Free Groups and Rank (6.3.1)

Dummit and Foote Abstract Algebra, section 6.3, exercise 1:

Let $\langle~a_1,...,a_n~\rangle=F_1$ and $\langle~b_1,...,b_m~\rangle=F_2$ be free groups. Prove that $F_1≅F_2$ if and only if $n=m$. Consider the case when the two's ranks are infinite, as well.

Proof: ($⇒$) Assume $n≠m$, and generally $n$ is of smaller cardinality than $m$. Let $φ$ be the isomorphism from $F_1$ to $F_2$. For an arbitrary element $x∈F_2$, we have $x=φ(a_1^{k_1}...a_n^{k_r})=φ(a_1)^{k_1}...φ(a_r)^{k_r}$, so that $F_2$ is generated by a number of elements of smaller than or equal cardinality to those of $F_1$, a contradiction. ($\Leftarrow$) Let there be a bijection between the generators of $F_1$ and the generators of $F_2$, operating on the index by $\pi$. Define a mapping $φ : F_1 → F_2$ by $φ(a_i)=b_{\pi(i)}$ and homomorphically extend it. For any two generators in $F_1$, we have:$$φ(a_i)^x=φ(a_j)^y⇒b_{\pi(i)}^x=b_{\pi(j)}^y⇒\pi(i)=\pi(j)⇒i=j⇒a_i=a_j$$Now observe: $$φ(a_1^{k_1}...a_r^{k_r})=φ(a_1^{j_1}...a_s^{j_s})⇒b_{\pi(1)}^{k_1}...b_{\pi(r)}^{k_r}=b_{\pi(1)}^{j_1}...b_{\pi(s)}^{j_r}⇒∀i[b_{\pi(i)}^{k_i}=b_{\pi(i)}^{j_i}]$$$$⇒∀i[φ(a_i)^{k_i}=φ(a_i)^{j_i}]⇒∀i[a_i^{k_i}=a_i^{j_i}]⇒a{\pi(1)}^{k_1}...a_{\pi(r)}^{k_r}=a_1^{j_1}...a_s^{j_s}$$So that $φ$ is an injective homomorphism. Similarly, we can construct an injective homomorphism from $F_2$ into $F_1$, therefore providing an isomorphism. This argument did not assume the finiteness of rank, merely the equivalency of cardinality, so is therefore applicable to infinite groups as well.$~\square$

Warning: Invalid proof and erroneous extension of the Bernstein-Schroeder theorem.