# Prime Group has no Proper Subgroups

## Theorem

A nontrivial group $G$ has no proper subgroups except the trivial group if and only if $G$ is finite and its order is prime.

## Proof

### Sufficient Condition

Suppose $G$ is finite and of prime order $p$.

From Lagrange's Theorem, the order of any subgroup of $G$ must divide the order $p$ of $G$.

From the definition of prime, any subgroups of $p$ can therefore only have order $1$ or $p$.

Hence $G$ can have only itself and the trivial group as subgroups.

$\Box$

### Necessary Condition

Suppose $G$ is not finite and prime.

Let the identity of $G$ be $e$.

Let $h \in G$ be an element of $G$ such that $h \ne e$.

Then $H = \gen h$ is a cyclic subgroup of $G$.

If $H \ne G$ then $H$ is a non-trivial proper subgroup of $G$, and the proof is complete.

Otherwise, $H = G$ is a cyclic group and there are two possibilities:

$(1): \quad G$ is infinite
$(2): \quad G$ is finite (and of non-prime order).

First, suppose $G$ is infinite.

From Infinite Cyclic Group is Isomorphic to Integers, $G$ is isomorphic to $\struct {\Z, +}$.

From Subgroups of Additive Group of Integers, $\struct {\Z, +}$ has proper subgroups, for example: $\gen 2$.

Because $G \cong \struct {\Z, +}$, then so does $G$ have proper subgroups, and the proof is complete.

Suppose $G$ is finite, and of order $n$ where $n$ is not prime .

Then:

$\exists d \in \N: d \divides n, 1 < d < n$

From Subgroup of Finite Cyclic Group is Determined by Order, $G$ has a proper subgroup of order $d$ and again, the proof is complete.

$\blacksquare$