4
$\begingroup$

This is part of practice midterm that i have been given ( our prof doesn't post any solutions to it) Id like to know whats right before i right the midterm on monday this was actually a 4 part question im post just 1 piece as a question in its own cause it was much too long.

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 $m$.

Suppose that $a$ has order $m$ 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.

$\endgroup$
10
$\begingroup$

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 $x=e$.

$\endgroup$
  • $\begingroup$ Both are great but i really like this answer =) $\endgroup$ – Faust Mar 17 '13 at 5:10
  • $\begingroup$ Nicely done +10 $\endgroup$ – mrs Mar 17 '13 at 5:29
5
$\begingroup$

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 $e$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.