Subgroup of Order 1 is Trivial

Theorem

Let $\struct {G, \circ}$ be a group.

Then $\struct {G, \circ}$ has exactly $1$ subgroup of order $1$: the trivial subgroup.

Proof

From Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $\struct {G, \circ}$.

Suppose $\struct {\set g, \circ}$ is a subgroup of $\struct {G, \circ}$.

From Group is not Empty, $e \in \set g$.

Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$.

That is, $\struct {\set e, \circ}$ is the only subgroup of $\struct {G, \circ}$ of order $1$.

$\blacksquare$

Sources

• 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts