Find the subgroup for this given group, $G$(Easy question) - direct product

Let the Group, $$G = Z_{16} \times Z_{60} \times Z_{72}$$.

Find the $$H = \{x \in G | 2x = (_{16}, _{60}, _{72}) \}$$

When I saw the above the first time, I thought the $$H = \langle8,30,36\rangle$$

But the answer is $$\langle8\rangle \times \langle30\rangle \times \langle36\rangle$$

This might be little silly question. But I want to exact reason or principle Why does the subgroup, $$H$$ having the form like that.

The group you propose is order $$2$$ only, because $$(8,30,36)+(8,30,36)=(16,60,72)=0$$. On the other hand, the elements of order $$2$$ in the group are precisely those $$(x,y,z)\in \mathbb{Z}/16\mathbb{Z}\times \mathbb{Z}/60\mathbb{Z}\times \mathbb{Z}/72\mathbb{Z}$$ such that $$2x\equiv 0 \pmod{16}$$, $$2y\equiv 0 \pmod{60}$$ and $$2z\equiv 0 \pmod{72}$$. So, $$x=0,8$$, $$y=0,30$$, and $$z=0,36$$. Making all ordered pairs of these $$(x,y,z)$$ we get $$8$$ elements, and in fact exactly $$\langle 8\rangle\times \langle 30\rangle\times \langle 36\rangle.$$
• Aren’t there $2^3=\color{red}8$ elements in $H$? – J. W. Tanner Sep 13 at 13:18
The following $$8$$ elements of $$G$$ are in $$H$$: $$(0,0,0), (0,0, 36), (0,30,0), (0, 30, 36), (8,0,0),(8,0,36),(8,30,0),$$ and $$(8,30,36)$$. You can easily check that each of these meets the criterion for membership in $$H$$. [Your initial thought had only $$(0,0,0)$$ and $$(8,30,36)$$.]