# Suppose $\pi_1(X) \cong A_5$. Compute homology groups of X

The full problem is... Let $$F$$ be the free group generated by $$a$$ and $$b$$ and $$R$$ be the smallest normal subgroup containing $$a^2, b^3, (ab)^5$$, then $$F/R$$ is isomorphic to the alternating group $$A_5$$. Construct a space $$X$$ with the fundamental group isomorphic to $$A_5$$. Compute the homology groups of $$X$$.

Here's what I think... By Hurewicz Theorem, $$H_1$$ is also $$A_5$$. Outside of that, I don't know how to use $$\pi_1$$ to calculate homology groups.

One idea that I have is trying to work Van Kampen's Theorem backwards to find a decomposition of X into spaces that could fit into a Mayer Vietoris sequence. However, I'm confused about how to find the normal subgroup in Van Kampen's theorem, which we mod out by to get $$\pi_1(X)$$. So I'm struggling to create the appropriate pushout square.

Does this approach seem reasonable? If so, could you please help me figure out how to put the pieces together. If not, do you have a hint of another approach?

• $A_5$ is simple, so Hurewicz says that $H_1$ vanishes. But you didn’t define $X$. Is, eg, $X$ a $K(A_5,1)$? Jun 22, 2020 at 19:32
• This is impossible to answer, many spaces that don't have the same homology have $\pi_1 = A_5$ Jun 22, 2020 at 20:07
• I voted to close yor question because it lacks details. What is $X$? Jun 22, 2020 at 23:14
• @Mindlack my apologies. I'm editing the question now. I was missing the line "construct a space $X$ with fundamental group isomorphic to $A_5$." My professor sent me this problem. Is it still impossible to solve? Jun 23, 2020 at 14:34
• Homology of $X$ (in degrees $\ge 2$) will depend on which $X$ you are using. Jun 23, 2020 at 15:08

Your question is no longer impossible. As people mentioned in the comments, there are many different spaces with $$\pi_1(X)\cong A_5$$, but your question is now framed in a way that suggests looking at a particular such space.

Using the presentation $$A_5\cong\langle a,b\mid a^2,b^3,(ab)^5\rangle$$ given in the problem statement, one can construct a 2-dimensional CW complex $$X$$ with $$\pi_1(X)\cong A_5$$ as follows: give $$X$$ a single $$0$$-cell $$e^0=*$$, 2 different $$1$$-cells $$e^1_a,e^1_b$$ with attaching maps $$\partial e^1_a,\partial e^1_b\to X^0=\{*\}$$ necessarily collapsing their boundaries to a point (i.e. the 1-skeleton $$X^1$$ is just a wedge of two circles), and 3 different $$2$$-cells $$e^2_{a^2},e^2_{b^3},e^2_{(ab)^5}$$ with attaching maps described by their subscript.

For example, the attaching map $$S^1\cong\partial e^2_{(ab)^5}\to X^1\cong S^1\vee S^1$$ corresponds to the element $$(ab)^5\in\pi_1(X^1)\cong\langle a,b\rangle$$ (i.e. its a loop that goes around the $$e^1_a$$ cell and then goes around the $$e^1_b$$ cell and then around the $$e^1_a$$ cell again and then around the $$e^1_b$$ cell again and so on 5 times in total).

Using Van Kampen, one shows that this space $$X$$ has $$A_5\cong\langle a,b\mid a^2,b^3,(ab)^5\rangle$$ as its fundamental group.

