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Let $G_{1}, G_{2}$ be two profinite groups, and $f: G_{1} \longrightarrow G_{2}$ is an continue injective homomorphism. Let $p$ be a prime number, then we get a pro-p completion $f^{p}: G_{1}^{p} \longrightarrow G_{2}^{p}$.

Is $f^{p}$ an injective homomorphism?

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    $\begingroup$ The homomorphism $f$ is continuous? Otherwise there is a trivial counterexample because $G_2^p$ can be $\\{1\\}$. $\endgroup$
    – user6976
    Jul 12, 2012 at 9:43

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Let $G_2$ be the group $A_5$, which is simple of order 60 and contains a subgroup $G_1$ of order 5. Let $f$ be the inclusion map. Then $G_1$ is its own pro-5 completion, whereas the pro-5 completion of $G_2$ is trivial, so $f^5$ is not injective.

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