Grushko theorem

Let F be a finitely generated free group, with n generators. Let G₁ and G₂ be two finitely presented groups. Suppose there exists a surjective homomorphism ${\displaystyle \phi :F\rightarrow G_{1}\ast G_{2}}$. Then there exists two subgroups F₁ and F₂ of F with ${\displaystyle \phi (F_{1})=G_{1}}$ and ${\displaystyle \phi (F_{2})=G_{2}}$, such that ${\displaystyle F=F_{1}\ast F_{2}.}$

Proof: We give the proof assuming that F has no generator which is mapped to the identity of ${\displaystyle G_{1}\ast G_{2}}$, for if there are such generators, they may be added to any of ${\displaystyle F_{1}}$ or ${\displaystyle F_{2}}$.

The following general results are used in the proof.

1. There is a one or two dimensional CW complex, Z with fundamental group F. By Van Kampen theorem, the wedge of n circles is one such space.

2. There exists a two complex ${\displaystyle X=X_{1}\cup X_{2}}$ where ${\displaystyle \{p\}=X_{1}\cap X_{2}}$ is a point on a one cell of X such that X₁ and X₂ are two complexes with fundamental groups G₁ and G₂ respectively. Note that by the Van Kampen theorem, this implies that the fundamental group of X is ${\displaystyle G_{1}\ast G_{2}}$.

3. There exists a map ${\displaystyle f:Z\rightarrow X}$ such that the induced map ${\displaystyle f_{\ast }}$ on the fundamental groups is same as ${\displaystyle \phi }$

For the sake of convenience, let us denote ${\displaystyle f^{-1}(X_{1})=:Z_{1}}$ and ${\displaystyle f^{-1}(X_{2})=:Z_{2}}$. Since no generator of F maps to identity, the set ${\displaystyle Z_{1}\cap Z_{2}}$ has no loops, for if it does, these will correspond to circles of Z which map to ${\displaystyle p\in X}$, which in turn correspond to generators of F which go to the identity. So, the components of ${\displaystyle Z_{1}\cap Z_{2}}$ are contractible. In the case where ${\displaystyle Z_{1}\cap Z_{2}}$ has only one component, by Van Kampen's theorem, we are done, as in that case, :${\displaystyle F=\Pi _{1}(Z_{1})\ast \Pi _{1}(Z_{2})}$.

The general proof follows by reducing Z to a space homotopically equivalent to it, but with fewer components in ${\displaystyle Z_{1}\cap Z_{2}}$, and thus by induction on the components of ${\displaystyle Z_{1}\cap Z_{2}}$.

Such a reduction of Z is done by attaching discs along binding ties.

We call a map ${\displaystyle \gamma$ :[0,1]\rightarrow Z} a binding tie if it satisfies the following properties

1. It is monochromatic i.e. ${\displaystyle \gamma ([0,1])\subseteq Z_{1}}$ or ${\displaystyle \gamma ([0,1])\subseteq Z_{2}}$

2. It is a tie i.e. ${\displaystyle \gamma (0)}$ and ${\displaystyle \gamma (1)}$ lie in different components of ${\displaystyle Z_{1}\cap Z_{2}}$.

3. It is null i.e. ${\displaystyle f\circ \gamma ([0,1])}$ is null homotopic in X.

Let us assume that such a binding tie exists. Let ${\displaystyle \gamma }$ be the binding tie.

Consider the map ${\displaystyle g:[0,1]\rightarrow D^{2}}$ given by ${\displaystyle g(t)=e^{it}}$. This map is a homeomorphism onto its image. Define the space ${\displaystyle Z'}$ as

${\displaystyle Z'=Z\coprod \!D^{2}/\!\sim }$ where :${\displaystyle x\!\!\sim y{\text{ iff}}{\begin{cases}x=y,{\mbox{ or }}\\x=\gamma (t){\text{ and }}y=g(t){\text{ for some }}t\in [0,1]{\mbox{ or }}\\x=g(t){\text{ and }}y=\gamma (t){\text{ for some }}t\in [0,1]\end{cases}}}$

Note that the space Z' deformation retracts to Z We first extend f to a function ${\displaystyle f'':Z\coprod \partial D^{2}/\!\sim }$ as

${\displaystyle f''(x)={\begin{cases}f(x),\ x\in Z\\p{\text{ otherwise.}}\end{cases}}}$

Since the ${\displaystyle f(\gamma )}$ is null homotopic, ${\displaystyle f''}$ further extends to the interior of the disc, and therefore, to ${\displaystyle Z'}$ . Let ${\displaystyle Z_{i}'=f'^{-1}(X_{i})}$ i = 1,2. As ${\displaystyle \gamma (0)}$ and ${\displaystyle \gamma (1)}$ lay in different components of ${\displaystyle Z_{1}\cap Z_{2}}$, ${\displaystyle Z_{1}'\cap Z_{2}'}$ has one less component than ${\displaystyle Z_{1}\cap Z_{2}}$.

