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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 $b^j = e?$

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marked as duplicate by Alex Wertheim, user223391, Chill2Macht, Daniel W. Farlow, Ramiro Jul 27 '16 at 3:12

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  • $\begingroup$ What can you say about the order of an element of a group? In particular, what can you say about the order of an element of the group $\langle a \rangle $ generated by $a$? $\endgroup$ – Devlin Mallory Jul 16 '16 at 20:44
  • $\begingroup$ What can you say about $k$ and $j$ wrt $|a|$ and $|b|$? $\endgroup$ – user223391 Jul 16 '16 at 20:46
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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.

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    $\begingroup$ It doesn't require gcds or Bezout, only the much simpler fact that $1$ is the only common divisor of coprime integers - see my answer. $\endgroup$ – Bill Dubuque Jul 17 '16 at 2:00
  • $\begingroup$ @Bill Dubuque: You're right.Lagrange's theorem is enough to get the result. $\endgroup$ – Dilemian Jul 17 '16 at 6:42
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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.

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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).

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