Cardinality of Linearly Independent Set is No Greater than Dimension

Theorem
Every linearly independent subset of $G$:
 * $(1): \quad$ has at most $n$ elements
 * $(2): \quad$ is contained in a basis for $G$
 * $(3): \quad$ is a basis for $G$ it contains exactly $n$ elements.

Proof
Let $H$ be a linearly independent subset of $G$.

Then by Size of Linearly Independent Subset is at Most Size of Finite Generator, $H$ has at most $n$ elements.

By Bases of Finitely Generated Vector Space have Equal Cardinality and Sufficient Conditions for Basis of Finite Dimensional Vector Space, $H$ is itself a basis iff it has exactly $n$ elements.

By hypothesis there is a basis $B$ of $G$ with $n$ elements.

Then $H \cup B$ is a generator for $G$.

So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $G$ such that $H \subseteq C \subseteq H \cup B$.