Group is Subgroup of Itself

Theorem

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


Then:

$\struct {G, \circ} \le \struct {G, \circ}$


That is, a group is always a subgroup of itself.


Proof

By Set is Subset of Itself, we have that:

$G \subseteq G$

Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.2$. Subgroups: Example $90$
  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Subgroups: Example $25$
  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36.1$: Subgroups
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $4$: Subgroups: Definition $4.1$
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