Fundamental group of a Cayley Graph used to Calculate Kernels of Group Morphisms

I'm studying Algebraic Topology and I have a question concerning the calculation of free generators for the kernel of a group morphism.

Suppose we have a regular based connected covering space $p : (E, e) \rightarrow \vee_{i = 1}^n S^1$. There is a corresponding short exact sequence $0 \rightarrow \pi_1(E, e) \stackrel{\alpha}{\rightarrow} F(a_1, ..., a_n) \stackrel{\beta}{\rightarrow} Q \rightarrow 0$, where $F(a_1, ..., a_n)$ is the free group on $n$ generators, and $Q$ is the cokernel. There is a right action $\cdot$ of $F_n$ on $Q$ induced by the morphism $F_n \rightarrow Q$. We can make a Cayley graph $C$ whose vertices are elements of $Q$ and whose edges are of the form $e : q \rightarrow q \cdot a_i$.

I'm wondering if the map $\pi_1(C) \rightarrow F(a_1, ..., a_n)$ is a kernel for $\beta$, and if so, how to show this. If so, we can recover free generators of $\pi_1(E, e)$ from $\beta$.

If this is true, then for example we could easily calculate the kernel of the map $F(a, b) \rightarrow S_3$ where $a \mapsto (1 \ 2)$ and $b \mapsto (1 \ 2 \ 3 )$ in terms of free generators.

I think I'm missing something obvious, but I'm a bit stuck. I've read most of May's chapter on covering spaces (hopefully this provides some context).

• What you are asking is unclear to me. You already have a kernel for $\beta$, namely $\alpha(\pi_1(E,e))$. And since $\alpha$ is injective, you can already easily calculate free generators of the kernel or, equivalently, of $\pi_1(E,e)$, because $E$ is a connected graph and you can use the same method you would use for any connected graph: choose a maximal tree $T \subset E$, and there is a generator for each edge in $E-T$. – Lee Mosher Mar 14 '17 at 2:20
• Sorry the question was put poorly. What I mean is that from $\beta$ we can recover free generators of $\alpha(\pi_1(E, e))$ using the Caley graph. In the example I gave, we don't know what $\pi_1(E, e)$ is, at least not right away. – Dean Young Mar 14 '17 at 13:55
• Okay, that makes more sense, I think I can write up an answer... – Lee Mosher Mar 14 '17 at 15:02

I hope I better understand now what you are asking. It's not so much that we are given the covering map. Instead we are given the quotient homomorphism. So your question could be reworded like this (please correct me if I'm still off base):

• Given a surjective homomorphism $$\beta : F(\alpha_1,...,\alpha_n) \to Q$$ can one construct a free basis for $N=\text{kernel}(\beta)$ using the graph $C$ in your question?

And the answer is yes, because the graph $C$ that you construct is the same as the labelled graph $E$, where $(E,e)$ is the pointed covering space of $(X = \vee_{i=1}^n S^1,x)$ corresponding to the subgroup $N$, each edge of $E$ being labelled by its image edge in $X$. (Thus there is a group isomorphism $\pi_1(E,e) \approx N$ described by the edge labelling of $E$, so one can produce a free basis of this group in the usual manner one does for the fundamental group of any graph, as outlined in my comment above.)

The reason this holds is because of a basic theorem of fundamental groups, which goes like this.

• Hypothesis: $p : (Y,y) \to (X,x)$ is a based connected regular covering map (with other hypotheses on $X$ that all hold for $X = \vee_1^n S^1$).
• Conclusion: The image of the induced injection $p_* : \pi_1(Y,y) \to \pi_1(X,x)$ is a normal subgroup $N < \pi_1(X,x)$ and there is a natural isomorphism between the quotient group $Q = \pi_1(X,x) \,/\, N$ and the deck transformation group of the map $p$ acting on $Y$.

Applying this to your situation, and using that $x$ is the unique vertex of $X$, we obtain a free action of $Q$ on $Y$ which is transitive on the vertex set $p^{-1}(x)$.

So if one identifies $p^{-1}(x) \approx Q$ using this action, and if one then attaches edges as you've described, then yes, the resulting labelled graph is isomorphic to $E$.

In the example you ask about, the number of vertices of $E$ would equal $6$ which is the order of the quotient group $S_3$, so the degree of the covering map $p : E \to X$ would equal $6$, so the number of edges would be $2 \cdot 6 = 12$, and the rank of the graph $E$ would therefore equal $12-6+1=7$. The edges would be labelled as you say, $6$ of them labelled $a$ and $6$ of them labelled $b$. Thus the kernel is a free subgroup of rank $7$ and index $6$. In $E$ one can choose a maximal tree consisting of $5$ edges and then use the remaining $7$ edges to write down a free basis of the kernel.