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I'm never sure about free groups whether a question is easy or not. It feels to me like this is impossible, but I couldn't come up with any argument.

If $F_n$ is a free group on $n$ generators, could there exist a non-trivial $N\triangleleft F_n$ such that $F_n/N \cong F_n$?

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No, the free groups are Hopfian because they are residually finite.

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    $\begingroup$ And free groups are residually finite because, by a simple application of the ping-pong lemma, they can be embedded into $SL_2(\mathbb{Z})$. $\endgroup$ Jul 14, 2011 at 22:37
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    $\begingroup$ @Alain: this is not the easiest proof. Consider $F_2= < a,b >$ and any reduced word $w=c_1c_2...c_n\ne 1$ in $F_2$, $c_i\in \{a,b,a^{-1}, b^{-1}\}$. The letters $a,b$ produce two partial permutations of the set $\{0,1,...,n\}$: $c_i$ takes $i-1$ to $i$. Complete these partial permutations to full permutations $\alpha, \beta$. And let $G$ be the (finite) subgroup of $S_{n+1}$ generated by $\alpha, \beta$. Let $\phi$ be the homomorphism $F_2\to G$ taking $a$ to $\alpha$, $b$ to $\beta$. Then the image of $w$ is not the trivial permutation because $\phi(w)$ takes $0$ to $n$. $\endgroup$
    – user6976
    Jul 14, 2011 at 23:59
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    $\begingroup$ @ Mark: I respectfully disagree. The ping-pong argument gives, in a completely elementary way, a faithful representation of $F_2$ into $SL_2(\mathbb{Z})$, which yields much more than residual finiteness. $\endgroup$ Jul 15, 2011 at 18:29
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    $\begingroup$ @Alain: this argument also gives more than residual finiteness. For example, it gives LERF. But of course being in $SL_2(\mathbb{Z})$ is also important. $\endgroup$
    – user6976
    Jul 15, 2011 at 19:00
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    $\begingroup$ I think Mark provided O. Schreier's original proof. Both proofs are quite elementary and of independent interest. Being of very different flavor (automaton-theoretic vs using matrix groups and elementary arithmetic), it's hard to call one easier (it's really a matter of taste). $\endgroup$
    – YCor
    May 1, 2020 at 10:12

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