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Is it true that if $F_a(X)$ is the free group with free basis $X$, and the same for $F_a(Y)$, and $X,Y$ have the same cardinality, then there is an isomorphism between $F_a(X)$ and $F_a(Y)$?

user73564
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2 Answers2

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Yes, this follows immediately from the universal property of a free group. If $\phi: X \to Y$ is a bijection, then composing it with inclusion $Y \subset F(Y)$ we get function $X \to F(Y)$ which uniquely determines a homomorphism $F(X) \to F(Y)$ that maps bijectively $X \subset F(X)$ to $Y \subset F(Y)$. Similarly, considering $\phi^{-1}: Y \to X$, we get a unique homomorphism $F(Y) \to F(X)$ that maps $Y \subset F(Y)$ to $X \subset F(X)$. Now the composition of these two is identity.

xyzzyz
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  • But why the composition of $\phi$ with inclusion determines a homomorphism instead of an ordinary map? – Hilder Vitor Lima Pereira Jun 25 '17 at 10:50
  • If we denote the inclusion by $i: Y \hookrightarrow F(Y)$, then indeed, $i \circ \phi: X \to F(Y)$ is only an ordinary map. However, the universal property of the free group $F(X)$ says precisely that any ordinary function $X \to G$, where $G$ is some group, gives rise to a unique homomorphism $F(X) \to G$, and that's how we get a homomorphism $F(X) \to F(Y)$ from an ordinary map $X \to F(Y)$. – xyzzyz Jun 26 '17 at 17:57
  • Yes, man. I have just started studying it and I had misunderstood the definition of the universal property. Shame on me. I am sorry. – Hilder Vitor Lima Pereira Jun 27 '17 at 11:04
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Hint: You know that there is a bijection $X \to Y$; how might you extend this to a group isomorphism?

Clive Newstead
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