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

Find the $H = \{x \in G | 2x = ([0]_{16}, [0]_{60}, [0]_{72}) \}$


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.

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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 $\langle 8\rangle\times \langle 30\rangle\times \langle 36\rangle.$

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  • $\begingroup$ Aren’t there $2^3=\color{red}8$ elements in $H$? $\endgroup$ – J. W. Tanner Sep 13 at 13:18
  • $\begingroup$ Lol oh boy. I guess it’s early :) $\endgroup$ – Antonios-Alexandros Robotis Sep 13 at 13:19
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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 $(8,30,36)$.]

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