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 $6$.

  • $\begingroup$ 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? $\endgroup$ – Menezio May 13 '20 at 9:57
  • 1
    $\begingroup$ You're right. I have reworked it a little bit. $\endgroup$ – 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 $H=\langle 2^{12},2^9\rangle = \langle 2^3\rangle = \langle 8\rangle$.


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