Cardinality of Linearly Independent Set 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 $G$ iff it contains exactly $n$ elements.

Proof

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

Then by Linearly Independent Subset of Finitely Generated Vector Space, $H$ has at most $n$ elements.

By Bases of Finitely Generated Vector Space and Basis of Vector Space is Linearly Independent and a Generator, $H$ 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 Linearly Independent Subset of Basis of Vector Space there exists a basis $C$ of $G$ such that $H \subseteq C \subseteq H \cup B$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 27$: Theorem $27.14$