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I need to prove that $(H,+)$ is a cyclic sub-group of $(\mathbb{Z},+)$ when $H=\{30x+42y+72z:x,y,z\in\mathbb{Z}\}$.

Attempt: First, $H\subseteq \mathbb{Z}$ since the addition of integer numbers, yields an integer number. Now, we show the closure of $H$ with ($+$). Therefore, let $a_1=30x_1+42y_1+72z_1$, $a_2=30x_2+42y_2+72z_2$. We shall show that $a_1+a_2\in H$. Thus: $$a_1+a_2=30x_1+42y_1+72z_1+30x_2+42y_2+72z_2=30(x_1+x_2)+42(y_1+y_2)+72(z_1+z_2)\in H$$

In addition, we should show the existence of the opposite number, for each number in $H$. Hence, let $H \ni h=30x_1+42y_1+72z_1$. We define $H \ni h^{-1}=30(-x_1)+42(-y_1)+72(-z_1)$, therefore, we have that: $$h+h^{-1}=30(x_1-x_1)+42(y_1-y_1)+72(z_1-z_1)=0+0+0=0$$

Therefore, $(H,+)$ is a sub-group of $(\mathbb{Z},+)$. The last is to show that $(H,+)$ is a cyclic group, so for that, we shall look at the Gcd of $(30,42,72)$, when $(30,42,72)=6$, so by the definition of being in $H$. $$H=\{30x+42y+72z:x,y,z\in\mathbb{Z}\}$$ So $30x+42y+72z=6(5x+7x+12z)=6k$, when $k\in \mathbb{Z}$ then 6 is a creator element of $H$, thus, $(H,+)$ is cyclic. In conclusion, we have that $(H,+)$ is a cyclotomic sub-group of $(\mathbb{Z},+)$.

  • I don't know If I got it right, and my way is correct, so I will be glad if you can tell me whether my attempt for that question is correct, or not.

Thanks!

Bill Dubuque
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Anon142
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  • What is a cyclotomic subgroup? Do you mean cyclic subgroup? In this case you can use a theorem ( I know you like this): every subgroup of a cyclic group is cyclic. – Dietrich Burde Feb 10 '21 at 15:46
  • @DietrichBurde Yes I am sorry. – Anon142 Feb 10 '21 at 15:48
  • @DietrichBurde I fixed it. – Anon142 Feb 10 '21 at 15:50
  • Good. So all you need to do is to verify that $H$ is a subgroup. Show that $x-y\in H$ for all $x,y\in H$ and you are done - subgroup axiom! You can combine $x+y\in H$ and $-y\in H$ this way. Obviously $0\in H$, too. – Dietrich Burde Feb 10 '21 at 15:52
  • @DietrichBurde I have showed that, no? can you look carefully? – Anon142 Feb 10 '21 at 16:44
  • No, it's your job to look carefully. You have shown $x+y\in H$, but that is unnecessary then if you show first $x-y\in H$. – Dietrich Burde Feb 10 '21 at 17:06
  • @DietrichBurde Yes i'm sorry! what about the proof that the subgroup is cyclic, is it correct? – Anon142 Feb 10 '21 at 17:12
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    It is unnecessary, too. Every subgroup of a cyclic group is cyclic. This is a Theorem, and you like Theorems. Did you find this in the meantime? And since $(\Bbb Z,+)$ is cyclic, we are done. – Dietrich Burde Feb 10 '21 at 17:12
  • @DietrichBurde didn't get it. Why is it unnecessary? In order to show that she is a subgroup, I need to see that she is closed to the + method, and that she has an opposite number for each number in the group. – Anon142 Feb 10 '21 at 17:17
  • @DietrichBurde ok got it. thanks! – Anon142 Feb 10 '21 at 17:23
  • It seems you are not yet familiar with said general result about subgroups of cyclic groups but instead seek a direct proof, so I closed it as a dupe of the familiar Bezout style proof. The cyclic group form is also a dupe. – Bill Dubuque Feb 10 '21 at 20:26
  • If you know a little group theory note that that "closed under subtraction" is simply the subgroup test $\ \ $ – Bill Dubuque Feb 10 '21 at 20:29
  • @BillDubuque since the gcd of three numbers is 6, then I can say that H is actually defined to be all integer number to hold $6x$ when $x\in \mathbb{Z}$ so-6 is an the element creator of $H$ so that $H$ with + is cyclic? – Anon142 Feb 11 '21 at 18:00
  • @BillDubuque btw, and I'm sorry for bothering you, can you tell me why subtraction, and not +? because I have a plus method in the subgroup. – Anon142 Feb 11 '21 at 18:02

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