# Quotients of finitely generated abelian groups

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 $$G/(H_1+H_2)\cong\Bbb Z/2\Bbb Z$$.