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Let $G=Z_2= \{1,2\}$, cyclic of order 2. Let $A = Z/4Z$ and suppose $G$ acts on $A$ by inversion. Show $|H^2(G,A)|=2$ by direct computation.

So consider $0\rightarrow C^0(G,A)\rightarrow C^1(G,A)\xrightarrow{d_1} C^2(G,A)\xrightarrow{d_2} C^3(G,A)$. For $f\in Z^2(G,A)$ we have $d_2(f)$ zero map. Now $d_2(f)(g,h,i)=gf(h,i)-f(gh,i)+f(g,hi)-f(g,h)$.

I get from $(g,h,i)=(1,2,1),(1,2,2),(2,1,1),(2,1,2)$ that $f(1,2)=f(2,1), f(1,1)=f(2,2), f(1,1)=-f(1,1), f(1,2)=-f(1,2)$. so $f:G\times G \rightarrow A$ has 4 choices, letting $f(1,2)=f(2,1)=s, f(1,1)=f(2,2)=t$, and $s,t$ take values in $\{0,2\}$. Then $|Z^2(G,A)|=4$

Then for $B^2(G,A)$ I get that $d_1(f)(g,h)=gf(h)-f(gh)+f(g)$. so $(1,1)\mapsto f(1)$, $(1,2)\mapsto f(1)$, $(2,1)\mapsto -f(1)$, $(2,2)\mapsto -f(1)$. Then $d_1(f)$ depends on $f(1)$. But we can find $f$ with $f(1)$ in any of $\{0,1,2,3\}$, which means $|B^2(G,A)|=4$.

But then the quotient is trivial.

Could you help me identify the bug? Is it the $d_1,d_2$, the $f$, or something else? I have verified again and again with no avail.

Jun Xu
  • 449

1 Answers1

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The mistake is in the computation by definition of $d_2(f)$. After correction, I have $B^2(G,A)=\{c_0, c_2, g_1, g_2\}$, where $c_i$ is constant on $I$, and $g_1(1,1)=g_1(1,2)=1$, $g_1(2,1)=g_1(2,2)=3$, and $g_2$ takes opposite values of $g_1$.

Then for $f\in Z^2(G,A)$, $f(1,1)=f(1,2)=-s$, $f(2,1)=s$, $f(2,2)=t$, where $(s,t)\in \{(0,0),(0,2),(1,1),(1,3),(2,0),(2,2),(3,1),(3,3)\}$.

Jun Xu
  • 449