Return to Answer

Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.

So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is nonzero

Let $G = C_4 \times C_2$ be the direct product of cyclic groups of orders $4$ and $2$, with the direct factors generated by elements $g$ and $h$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.

I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.

Consider the group defined by the following presentation. (I am putting it in Magma format for ease of cutting and pasting.)

E := Group< g, h, a, b | g^4=a*b, h^2=a^2=b^2=1, a^g=b, b^g=a,
                         a^h=a, b^h=b, a*b=b*a, h^g=h*a >;

You can check that $|E|=32$, and you can see directly from the presentation that it is an extension of $M$ by $G$ with the prescribed induced module action of $G$ on $M$.

Now the extension corresponding to the image of the corresponding element of $H^2(G,M)$ under the induced map $H^2(G,M) \to H^2(G,T)$ is

  Group< g, h, t | g^4=1, h^2=t^2=1, t^g=t, t^h=t, h^g=h*t >;

which defines a nonabelian group of order $16$, so it cannot be the split extension of $T$ by $G$.

Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.

So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is nonzero

Let $G = C_4 \times C_2$ be the direct product of cyclic groups of orders $4$ and $2$, with the direct factors generated by elements $g$ and $h$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.

I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.

Consider the group defined by the following presentation. (I am putting it in Magma format for ease of cutting and pasting.)

E := Group< g, h, a, b | g^4=a*b, h^2=a^2=b^2=1, a^g=b, b^g=a,
                         a^h=a, b^h=b, a*b=b*a, h^g=h*a >;

You can check by computer that $|E|=32$ (it is $\mathtt{SmallGroup}(32,7))$, and you can see directly from the presentation that it is an extension of $M$ by $G$ with the prescribed induced module action of $G$ on $M$.

Now the extension corresponding to the image of the corresponding element of $H^2(G,M)$ under the induced map $H^2(G,M) \to H^2(G,T)$ is

  Group< g, h, t | g^4=1, h^2=t^2=1, t^g=t, t^h=t, h^g=h*t >;

which defines a nonabelian group of order $16$, so it cannot be the split extension of $T$ by $G$.

There is a similar example with $G$ dihedral of order $8$.

Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.

So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is not surjective.

Let $G = \langle g,h\rangle$ be the dihedral group of order $8$ with $g=(1,2,3,4)$ and $h=(1,3)$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.

I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.

Consider the group extension of $T$ by $G$ in which $g^4$ is equal to the element $1$ of $T$, and $h^2$ is equal to $0$. This extension is isomorphic to the dihedral group of order $16$.

The corresponding element of $H^2(G,T)$ cannot be in the mage of $H^2(G,M)$, because if it was then, ion the corresponding extension of $M$ by $G$, we would have $g^4 = (1,0)$ or $(0,1)$. But that's impossible because $g$ does not centralize these module elements.

Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.

So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is nonzero

Let $G = C_4 \times C_2$ be the direct product of cyclic groups of orders $4$ and $2$, with the direct factors generated by elements $g$ and $h$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.

I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.

Consider the group defined by the following presentation. (I am putting it in Magma format for ease of cutting and pasting.)

E := Group< g, h, a, b | g^4=a*b, h^2=a^2=b^2=1, a^g=b, b^g=a,
                         a^h=a, b^h=b, a*b=b*a, h^g=h*a >;

You can check that $|E|=32$, and you can see directly from the presentation that it is an extension of $M$ by $G$ with the prescribed induced module action of $G$ on $M$.

Now the extension corresponding to the image of the corresponding element of $H^2(G,M)$ under the induced map $H^2(G,M) \to H^2(G,T)$ is

  Group< g, h, t | g^4=1, h^2=t^2=1, t^g=t, t^h=t, h^g=h*t >;

which defines a nonabelian group of order $16$, so it cannot be the split extension of $T$ by $G$.

Example was incorrect and has been deleted.

Here is a new attempt at an example. I prefer to denote the $1$-dimensional module for $G$ over the field of order $2$ with trivial action by $T$ rather than by ${\mathbb Z}_2$, which is used with too many different meanings.

So now we are just looking for an example in which the induced map $H^2(G,M) \to H^2(G,T)$ is not surjective.

Let $G = \langle g,h\rangle$ be the dihedral group of order $8$ with $g=(1,2,3,4)$ and $h=(1,3)$, and define $\pi$ by $\pi(g) = 1$ and $\pi(h) = 0$.

I prefer to describe the example in terms of group extensions rather than cocycles, but I can calculate a corresponding $2$-cocycle if you like.

Consider the group extension of $T$ by $G$ in which $g^4$ is equal to the element $1$ of $T$, and $h^2$ is equal to $0$. This extension is isomorphic to the dihedral group of order $16$.

The corresponding element of $H^2(G,T)$ cannot be in the mage of $H^2(G,M)$, because if it was then, ion the corresponding extension of $M$ by $G$, we would have $g^4 = (1,0)$ or $(0,1)$. But that's impossible because $g$ does not centralize these module elements.