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Math Help - existence of a subgroup

  1. #1
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    existence of a subgroup

    Let G be a finite group in which (ab)^{p}=a^{p}b^{p} for some prime p dividing the order of G. Prove that if P is the p-sylow subgroup of G, then there exists a normal subgroup Nof G with P \cap N=(e) and PN=G.


    All i can see is that the mapping x \mapsto x^{p} is a homomorphism and the set S=\{ x^{p} \ | \ x \in G\} and K=\{ x^{p-1} \ | \ x \in G\} are normal and have trivial intersection.
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    Let G be a finite group in which (ab)^{p}=a^{p}b^{p} for some prime p dividing the order of G. Prove that if P is the p-sylow subgroup of G, then there exists a normal subgroup Nof G with P \cap N=(e) and PN=G.


    All i can see is that the mapping x \mapsto x^{p} is a homomorphism and the set S=\{ x^{p} \ | \ x \in G\} and K=\{ x^{p-1} \ | \ x \in G\} are normal and have trivial intersection.
    let |P|=p^m. define f: G \longrightarrow G by f(g)=g^{p^m}. since (ab)^{p^m}=a^{p^m}b^{p^m}, for all a,b \in G, the map f is a group homomorphism. the kernel of f is a p-subgroup of G which contains P and thus

    \ker f=P. now let N=\text{Im}(f). see that N is a normal subgroup of G and P \cap N=\{1 \}. hence, since G/P \cong N, we have |PN|=|P||N|=|G|, and therefore PN=G.
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