Timeline for Is the free product of arbitrarily many copies of `${\mathbb{Z}}$` and `${\mathbb{Z}}/2$` residually nilpotent?
Current License: CC BY-SA 3.0
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| Feb 3, 2012 at 5:27 | comment | added | user6976 | @Gao: Because the free group embeds into the congruence subgroup of $SL(2,\mathbb{Z})$ corresponding to any prime $p$. | |
| Feb 3, 2012 at 4:53 | comment | added | Gao 2Man | Why is a free group residually (finite p-)group? Is there a suitable reference for this fact? | |
| Oct 11, 2011 at 3:09 | vote | accept | CommunityBot | ||
| Oct 11, 2011 at 3:07 | comment | added | user2529 | Thank you Mark so much for your detailed answers and proofs. | |
| Oct 10, 2011 at 23:14 | history | edited | user6976 | CC BY-SA 3.0 |
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| Oct 10, 2011 at 12:57 | history | edited | user6976 | CC BY-SA 3.0 |
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| Oct 9, 2011 at 23:25 | history | edited | user6976 | CC BY-SA 3.0 |
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| Oct 9, 2011 at 10:09 | history | edited | user6976 | CC BY-SA 3.0 |
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| Oct 9, 2011 at 9:53 | history | answered | user6976 | CC BY-SA 3.0 |