# 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 $$\phi(G)= N$$

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.

• Its still somehow related to Burnside's normal p-complement theorem.. I am looking for a solution which only uses theory built in book of N. Herstein. – Lord KK Jun 16 at 19:22
• @LordKK I don't know to what extent this book can be used to complete this proof. I'm convinced that the focal subgroup theorem in this case can be proved using elementary statements, I've been trying this with no result so far. Maybe it could be the subject of a new question? – Marc Bogaerts Jun 16 at 22:06
• I can prove that each coset of $P$ has a unique element whose order is coprime to $p$ but can't prove they form a subgroup, despite the fact that all the examples I tested are affirmative – Marc Bogaerts Jun 20 at 15:26

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.

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 $$ \not\subset P$$, 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 $$H$$.

• Sorry, but this answer is wrong. In general $H$ need not even be a group. And your proof of $H$ being subgroup is also faulty. "By Lagrange's theorem we know that $o(a⋅b∣p^n$ , which implies $o(a)⋅o(b)∣p^n$ " Though it is not true in general. – Lord KK Jun 15 at 6:43
• I might be wrong, but I think in this case H is a group precisely because you can decompose elements into the $q \cdot h$ form, which is not possible in general. Also the second part relies on $p$ being prime, although it's not true in general. – Juan Sebastian Lozano Jun 15 at 6:47
• That step is still not convincing. