# The subgroups of a cyclic group

We've got $$G=U(\mathbb Z/(27)\mathbb Z)=\langle 2 \rangle$$ a cyclic group, and $$H=\langle -8, -1 \rangle$$ a subgroup of $$G$$. I've calculated all the subgroups of $$G$$. Now I have to indentify $$H$$ with a subgroup of $$G$$, without calculating all the elements of $$H$$.

So I think that I can see clearly that $$H$$ is equal to the subgroup $$\langle 8 \rangle =\{8,10,-1,-8,-10,1\}$$, but as the problem says that I can't calculate all the elements of H to solve this problem, I don't know how can I justify that H is equal to $$\langle 8 \rangle$$. How can I do it?

$$G\cong\Bbb Z_{18}$$. The subgroups are all cyclic, of orders $$1,2,3,6,9$$ and $$18$$. But $$H$$ has an element of order two ($$-1$$). Thus its order is even. But the order of $$H$$ is greater than $$2$$, since it contains $$1,-1,-8$$.

Thus it's the subgroup of order $$6$$, or $$G$$ itself.

It is not $$G$$ though: Note that we have $$|8|=|2^3|=18/(18,3)=18/3=6$$, $$|-8|=|2^{12}|=18/(12,18)=3$$. And since $$G$$ is abelian, for any two elements $$a,b\in G$$, we have $$|ab||\operatorname{lcm}(|a|,|b|)$$. Thus the maximum order of an element of $$H$$ is $$6$$.

• The order of $-8$ seems to be not equal to the order of $8$. $8=2^3$ has multiplicative order $6$ , but $-8 = 2^{12}$ has multiplicative order $3$. And sorry but I don't understand why $2\notin H$. Can you please explain more? – Menezio May 13 '20 at 9:57
• You're right. I have reworked it a little bit. – Chris Custer May 13 '20 at 15:03

We have $$|G|=27-9=18$$, hence $$2^9 = -1$$.

Now $$H=\langle -8,-1 \rangle = \langle -2^3, 2^9\rangle = \langle 2^9\cdot 2^3, 2^9\rangle = \langle 2^{12},2^9\rangle$$.

Now thanks to the Bezout Lemma we have $$H=\langle 2^{12},2^9\rangle = \langle 2^3\rangle = \langle 8\rangle$$.