Let $G=\mathbb{Z}^n\times F$ be a finitely generated abelian group, where $F$ is finite, and let's assume that the order of $F$ is odd. Let $H_1,H_2\leq G$ be two subgroups of $G$ such that the quotients $G/H_1$ and $G/H_2$ don't have elements of finite even order, i.e. their torsion subgroups have odd order.

Is it true that $G/(H_1+H_2)$ still has no element of finite even order?

Is it already visible from e.g. the (finitely many) geneators of $H_i$ that in the quotient $G/H_i$ there cannot be elements of even order?


How about $G=\Bbb Z^2$, $H_1=\left<(1,0)\right>$ and $H_1=\left<(1,2)\right>$? Then $G/H_1\cong G/H_2\cong\Bbb Z$ but $G/(H_1+H_2)\cong\Bbb Z/2\Bbb Z$.

  • $\begingroup$ Ah, you are right! It was simpler than I thought. Thanks! $\endgroup$ – user446046 Mar 2 '19 at 11:42

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