# Two generators of $F_2$ must be free?

Suppose we are given $a, b \in F_2$, that happen to generate it. Then must they be free generators? That is, there is no non-trivial reduced word on $a^{\pm 1}, b^{\pm 1}$ defining the identity. Equivalently, $a, b$ satisfy the universal property defining $F_2$.

I tried attacking this the first way: suppose $u$ and $v$ are free generators, then if I take a word on $a$ and $b$ I can rewrite it in terms of $u$ and $v$. But how do I know that if I took a non-trivial reduced word on $a, b$ defining the identity, it does not become trivial after reduction with $u$ and $v$?

I also tried to use the universal property. Then we get that if $a$ and $b$ are not free, $F_2$ is isomorphic to one of its propert quotients. Maybe this is not the case for $F_2$ specifically, but it may be for other groups, so no contradiction in sight once again.

I know this is true because it is used in a class I am taking. To prove that $SL_2(\mathbb{Z})$ is not free, the professor just exhibited two generators and a non-trivial relation.

• Can you give an example of non-free generators? – freakish Jun 26 '18 at 8:37
• $a, b, a^{-1}$ are non-free generators. The claim is that you cannot have two non-free generators... – frafour Jun 26 '18 at 8:39
• I assume these generators are non-free because $a^{-1}$ is evaluable from $a$. So if I remove $a^{-1}$ I still get generators. So is the set of free generators the same as the minimal set of generators? Then obviously $\mathbb{F}_2$ cannot have a minimal set of generators with only 1 element. Otherwise it would be $\mathbb{Z}$. – freakish Jun 26 '18 at 8:40
• @freakish: They are non-free because there is a nontrivial word in three letters that reduces to the identity when you plug in $a, b, a^{-1}$ for the three letters. – Hurkyl Jun 26 '18 at 9:03
• @Hurkyl Yes, then I'm asking what's the difference between "non-free" and "minimal"? – freakish Jun 26 '18 at 9:04

Your question is equivalent to the following:

Is every surjective group homomorphism $F_2 \to F_2$ necessarily injective?

Groups with this property are called Hopfian groups: see the Wikipedia article .

In particular, it is stated that $F_n$ is Hopfian for all $n$: proof.

Thus, the answer to your question is yes: every set of 2 generators of $F_2$ is free.

• Wait, the linked proof shows that $F_n$ is residually finite, but how do we know that finitely generated residually finite groups are Hopfian? – Vincent Jun 26 '18 at 9:08
• @Vincent sorry, I pasted the wrong link. Thanks for having alerted me. – Crostul Jun 26 '18 at 9:11
• Just went through the proof, very clear. Thanks a lot! – frafour Jun 26 '18 at 9:17
• math.stackexchange.com/questions/2447820/… – Crostul Jun 26 '18 at 9:22
• As you see, this is not specific to $n=2$. However there's a more specific fact for pairs: in a free group, any two elements freely generate a free group, unless they commute (which is easily checkable, and actually holds iff they have a common root). – YCor Jun 26 '18 at 16:04