We now turn to the homology of $$X$$. As Mindlack mentioned, since $$A_5$$ is simple (and nonabelian), $$H_1(X)\cong\pi_1^{\mathrm{ab}}(X)\cong A_5/[A_5,A_5]=0$$. Since $$X$$ is a 2-dimensional CW-complex we also know $$H_k(X)=0$$ for all $$k\ge3$$, so we're just left with $$H_2(X)$$. For this, we use the cellular chain complex

$$0\to H_2(X^2,X^1)\to H_1(X^1, X^0)\to H_0(X^0)\to 0,$$

which, in this case, looks like

$$0\to\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}\to\mathbb Z\to 0,$$

so we're interested in calculating the map $$\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}$$ above. This map sends $$[e^2_{a^2}]\mapsto c[e^1_a]+d[e^1_b]$$ where $$c$$ is the degree of the map

$$S^1\cong\partial e^2_{a^2}\to X^1\to X^1/(X^1\setminus e_a^1)\cong S^1$$

and $$d$$ is the degree of the map

$$S^1\cong\partial e^2_{a^2}\to X^1\to X^1/(X^1\setminus e_b^1)\cong S^1.$$

Since $$e^2_{a^2}$$ is attached via the word $$a^2$$ (i.e. it loops around $$e_a^1$$ twice), we see that $$c=2$$ and $$d=0$$.

One can similarly determine the images of $$e^2_{b^3}$$ and $$e^2_{(ab)^5}$$. In the end, you get that the map $$\mathbb Z^{\oplus 3}\to\mathbb Z^{\oplus 2}$$ is given, using the ordered bases $$\{e^2_{a^2},e^2_{b^3},e^2_{(ab)^5}\}$$ and $$\{e^1_a,e^1_b\}$$, by the matrix

$$\begin{pmatrix} 2 & 0 & 5\\ 0 & 3 & 5 \end{pmatrix}$$

From the cellular chain complex, the second homology group $$H_2(X)$$ is given by the kernel of this matrix, so we see that

$$H_2(X)\cong\mathbb Z$$

with explicit generator $15[e^2_{a^2}]+10[e^2_{b^2}]-6[e^2_{(ab)^5}]$$. • thank you. How do you know how to break$X$up into 2 sets to use Van Kampen's Theorem? I'm not sure what those sets would be. Also you have a SES$0 \rightarrow H_2(X^2,X^1) \rightarrow H_1(X^1,X^0)\rightarrow H_0(X^0)\rightarrow 0$. Is there a name for this sequence? It looks similar to the relative long exact sequence I know, but that one is$H_n(A)\rightarrow H_n(X) \rightarrow H_n(X,A)\rightarrow \dots$Jun 25, 2020 at 17:46 • How is the kernel of the matrix$\mathbb{Z}$? If I call the matrix$A$, then I solved$AX=0$for general 3x1 matrix$X$with$x_1,x_2,$and$x_3$in the first, second, and third rows respectively. I got a 3x1 matrix with$3/2 x_2, x_2,$and$x_3$as my solution. I've only used boundary maps in the past, not matrices. Either way I struggle a bit Jun 25, 2020 at 18:04 • For applying van Kampen, looking at Proposition 1.26 in Hatcher would probably be more clear than what I can say in a comment. Intuitively, at least,$\pi_1(X)$is generated by the loops$e^1_a,e^1_b$(if part of a loop lands in the interior of a 2-cell, you can homotope it to the boundary) and the boundary of any 2-cell is nullhomotopic, so$\pi_1(X)$should at least be a quotient of$A_5$. There are no other obvious relations, so hopefully it is believable that it is exactly$A_5$. The sequence I use is called the "cellular chain complex"... Jun 25, 2020 at 21:05 • it is a useful alternative to the singular chain complex when calculating homology of a CW-complex. If you don't want to appeal to it, you can look at the LES of the pair$(X,X^1)$which incluces$0\to H_2(X)\to H_2(X,X^1)\to H_1(X^1)\to0$. Since$ H_1(X^1)\cong\mathbb Z^2$and$ H_2(X,X_1)= H_2(X^2,X^1)\cong\mathbb Z^3$, you still get that$ H_2(X)\cong\mathbb Z$. The kernel of the matrix is tuples$(x_1,x_2,x_3)$with$2x_1=-5x_3=3x_2$. Since$x_1,x_2,c_3\in\mathbb Z$, this is free abelian of rank 1 with generator$(x_1,x_2,x_3)=(15,10,-6)\$. Jun 25, 2020 at 21:19