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Does every normal subgroup has complement in free groups? What about free abelian groups i.e. Is free abelian gorup complemented group?

Definition: If there exist a subgroup K such that HK = G and H ∩ K = {e}, then K is called complement of H. Here e is identity of G

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    $\begingroup$ What do you mean by the complement of a normal subgroup? In which sense? $\endgroup$ – Malkoun Jun 13 '16 at 8:45
  • $\begingroup$ ok. What is the complement of $2\mathbb{Z}$ in $\mathbb{Z}$? $\endgroup$ – Malkoun Jun 13 '16 at 9:22
  • $\begingroup$ @Malkoun okay I got it. complement of 2Z doesn't exist because any other subgroup intersect it. $\endgroup$ – Sushil Jun 13 '16 at 9:37
  • $\begingroup$ I am glad I could help! $\endgroup$ – Malkoun Jun 13 '16 at 9:39
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I know that if |G|=p^n where p is a prime number and G is a ableian group and assume H«G that H is cyclic with maximal order in G then H is complement in G we can prove this with sylow's theorms

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  • $\begingroup$ I don't think I see how this is relevant to the OP's question. Is there some connection? $\endgroup$ – pjs36 Jun 14 '16 at 4:55

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