of both $H$ and $K$, and then applying Lagrange's Theorem.
