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I had asked this question on MathStackExchange but did not get any response. It will be very helpful to have an answer:

I have some doubts from the book 'Homological algebra' by Cartan and Eilenberg. On page 253, the authors describe a resolution as follows:

Fix a presentation of the group $$Q_{4t}=\langle x,y:x^t=y^2,xyx=y\rangle.$$

Let $\Sigma=\mathbb Z[G]$, and $a_p,b_p,b_p',c_p,c'_p,e_p$ be a list of abstract generators. Consider \begin{align*} R_{4p}=\Sigma a_p,\,R_{4p+1}=\Sigma b_p+\Sigma b_p',\, R_{4p+2}=\Sigma c_p+\Sigma c_p',\,R_{4p+3}=\Sigma e_p, \end{align*} and the maps \begin{align*} \begin{aligned} d a_p &= (1+x+\cdots+x^{2t-1})\, e_{p-1},\\ d b_p &= (x-1)a_p,\\ d b_p' &= (y-1)a_p, \end{aligned} \qquad \begin{aligned} d c_p &= (1+x+\cdots+x^{t-1})b_p-(y+1)b_p',\\ d c_p' &= (xy+1)b_p + (x-1)b_p',\\ d e_p &= (x-1)c_p - (xy-1)c_p'. \end{aligned} \end{align*}

Then they say that "the verification that $d^2$=0 is straightforward." I am now trying to prove that $d^2=0$. Here are a few calculations:

\begin{align*} 1. d^2(b)&=d((x-1)a)=(x-1)da=(x^{2t}-1)a=0,\\ 2. d^2(c)&= d((xy+1)b+(x-1)b')=(xy+1)db+(x-1)db'\\ &=(xy+1)(x-1)a+(x-1)(y-1)a\\&=(xyx+x-xy-1+xy-x-y+1)a=0,\text{ since } xyx=y\\ 3. d^2(c')&=d((xy+1)b+(x-1)b')=(xy+1)(x-1)a+(x-1)(y-1)a\\ &=(xyx+x-xy-1+xy-x-y+1)a=0\\ 4. d^2(e)&= d((x-1)c-(xy-1)c')\\ &=(x-1)(1+x+\ldots+x^{t-1})b - (x-1)(y+1)b'-(xy-1)(xy+1)b -(xy-1)(x-1)b'\\ &=(x^t-1-(xyx)y+xy-xy+1)b - (xy-y+x-1+xyx-x-xy+1)b'\\ &=0, \text{ since } x^t=y^2, xyx=y. \end{align*} Now when we compute $d^2a$, we get \begin{align*} d^2a&=d((1+x+\cdots+x^{2t-1})e)\\ &=(1+x+\cdots+x^{2t-1})(x-1)c-(1+x+\cdots+x^{2t-1})(xy-1)c'\\ &=0-(1+x+\cdots+x^{2t-1})(xy-1)c'. \end{align*}

My question is: How do we show that $d^2a=0$? The same problem also occurs while computing $d^2b'=(y-1)(1+x+\cdots+x^{2t-1})$.

Since $$(y-1)(1+x+\cdots+x^{2t-1})=(1+x+\cdots+x^{2t-1})(y-1)=(1+x+\cdots+x^{2t-1})(xy-1),$$ because of $x^tyx^t=y$ and $(1+x+\cdots+x^{2t-1})x=(1+x+\cdots+x^{2t-1})$, it is enough to prove $d^2b'=0$. My final goal is to compute group cohomology with the help of this resolution. Thanks in advance.

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    $\begingroup$ I don't think $da_p$ is correct. For, if you apply $Hom_\Sigma(-,\mathbb{Z})$ to the complex $R_\ast$, then $d^\ast:Hom_\Sigma(R_3,\mathbb{Z})\to Hom_\Sigma(R_4,\mathbb{Z})$ is multiplication by $2t$. This leads to $H^4(Q_{4t},\mathbb{Z})=\mathbb{Z}/2t$, while the correct cohomology in degree 4 is $\mathbb{Z}/4t$. I suppose $da_p$ should be the norm element of $\Sigma$. BTW: Doesn't your presentation of $Q_{4t}$ miss the relations $x^{2t}=y^4=1$? $\endgroup$ Commented 13 hours ago
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    $\begingroup$ Indeed, in the book the formula C&E write is «da_p=Ne_{p-1}». If one backtracks from that point to where $N$ is last defined, it is defined as the norm element of the cyclic group generated by $x$, but that is just another example: $N$ is actually defined as the norm of the group of whatever group one works with in Section 1 (whose title is, appropriatedly, Norms :-) ), so that is what is meant here. $\endgroup$ Commented 12 hours ago
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    $\begingroup$ @tj_ Regardless of correctness of a formula for the differential, this two relator presentation indeed gives the generalized quaternion group: because $Yx^t y = x^{-t}$, and $x^t = y^2$ commutes with $y$, we have $x^{t} = x^{-t}$. $\endgroup$ Commented 12 hours ago
  • $\begingroup$ @MarianoSuárez-Álvarez Thank you very much. I will put this question off as there are downvotes saying it does not adhere to the community guidelines. $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @tj_, you are right, it should be the norm element. I started in the middle of the book to practice some calculations. In the previous example, it was $N=\sum\limits_{i=0}^{|x|} x^i$. So I was calculating with that. $\endgroup$ Commented 11 hours ago

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As you noticed, $x$ acts (by multiplication) on the subgroup of $\mathbb ZG$ generated by $(1 + x + \dots + x^{2t-1})$ as the identity. Therefore, $x^k$ does the same, so $(1 + x + \dots + x^{2t-1})(1 - x^k) = 0$ for any $k$. Finally, $(1 - xy) = (1 - x^{t+1})$.

Computations in groups rings were one of the most fruitful sources of computational mistakes even in published papers, so I'd not worry much if it feel hard. It is; but, luckily, not so much in this particular case.

I made a bold assumption that the question was correctly representing an exercise from Cartan-Eilenberg, but nevertheless the general reasoning still stands.

Every finite group ring has two notable ideals.

One ($N$, norm ideal) is generated by $\sum_{g \in G} g$; the other ($\Delta$, augmentation ideal) is the kernel of $\mathbb Z[G] \to \mathbb Z$.

Because augmentation ideal is characterised by the property that its elements have sum of coefficients equal to zero, and every group element acts as identity on the norm ideal, we have $N \Delta = 0$.

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  • $\begingroup$ In my opinion the formula for $da_p$ in the OP is wrong. See my comment above. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ @Denis T thank you for pointing out the mistake and the explanation. I will put this question off as there are downvotes saying it does not adhere to the community guidelines. $\endgroup$ Commented 11 hours ago

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