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Let $G$ be an abelian group. Suppose that $a$ is in $G$ and has order $m$ (such that $m$ is finite) and that the positive integer $k$ divides and $b$ has order $n$ with $\gcd(m,n)$=1. Prove that $\langle a\rangle \cap \langle b\rangle = \{e\}$.

since m and n are relatively prime this should be easy we know $a^{m} =e=b^{n}$ we know $\langle a\rangle$ is a cyclic subgroup of g and that $\langle b\rangle$ is a cyclic subgroup of g so $\langle a\rangle \cup \langle b\rangle $ is a subset of G. since G is abelian this must a cyclic group or its not a subgroup of G so $\langle b\rangle$ must be a subset of $\langle a\rangle$ or $\langle a\rangle$ must be a subset of $\langle b\rangle$ since the element $a^{m-1}$ cannot equal $b^{n-1}$ and both are generations for a and b respectively we have reached a contradiction.

Clemens Bartholdy
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Faust
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2 Answers2

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The order of $\langle a\rangle$ is $n$, the order of $\langle b\rangle$ is $m$. Now if $x$ belongs to the former, its order divides $n$. And if it belongs to the latter, its order divides $m$. So if it belongs to both, its order divides $\gcd(n,m)=1$. So $x=e$.

Julien
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Note that if $x\in H$, then the order of $x$ divides the order of $H$.

if $x\in \langle a\rangle\cap\langle b\rangle$, then, the order of $x$ divides both order of $a$ and order of $b$, which must be $1$.

Therefore $x$ can only be the identity $e$.

NECing
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