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Sep 27, 2023 at 5:42 vote accept Jordi Daura
Sep 26, 2023 at 18:26 comment added Ian Agol @user513804: yes, I think that’s right. At first I thought I could answer the question in general, until I looked up the cohomology of infinite lens spaces and realized it’s not generated in dim 1 for odd order. Still the restriction of H^2 to cyclic subgroups should be surjective which might be useful.
Sep 26, 2023 at 16:50 comment added Jordi Daura Thanks for the reply. Your answer gives more evidence that $Inn(\pi_1)$ has always elements of infinite order. The only step in your proof that cannot be generalized to the case of $Inn(\pi_1)$ being a $p$-group is the map to $H^1(<a>,\mathbb{F}_p)$ being non-trivial where $a$ has order $p$, because of the ring structure of $H^1(<a>,\mathbb{F}_p)$, right?
Sep 26, 2023 at 1:20 history answered Ian Agol CC BY-SA 4.0