# Property of free group from categorical definition

I'm proving some properties about the free groups using the categorical definition, which is the following:

Let $X$ be a set, $L$ a group and $i:X\to L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:X\to H$ there exists a unique group homomorphism $\varphi:G\to H$ such that $\varphi\circ i=f$.

(1) $i$ must be injective and that

(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $\varphi:L_1\to L_2$ such that $\varphi\circ i_1=i_2$.

Now I'm stuck with the following:

If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1\cong L_2$.

I tried using a biyection $g:X\to Y$ and (2) to find $\varphi:L_1\to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?

• I'm going to presume something along the lines of 'Note that there must be a similar $\phi : L_2\mapsto L_1$ and prove their compositions $\phi\circ\varphi$ and $\varphi\circ\phi$ are natural isomorphisms?' – Steven Stadnicki Jul 31 '18 at 17:38

Fix a bijection $g: X\rightarrow Y$ with inverse $h: Y\rightarrow X$. Then $f_1= i_2\circ g: X\rightarrow L_2$ defines a $\varphi_1: L_1\rightarrow L_2$ and similarly $f_2= i_1\circ h: Y\rightarrow L_1$ defines a $\varphi_2: L_2\rightarrow L_1$. The composition $\varphi_2\circ \varphi_1: L_1\rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: X\rightarrow X$. As this map "extends" uniquely to a homomorphism $L_1\rightarrow L_1$ and the identity $id_{L_1}$ is obviously such a homomorphism, we have $\varphi_2\circ \varphi_1= id_{L_1}$. Similarly $\varphi_1\circ \varphi_2= id_{L_2}$, thus $\varphi_1$ is the desired isomorphism.

• To "extend" the map to a homomorphism $L_1\to L_1$ do I have to assume that $L_1=\langle{X}\rangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group. – Javi Jul 31 '18 at 18:14
• No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way. – A. Pongrácz Jul 31 '18 at 18:18
• I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$. – Musa Al-hassy Aug 1 '18 at 15:12

While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.

That $L, 𝒾$ is the free group for $X$ means that there is a way to “extend” maps from $X$ to homomorphisms from $L$,

$$⟨\_⟩ \,:\, (X → H) \,⟶\, (L → H)$$

Such that this is indeed an extension when restricted to the elements of $X$,

$$(0) \qquad ϕ ∘ 𝒾 = f \quad≡\quad ϕ = ⟨f⟩$$

In (0) taking $ϕ$ to be $⟨f⟩$ yields

$$(1) \qquad\qquad\qquad\qquad ⟨f⟩ ∘ 𝒾 = f$$

In (0) taking $ϕ, f$ to be $id, i$ yields

$$(2) \qquad\qquad\qquad\;\;\;\;\qquad id = ⟨𝒾⟩$$

Now suppose we are given $g : X ≅ Y$ where $X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.

Indeed we can show this to be true with a simple calculation: --omitting the subscripts--

  ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡{ Using (2) with an aim to utilising (0) }
⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡{ Now in a position to utilise (0) }
⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡{ Inverses and identity laws }
𝒾 = 𝒾
≡{ Reflexivity of equality }
true

• Thanks for your approach :) – Javi Aug 1 '18 at 19:30