Group is Subgroup of Itself
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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 $4.1$