The binding tie is constructed in two steps.

Step 1: Constructing a null tie:

Consider a map ${\displaystyle \gamma ':[0,1]\rightarrow Z}$ with ${\displaystyle \gamma '(0)}$ and ${\displaystyle \gamma '(1)}$ in different components of ${\displaystyle Z_{1}\cap Z_{2}}$. Since ${\displaystyle f_{\ast }}$ is surjective, there exits a loop ${\displaystyle \!\lambda }$ based at γ'(1) such that ${\displaystyle \!f(\gamma ')}$ and ${\displaystyle \!f(\lambda )}$ are homotopically equivalent in X. If we define a curve ${\displaystyle \gamma$ :[0,1]\rightarrow Z} as ${\displaystyle \gamma (t)=\gamma '\ast \lambda (t)}$ for all ${\displaystyle t\in [0,1]}$, then ${\displaystyle \!\gamma }$ is a null tie.

Step 2: Making the null tie monochromatic:

The tie ${\displaystyle \!\gamma }$ may be written as ${\displaystyle \gamma _{1}\ast \gamma _{2}\ast \cdots \ast \gamma _{m}}$ where each ${\displaystyle \gamma _{i}}$ is a curve in ${\displaystyle Z_{1}}$ or ${\displaystyle Z_{2}}$ such that if ${\displaystyle \gamma _{i}}$ is in ${\displaystyle Z_{1}}$, then ${\displaystyle \gamma _{i+1}}$ is in ${\displaystyle Z_{2}}$ and vice versa. This also implies that ${\displaystyle f(\gamma _{i})}$ is a loop based at p in X. So,

${\displaystyle [e]=[f(\gamma )]=[f(\gamma _{1})]\ast \cdots \ast [f(\gamma _{m})]}$

Hence, ${\displaystyle [f(\gamma _{j})]=[e]}$ for some j. If this ${\displaystyle \!\gamma _{j}}$ is a tie, then we have a monochromatic, null tie. If ${\displaystyle \!\gamma _{j}}$ is not a tie, then the end points of ${\displaystyle \!\gamma _{j}}$ are in the same component of ${\displaystyle Z_{1}\cap Z_{2}}$. In this case, we replace ${\displaystyle \!\gamma _{j}}$ by a path in ${\displaystyle Z_{1}\cap Z_{2}}$, say ${\displaystyle \!\gamma _{j}'}$. This path may be appended to ${\displaystyle \!\gamma _{j-1}}$ and we get a new null tie

${\displaystyle \gamma ''=\gamma _{1}\ast \cdots \ast \gamma _{j-1}'\ast \gamma _{j+1}\cdots \gamma _{m}}$, where ${\displaystyle \!\gamma _{j-1}'=\gamma _{j-1}\ast \gamma _{j}'}$.

Thus, by induction on m, we prove the existence of a binding tie.

Suppose that ${\displaystyle G=A*B}$ is generated by ${\displaystyle \{g_{1},g_{2},\ldots ,g_{n}\}}$. Let ${\displaystyle F}$ be the free group with ${\displaystyle n}$-generators, viz. ${\displaystyle \{f_{1},f_{2},\ldots ,f_{n}\}}$. Consider the homomorphism ${\displaystyle h:F\rightarrow G}$ given by ${\displaystyle h(f_{i})=g_{i}}$, where ${\displaystyle i=1,\ldots ,n}$.

By the lemma, there exists free groups ${\displaystyle F_{1}}$ and ${\displaystyle F_{2}}$ with ${\displaystyle F=F_{1}\ast F_{2}}$ such that ${\displaystyle h(F_{1})=A}$ and ${\displaystyle h(F_{2})=B}$. Therefore, ${\displaystyle {\text{Rank }}(A)\leq {\text{Rank }}(F_{1})}$ and ${\displaystyle {\text{Rank }}(B)\leq {\text{Rank }}(F_{2})}$. Therefore, ${\displaystyle {\text{Rank }}(A)+{\text{Rank }}(B)\leq {\text{Rank }}(F_{1})+{\text{Rank }}(F_{2})={\text{Rank }}(F)={\text{Rank }}(A\ast B).}$