Non trivial subgroups in cyclic group G

Question

Following on from my amazing journey on better understanding group theory and number theory, I know want to understand how one finds non-trivial subgroups in a cyclic group G

So for example, I want to find five non-trivial subgroups from the following cyclic group $G = \mathbb{Z}_{97}^{\times}$

The order of $G = \mathbb{Z}_{97}^{\times}$ is of course 96 as 97 is prime and thus we do 97-1=96.

Now we also know that the factors of 96 are {1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96}.

The question is what is a non-trivial group and how would I find non-trivial groups, five at least?

Answer 1

Let $G=\Bbb{Z}_{97}^{\times}$. Consider $2 \in G$.

What is the order of $2$ in $G$?

The possible orders are divisors of $96$, namely $\{1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96\}$. Of course the order is not $1$. But instead of checking each power separately, we can try a different approach. Note that $96=2^5 \cdot 3$. So we will determine if $$2^{96/2}=2^{48}\equiv 1 \pmod{97} \qquad \text{ or} \qquad 2^{96/3}=2^{32}\equiv 1 \pmod{97}.$$ If neither of these holds, then the order of $2$ will be $96$, hence it will be a generator. Unfortunately, (you can try it) $2^{48} \equiv 1 \pmod{97}$. This means the order of $2$ is $\leq 48$. In fact, we can show that it is $48$. So $\langle 2 \rangle$ will be a subgroup of order $48$.

Now try the same approach with $5$ instead of $2$. You will see that $$5^{96/2}=5^{48}\not\equiv 1 \pmod{97} \qquad \text{ and} \qquad 5^{96/3}=5^{32}\not\equiv 1 \pmod{97}.$$

This shows that order of $5$ is $96$ and hence it is a generator of $G$.

Now what will be the order of say $5^3$? Suppose it is $t$, then $$5^{3t} \equiv 1 \pmod{97}.$$ This means $96 \mid 3t$. The smallest such $t$ is $32$. So the subgroup $$\langle 5^3 \rangle=\{(5^3)^k \, | \, k =1,2, \ldots\}$$ has order $32$. Likewise we can show that the order of $5^8$ is $12$. So the subgroup $$\langle 5^{8} \rangle=\{(5^{8})^k \, | \, k =1,2, \ldots\}$$ has order $12$ and so on.

Now use different powers of $5$ to generate subgroups of different orders. There are results which can make things go fast but I am deliberately avoiding those here so that you can develop your own understanding first.

Hopefully you can pick up from here.

Answer 2

A subgroup is defined to be "proper" if it is not the whole group. A proper subgroup is non-trivial if it is not the subgroup of size $1$ consisting of only the identity element.

Since $G$ is cyclic of order $96$, there is an element $a \in G$ such that $G = \{a^n ~ \vert ~ 1 \leq n \leq 96 \}$. Find a value of $a$ that works (there will be more than one). Then if $ 1 \lt m \text{ and } m \vert 96$, the set $H=\{(a^m)^k ~ \vert ~ 1 \leq k \leq \frac{96}{m} \}$ is a nontrivial subgroup of $G$.

You will find it a useful exercise to prove that $H$ as I have defined it is in fact a subgroup, and that $H$ is proper $ \iff \gcd (m, 96) \gt 1$. It should be clear that $H$ is non-trivial unless $96 \vert m$.

Answer 3

Let's try a smaller group. Say, $\Bbb Z_{11}^×$.

The order is $10$ and it's cyclic. Let's get a generator. Since $2^5\cong{10}\pmod{11}$ and $2^2\cong4\pmod{11}$, the result is that $2$ is a generator (since $2$ does not have order $1,2$ or $5$).

Now all the subgroups will be of the form $\langle 2^k\rangle$ for some $k$.

In fact, $\vert\langle 2^k\rangle \vert=\dfrac{10}{\operatorname {gcd}(10,k)}$.

The situation is just the same for $\Bbb Z_{97}^×$. The finding of a generator is just a little more cumbersome, because of the size of the group.

For any $\Bbb Z_p^×$, the situation is the same.