Recently I have encountered a problem concerning the property of slant product in group cohomology. The problem is as follows: Consider a finite group G (can have anti-unitary operations). And there is a center $Z_N$ of G generated by group element $g$, where the center is not a subgroup of any other center. Then we can construct the following sequence utilizing the slant product: $$H^3(G,U(1))\xrightarrow{i_g^3}H^2(G,U(1))\xrightarrow{i_g^2}H^1(G,U(1)).$$ where slant products are as follows:

for a 3-cocycle $\omega(a,b,c)$, $$i_g^3\omega(a,b)=\omega(g,a,b)\omega(a,b,g)/\omega(a,g,b),$$

for a 2-cocycle $x(a,b)$, $$i_g^2x(a)=x(g,a)/x(a,g)$$.

It is apparent that $im(i_g^3)\subset ker(i_g^2)$. But is it true that $ker(i_g^2)=im(i_g^3)$?

Recently I have encountered a problem concerning the property of slant product in group cohomology. The problem is as follows: Consider a finite group G (can have anti-unitary operations). And there is a center $Z_N$ of G generated by group element $g$. Then we can construct the following sequence utilizing the slant product: $$H^3(G,U(1))\xrightarrow{i_g^3}H^2(G,U(1))\xrightarrow{i_g^2}H^1(G,U(1)).$$ where slant products are as follows:

for a 3-cocycle $\omega(a,b,c)$, $$i_g^3\omega(a,b)=\omega(g,a,b)\omega(a,b,g)/\omega(a,g,b),$$

for a 2-cocycle $x(a,b)$, $$i_g^2x(a)=x(g,a)/x(a,g)$$.

It is apparent that $im(i_g^3)\subset ker(i_g^2)$. But is it true that $ker(i_g^2)=im(i_g^3)$?