# Tour:No proper nontrivial subgroup implies cyclic of prime order

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WHAT YOU NEED TO DO: Understand, thoroughly, the statement and proof of this theorem.

## Statement

If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order. In other words, it is generated by a single element whose order is a prime number.

## Converse

Further information: Prime order implies no proper nontrivial subgroup

The converse is clearly true from Lagrange's theorem. Namely, if we have a cyclic group of prime order $p$, any subgroup must have order either 1 or $p$, by Lagrange's theorem. The only possible subgroup of order $p$ is the whole group, and the only possible subgroup of order 1 is the trivial subgroup.

Note that this also shows something stronger: any group of prime order, since it has no proper nontrivial subgroup, must be cyclic of prime order.

## Proof

Given: A nontrivial group $G$, such that the only subgroups of $G$ are the trivial subgroup, and $G$ itself

To prove: There exists an element $g \in G$ such that $\langle g \rangle = G$, and the order of $g$ is a prime number. In particular, $G = \{ e, g, g^2, \dots, g^{p-1} \}$ with $g^p = e$

Proof: Since $G$ is nontrivial, there exists $g \ne e$ in $G$. Then, consider the subgroup generated by $g$. This is a nontrivial subgroup, hence, by assumption, $\langle g \rangle = G$.

Now there are two possibilities. First, that $g$ has infinite order: no positive power of $g$ is trivial. In this case, the group $G$ is isomorphic to (i.e., can be identified with) the group $\mathbb{Z}$, under the identification $n \mapsto g^n$. In particular, the subgroup generated by $g^2$, which corresponds to the even integers, is a proper nontrivial subgroup, leading to a contradiction.

Hence, $g$ cannot have infinite order, so let $n$ be the order of $g$. Then, $g^n = e$. Suppose that $n$ is composite. Then, $G$ is isomorphic to the cyclic group of order $n$, under the identification $a \mapsto g^a$. Pick a positive integer $d \ne 1,n$ such that $d | n$. Then the subgroup generated by $g^d$ is a proper nontrivial subgroup of $G$ (corresponds to the multiples of $d$ mod $n$). More precisely, it is a subgroup of order $n/d$, because the order of $g^d$ is $n/d$. This is again a contradiction.

Hence, $n$ must be a prime, so $G$ is cyclic of prime order (as desired).