When the action on cohomology is trivial?

Suppose $$G$$ is a group acting freely on a topological space $$X$$. Take an element $$g$$ of $$G$$. My question is: If the induced action of $$g$$ on cohomology with $$\Bbb Z$$-coefficient is trivial then when does it follow that induce map on cohomology with $$\Bbb Z_2$$-coefficient is also trivial?

• A trivial sufficient condition: when $H^*(X,Z)$ is torsion-free. – user43326 Nov 24 '18 at 14:53
• In that case $H^*(X,Z/2)$ is naturally isomorphic to $H^*(X,Z)\otimes Z/2$, so $g_{Z/2}^*=g_Z^*\otimes Z/2$. – user43326 Nov 27 '18 at 17:19

Consider the group $$A=\mathbb{Z}\times\mathbb{Z}/2$$ and the automorphism $$f(n,m)=(n,n+m)$$. This gives an automorphism $$g=Bf$$ of the classifying space $$X=BA$$. Standard methods show that $$H^*(X;\mathbb{Z}/2)=E[x]\otimes(\mathbb{Z}/2)[y]$$ with $$|x|=|y|=1$$ and $$g^*(x)=x$$ and $$g^*(y)=x+y$$, so in particular $$g^*$$ is not the identity. On the other hand $$H^*(X;\mathbb{Z})=E[x]\otimes\mathbb{Z}[z]/(2z)$$ with $$|z|=2$$. Here $$g^*(z)$$ must be $$z$$, because there is no other nonzero element in $$H^2(X)$$ that it could be, and $$g^*(x)=x$$ because $$x$$ comes from $$H^1(B\mathbb{Z})$$. So $$g^*$$ is just the identity on integral cohomology.
If we want to make sure that the group $$\langle g\rangle$$ acts freely, we can replace $$X$$ by $$\mathbb{R}\times X$$ and $$g$$ by $$(t,x)\mapsto (t+1,g(x))$$.
• What is $E[x]$? – Qfwfq Nov 22 '18 at 10:34
• $E[x]$ is the exterior algebra on the generator $x$, which is just $\mathbb{Z}\oplus\mathbb{Z}x$ and is the cohomology of $B\mathbb{Z}=S^1$. – Neil Strickland Nov 22 '18 at 11:48
• I guess if we set $f(n,m)=(n+m,m)$ instead, we get an action of $Z/2$ on $BA$, so you get an example with $G$ finite. – user43326 Dec 1 '18 at 18:04