If $p$ is prime then the additive group $\mathbb{Z}_p$ has no proper non-trivial subgroup. If $p$ is prime then the additive group $\mathbb{Z}_p$ has no proper non-trivial subgroup.
I need to use the theorem that any subgroup of a cyclic group is cyclic.
My thoughts:
Let $H$ be a subgroup of $\mathbb{Z}_p$.
Then there's  $a$ $\in H$. Since $\mathbb{Z}_p$ is cyclic, $H$ is cyclic by the theorem. So $H =\langle a\rangle$.
From here, what I know is that  $\mathbb{Z}_p$ can be generated by any element of it since $p$ is prime.
But I don't know how to prove this part.
 A: Let $H$ be a subgroup of $\mathbb Z_p$. By your theorem $H$ is cyclic.
Instead of considering a random $a\in H$, we choose $a$ such that it is a generator of $H$, i.e. $\langle a \rangle =H$.
Let $g$ be a generator of $\mathbb Z_p$. Since $a \in \mathbb Z_p$, we can write $a = g^k$ for some $k \in \mathbb N$.
If $p \mid k$, $a = e$, and $\langle e \rangle = \{e\}$, which is the trivial subgroup.
Suppose $p \nmid k$. We prove that $\langle a \rangle = \mathbb Z_p$ by showing that $\{a, a^2, \dots, a^p\} \subseteq \langle a \rangle$ and $a, a^2, \dots, a^p$ are distinct.
Suppose not. Then $a^i = a^j$ for some $i < j$.
Then $e=a^{j-i}=g^{k(j-i)}$.
This shows that $p \mid k(j-i)$, and by $p \nmid k$ and Euclid's Lemma $p \mid (j-i)$.
But $1 \le j - i \le p - 1$. This is a contradiction.
Hence $a, a^2, \dots, a^p$ are $p$ distinct elements of $\langle a \rangle$.
But $|\mathbb Z_p| = p$ and $|\langle a \rangle|\ge p$. This forces $\mathbb Z_p = \langle a \rangle$.
Therefore $\mathbb Z_p$ contains no non-trivial proper subgroups.
A: The cleanest way is to use Lagrange's theorem, that the order of any subgroup of a finite group divides the order of the group.
Since a prime has no divisors, we are done.
Btw, this shows $\Bbb Z_p$ is simple.
