DEFINITIONS:
A word in $X ∪ X^{−1}$ is an ordered set of $n ∈ N ∪ {0}$ elements, each from $X ∪ X^{−1}$, with repetitions allowed. We write a word in the following way: $w = x_{\lambda_1}^{\epsilon_1} · · · x_{\lambda_n}^{\epsilon_n}$, where $i = ±1$. The number n is the length of the word.
Two words $u, v$ in $X ∪ X^{−1}$ are said to be adjacent if there exists words $w, w' \in X ∪ X^{−1}$ and $a ∈ X ∪ X^{−1}$
such that either (1) $u = ww'$ and $v = waa^{−1}w'$; or (2)
$v = ww'$ and $u = waa^{−1}w'$. If u, v are adjacent then we write $u ∼ v$.
Let $u, v$ be words in $X ∪ X^{−1}$. We say that $u, v$ are equivalent if there exists a finite sequence of words $w_1, · · · , w_n$ such that
$u = w_1 ∼ w_2 ∼ · · · ∼ w_n = v$. We denote equivalence of words by the notation $u ≈ v$.
We denote the equivalence class of a word $w$ by $[w]$.
Define $[u][v] = [uv]$.
A free group on the non-empty set $X$ is the set of equivalence classes of words in $X ∪ X^{−1}$.
We define the rank of $F(X)$ to be $|X|$.
I know that this is probably quite obvious, but I wanted to check/improve my reasoning.
A free group of rank $n$ is generated by $n$ elements. Therefore, since $k \le n$, we can generate a free group from $k$ of $F_n$'s generating elements.
Is this correct? How could I make it more rigorous if so?
Best Answer
Yes, your intuition is correct.
For a rigorous proof, it's enough to show that if $X\subseteq Y$, then $F(X)$ is a subgroup of $F(Y)$, which is quite immediate from the construction.