$F_n$ cannot be generated by fewer than $n$ elements

Question

Let $F(S)$ be the free group generated by the set $S$ with $|S|=n$. I need to show that there is no generating set $T$ of $F(S)$ with $|T|<n$.

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So far I noticed that if $|T|$ generated $F(S)$, then we could use a surjective map $\phi:S\to T$. Due to the universal property, this extends to a homomorphism $\phi:F(S)\to F(S)$ with nontrivial kernel. Then $F(S)=\langle \phi(S)\mid\ker \phi\rangle$.

When showing the existence of free groups, one gets that $F(S)=\langle S\mid\_ \ \rangle$. But I don't know how to show that this gives a contradiction.

Answer 1

At the risk of being accused of completely ignoring your approach, here is a solution that comes directly from the definition of a free group. In one of your comments, you said you were looking for a solution that comes directly from the definition. But even so, this is essentially the same solution as J.-E. Pin's, and as the solution suggested by by deyore in comments.

According to the definition of $F(S)$, for any group $G$, any map $S \to G$ extends uniquely to a homomorphism $F(S) \to G$. So there is a bijection between the set of maps $S \to G$ and the set of homomorphisms $F(S) \to G$.

So let $G$ be the cyclic group of order $2$. Then the number of maps from $S$ to $G$ is $2^{|S|}$. On the other hand, if $F(S)$ is generated by $T$, then any homomorphism from $F(S)$ to $G$ is uniquely determined by the images of the elements in $T$, and there are at most $2^{|T|}$ possible images, so $2^{|T|} \le 2^{|S|}$ and hence (since $S$ is finite), $|T| \le |S|$.

Answer 2

Hint. Let $F_n$ be a free group with $n$ generators $a_1, \ldots, a_n$. Let $h: F_n \to (\mathbb{Z}/2\mathbb{Z})^n$ be the map defined by $h(a_i) = (0, \ldots, 0, 1, 0, \ldots, 0)$, where the unique $1$ occurs in $i$-th position. Verify that $h$ extends uniquely to a surjective group morphism. Now suppose that $F_n$ admits $m < n$ generators $g_1, \ldots, g_m$. Then $h(g_1), \ldots, h(g_m)$ would generate $(\mathbb{Z}/2\mathbb{Z})^n$. Is that possible?