Topics in Algebra - N. Herstein Exercise from Section 2.12, Question 16 (Page 103) Please help me with this Herstein exercise (Page 103,Sec 2.12, Ques 16).
\begin{array} { l } { \text { If } G \text { is a finite group and its } p \text { -Sylow subgroup } P \text { lies in the center of } } \\ { G , \text { prove that there exists a normal subgroup } N \text { of } G \text { with } P \cap N = (e)} \\ {  \text {and } P N = G . }  \end{array}
I got to know about more general theorems like Schur-Zassenhaus Theorem or Burnside's normal p-complement theorem from which this can be deduced as corollary. But, I want a solution which just uses theory built in Herstein's book. 
The question just before this is 
\begin{array} { l } { \text { Let } G \text { be a finite group in which } ( a b ) ^ { p } = a ^ { p } b ^ { p } \text { for every } a , b \in G , } \\ { \text { where } p \text { is a prime dividing } o ( G ) \text { . Prove } } \\ { \text { (a) The } p \text { -Sylow subgroup of } G \text { is normal in } G \text { . } } \\ { \text { (b) If } P \text { is the } p \text { -Sylow subgroup of } G , \text { then there exists a normal } } \\ { \text { subgroup } N \text { of } G \text { with } P \cap N = ( e ) \text { and } P N = G \text { . } } \\ { \text { (c) } G \text { has a nontrivial center. } } \end{array}
I have solved it by first proving, for $p^n|o(G)$ and $p^{n+1} \not| o(G)$, $$P=\{x\in G : x^{p^n}=e\}$$ is unique $p-Sylow$ sugroup of G and then taking a homomorphism $\phi:G\to G$ defined by $\phi(g)=g^{p^n}$, where $p^n$ is order of $p-Sylow$ subgroup of G. Then $\phi(G)= N$ 
 A: Let $G' = [G,G]$ be the derived subgroup of $G$. Let $\pi$ be the natural projection of $G$  onto the quotient $Q = G/G'$ (also called the abelization of $G$). The Abelian group $Q$  splits as a direct product $Q = \pi(P) \times M$. The group you are looking for is $N = \pi^{-1}(M)$. The fact that $P \cap N = \{e\}$ derives from the Focal subgroup theorem stating that $P \cap G' = P_0 = \{x^{-1}y \mid x \in P,\exists g\in G, y = g^{-1}xg\}$, which in this case is the trivial group.
A: Since your question was posted before I answered it there. The answer makes no use of other theorems than elementary mathematics. An other approach could be to embed the group into the wreath product of $P$ with the permutation group $G/P$, but I don't know if Herstein's book covers this.
A: Let $q = o(G) / p^k$, where $p^k$ divides $o(G)$ but $p^{k + 1}$ does not divide $o(G)$.
Define $\phi:G \to G$ by $\phi(a) = a^q$.  Then $o(\phi(a))$ divides $p^k$.  As the $p$-Sylow subgroup is normal (it is in the center of $G$), it includes all elements whose order divides $p^k$.
Thus $\phi(a)$ is in the $p$-Sylow subgroup.  The $p$-Sylow subgroup is Abelian, as it is in the center of $G$.  Thus $a^q \cdot b^q = (ab)^q$.  Thus $\phi$ is a homomorphism.  The desired set $N$ is the kernel of $\phi$.
A: The basic idea is that you know that because $P$ lies in the center it is the only $p$-Sylow subgroup. That fact is enough to tell you that every element can be decomposed into a "$P$ part" and a "not $P$ part", and each "part" is a subgroup. Here is how that decomposition works:
$G$ is a finite group, therefore it is finitely generated, and so we can write down the generators of $G$, let them be $\mathcal{A} = \{a_1, \dots, a_n\}$. Now, let $ \mathcal{P} = \{p_1, \dots, p_m\} \subset \mathcal{A}$ such that $<\mathcal{P}> = P$. Now let $\mathcal{H}$ be the set of all words in $G$ that do not contain any letters (generators) from $\mathcal{P}$, and let $H = <\mathcal{H}>$. Because $P$ commutes with every element in $G$, it is clear that every word $g \in G$ can be decomposed into a pair $(h,q) \in H\times P$ s.t. $g = h \cdot p$. What's left to prove is that $H$ is a subgroup. 
Let $a,b \in H$ be arbitrary non-identity elements (the identity case is easy). We must show that $a \cdot b^{-1} \in H$. By definition, we must simply show that $a\cdot b^{-1} \in <\mathcal{H}>$. We proceed by contradiction: assume that  $a\cdot b^{-1} \in <\mathcal{A} \setminus \mathcal{H}>$. By Lagrange's theorem we know that $p|gcd(o(a),o(b^{-1}))$. Without loss of generality we take $p \mid o(a)$. Since $a \in H$ that means that $<a> \not\subset P$, which means that the index of $P$ is divisible by $p$, which is a contradiction to the maxamality of a $p$-Sylow subgroup. This shows that it must be the case that $a\cdot b^{-1} \in G\setminus P$, as desired. This is a normal subgroup because every element is decomposed into being either in H or in the center of $G$, which is enough for the orbit of conjugation to be $H$. 
