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I am trying to prove that the quotient group $(\mathbb{Z} \oplus \mathbb{Z})/(\mathbb{Z} \cdot (11,13))$ is torsion-free.

I know how I have to show that the only element in the group that has finite order is the identity, but I do not know where to start. I was trying to do contradiction, but did not make much progress.

I was also wondering if this can be generalized to other primes in $\mathbb{Z}$ other than 11,13?

Gertrude Phyllis
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2 Answers2

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@Lord Shark's argument goes through for $(u,v)$ with $u$ and $v$ coprime. See this. Thus, the quotient is isomorphic to $\mathbb Z$...

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In effect you need to show that if $r(m,n)\in\Bbb Z(11,13)$ for $r\ne0$ then $(m,n)\in\Bbb Z(11,13)$.

You have $rm=11t$ and $rn=13t$ for some $t\in\Bbb Z$. As $1=6\times 11-5\times 13$ (an instance of the Bezout identity) then $$t=6(11t)-5(13t)=6rm-5rn$$ so that $r\mid t$. Then $(m,n)=(t/r)(11,13)$.

Angina Seng
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