Im struggling with the following question:
"Suppose that $X$ is a topological space admitting a cell decomposition whose only non-zero homology groups are $$ H_0(X) = \mathbb{Z} \oplus \mathbb{Z}, H_1(X) = (\mathbb{Z}/2), H_2(X) = \mathbb{Z}, H_3(X) = \mathbb{Z} \oplus (\mathbb{Z}/3). $$ Show that any cell decomposition of X has at least 8 cells."
I assumed that we had $a$ 0-cells, $b$ 1-cells , $c$ 2-cells and $d$ 3-cells. So constructed a chain group $$ 0\rightarrow\mathbb{Z}^d\xrightarrow{\partial_3}\mathbb{Z}^c\xrightarrow{\partial_2}\mathbb{Z}^b\xrightarrow{\partial_1}\mathbb{Z}^a\rightarrow 0 $$ So obviously we can see from the homologies that we must have $a\geq 2$, $b\geq 1$, c$\geq 1$ and $d\geq 2$. I then attempted to show that if $a=2$ or $d=2$ we would get a contradiction and then this would give us that $a+b+c+d\geq 8$, but I struggled to do so. Any help or advice would be appreciated.
The rank $b_2$ of the image of $\partial_2$ is not trivial since $H_1(X)$ has torsion. This implies that $c\geq 2$. Since $H_3(X)$ has torsion, the image of $\partial_4$ is not trivial, we deduce that $e\geq 1$ where $e$ is the number of $4$ cells.