Let $F_a$ be the group which is freely generated by $a$ elements. How to show that there is a homomorphism from $F_a$ onto $F_b$ if and only if $b\le a$?

I was thinking one possibility is if $F_a$ generated freely by $\{x_1, \dots , x_a\}$ and $F_b$ generated freely by $\{y_1, \dots , y_b\}$ then consider map which says $\phi(x_i)=y_i$ for $1\le i \le b$ and $y_1$ otherwise. I do not know if this works or even how to proceed. How to show this fact?

Suppose $\;n=a<b\;$ and further suppose $\;\phi:F_a\to F_b\;$ is an epimorphism.

So $\;\phi(F_a)=F_b\implies F_b=\langle\,\phi(x_1),..,\phi(x_n)\,\rangle\;$, which of course is impossible as $\;F_b\;$ cannot be generated, free or not, by less than $\;b\;$ elements

  • you're just re stating the question differently – Noam Kolodner Aug 23 '17 at 15:05

Hint: Let $f : F_a \to F_b$ be a surjective homomorphism. Look at $F_a / ker(f) \simeq F_b$. What do you know about the generators of the quotient?

Your map works! Just check that it is indeed a homomorphism and surjective.

For the other direction, argue by contradiction. A homomorphism is surjective iff its surjective on the generators. If $a<b$ then $F_b$ is generated by $(\phi(x_j))_{j=1}^a$. Contradiction!

  • I think showing surjectivity is straighforward..but how to show it is a homomorphism? – Leo Mar 2 '16 at 21:45
  • Show $\phi(xx') = \phi(x)\phi(x')$ for reduced words $x,x'$. The reduced words are just finite strings of the $x_j's$. – GiantTortoise1729 Mar 2 '16 at 23:03

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