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How can I prove that a free group is residually nilpotent group.

Definition- A group G is residually nilpotent if for every non-trivial element $g$ there is a homomorphism $h$ from G to a nilpotent group such that $h(g)\neq e$.

I have done that free groups are residually finite from bogopolski book, but proof for residually nilpotent. Which nilpotent group should I map my free group here, so that definition satisfies.

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  • $\begingroup$ It follows from the fact that the intersection of the terms in the lower central series of $G$ is trivial. $\endgroup$ – Derek Holt Apr 26 '15 at 12:34
  • $\begingroup$ Yes that is an equivalent condition, but then how to prove that for some free group. I thought it will be easier to go by definition. $\endgroup$ – Bhaskar Vashishth Apr 26 '15 at 12:49
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    $\begingroup$ I think you look it up in a textbook! For example it is 6.1.10 in "A Course in the Theory of Finite Groups" by Derek Robinson . I found that online. Theorem 6.1.9 by Iwasawa shows how to construct a homomorphism from a free group to a $p$-group (for any given prime $p$) that maps any given element nontrivially. $\endgroup$ – Derek Holt Apr 26 '15 at 13:26
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There is a classical method using Magnus homomorphism. There is an injective map from the free group with gen. from X into the ring of power series with non-commuting variables from X, sending every gen. x to 1+x. It is easy to show that an element of the n-th term of the LCS is sent by this hom. to a power series that has a free coefficient 1 and the following non-zero coefficient is only of degree greater or equal to n. It finishes the argument

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  • $\begingroup$ Welcome to Math.SE! You can use MathJax to format your posts. $\endgroup$ – fonini May 17 '16 at 5:09

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