Homology of hexagon My question has 4 parts, it is pretty long. Hence I would appreciate even the tiniest hint / comment regarding how to do this question. Thanks!
Part 1) Compute $\pi_1(X)$.
Part 2) Compute the integral homology $H_* (X;\mathbb{Z})$. 
Part 3) Compute the mod 2 homology $H_*(X;\mathbb{Z}/2\mathbb{Z})$.
Part 4) Compute $H_*(X,\mathbb{Z}/p\mathbb{Z})$ for an odd prime $p$. 

Consider $X$, the quotient space of the hexagon with identification on its edges as follows.
Part 1) Compute $\pi_1(X)$.
My solution: $\pi_1(X)=\langle a,b,c\mid a^2b^2c^2\rangle$. Is that correct? I did it using van Kampen's Theorem.
Part 2) Compute the integral homology $H_* (X;\mathbb{Z})$. 
My attempt: I managed to compute $H_0(X,\mathbb{Z})=\mathbb{Z}$, based on the following $\Delta$-complex.
However, I am stuck with $H_1(X,\mathbb{Z})$. I calculated $\partial a=\partial b=\partial c=0$, hence concluded that $\ker \partial_1=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$.
Then, I calculuated $\partial_2 (S_1)=a+e-d$, ..., $\partial_2 (S_6)=h-g+b$, and concluded that $\text{Im}\ \partial_2 =\mathbb{Z}^{\oplus 6}$ which seems to be wrong..
Thanks a lot once again.

Part 3) Compute the mod 2 homology $H_*(X;\mathbb{Z}/2\mathbb{Z})$.
Part 4) Compute $H_*(X,\mathbb{Z}/p\mathbb{Z})$ for an odd prime $p$. 
I am totally stuck with parts 3 and 4.
 A: For $\textbf{part 2}$:
The linear algebra is a bit messy, but you should get that $\ker \partial_1 =  \langle a,b,c,d-i,d-h,f-g,e-f,d-e \rangle$
and $\text{Im } \partial_2 = \langle 2a-d+f,2b-f+h,2c+d-h, -a+c+2d-e-i, -d+2e-f,a-b-e+2f-g \rangle$
But, what's important is that $\partial_2(S_1 + S_2 + \ldots + S_6) = (a+e-d) + (a+f-e) + (b+g-f) + (b+h-g) + (c+i-h) + (c+d-i) = 2a + 2b + 2c$ 
which you see from the first three generators of the image of $\partial_2$.  
After some algebra, a lot of things in the quotient cancel out and we're left with:
\begin{align*}
H_1(X;\mathbb{Z}) &= \ker \partial_1 /\text{Im } \partial_2 \\
&= \langle a,b,c \rangle / \langle 2a + 2b + 2c\rangle \\
&= \langle a,b,a+b+c \rangle / \langle 2a + 2b + 2c \rangle \cong \mathbb{Z}^2 \oplus \mathbb{Z}/2\mathbb{Z}
\end{align*}
So, you weren't too far off for $H_1$.  It seems like you got $H_2$ and $H_0$ but here's a quick run-through:
$\partial_0 = 0$ and $\text{Im }\partial_1$ is generated by $v-w$ so 
$H_0(X;\mathbb{Z}) = \ker \partial_0/ \text{Im }\partial_1 = \mathbb{Z}^2 / \mathbb{Z} = \mathbb{Z}$.
$\partial_2$ is injective so $H_2 = 0$.
Now, for $\textbf{part 3}$:
We're just changing coefficients so $H_0(X;\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$.  For $H_1(X;\mathbb{Z}/2\mathbb{Z}),$ $\text{Im } \partial_2$ vanishes so we get three generators for $H_1$, meaning $H_1(X;\mathbb{Z}/2\mathbb{Z}) = (\mathbb{Z}/2\mathbb{Z})^3$.  For $H_2$, note that $\ker \partial_2 = \langle S_1 + S_2 + \ldots + S_6 \rangle$ so we got a generator that we didn't have beforehand.  So, $H_2(X;\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$.
That should help out with part $4$ too.
A: Your part 1) is correct, however there is a justification that you failed to mention: all six vertices of the hexagon are identified to a single vertex in the quotient space. This needs to be said or else your Van Kampen application would be incorrect (I do see that you indicated this pretty clearly in your picture).
For part 2), the six vectors you get need not be linearly independent. So you really need to get a basis (over $\mathbb{Z}$) for their span (over $\mathbb{Z}$). Do this using linear algebra and matrices (over $\mathbb{Z}$).
Parts 3) and 4) should be easier than part 2), because the coefficients are fields, and linear algebra over fields is easier than linear algebra over $\mathbb{Z}$.
