0

I'm trying to prove the free group on 3 generators is not isomorphic to the free group on 2 generators. I have that there are many injections $F_3 \hookrightarrow F_2$ and of course $F_2 \hookrightarrow F_3$, which implies there is a bijection between the two, and I suppose the bijection is not going to be homomorphic and hence not an isomorphism. But I'm not sure how to prove this: I can explicitly show various injective homomorphisms $F_3 \hookrightarrow F_2$ but how do you prove that all such maps between the two free groups are not isomorphisms?

I've seen the post here, a rather long discussion that isomorphism implies equal cardinality. This question is the converse which is not true: equal cardinality does not imply isomorphism.

BBrooklyn
  • 224
  • 4
    Consider their (respective) abelianizations. – lulu Jun 13 '22 at 14:52
  • We have $|X|=2\neq 3=|Y|$. – Dietrich Burde Jun 13 '22 at 14:53
  • Actually both groups are infinite, they differ in the number of generators. But by showing injections both ways the cardinality of both groups is equal by Schroeder Bernstein. – BBrooklyn Jun 13 '22 at 14:58
  • @lulu is there a simpler approach? This was a bonus problem in week 1 of an old course and all we had at this point was basic group theory (Artin ch. 2) and the definition of a free group. – BBrooklyn Jun 13 '22 at 14:59
  • 3
    BBrooklyn, the answer by "bof" at the duplicate is very elementary:"Let $G$ be some group of order $n$ where $1\lt n\lt\aleph_0$. If $F_X\cong F_Y$, then the number of homomorphisms from $F_X$ to $G$ is equal to the number of homomorphisms from $F_Y$ to $G$, that is, $n^{|X|}=n^{|Y|}$. If $\min(|X|,|Y|)$ is finite, it follows from $n^{|X|}=n^{|Y|}$ that $|X|=|Y|$. If $|X|$ and $|Y|$ are both infinite, then $|X|=|F_X|=|F_Y|=|Y|$." – Dietrich Burde Jun 13 '22 at 15:11
  • 1
    What lulu suggested is this duplicate. We have $(F_2){ab}\cong \Bbb Z^2$ and $(F_3){ab}\cong \Bbb Z^3$, so they are not isomorphic. This is not much simpler than the above argument, I think. – Dietrich Burde Jun 13 '22 at 15:15
  • 1
    You are confusing what "equal cardinality" means in the post you link to. It's the cardinality of a free generating set, not of the groups itself. Equal cardinality of free generating set definitely implies isomorphism. – Arturo Magidin Jun 13 '22 at 16:02

0 Answers0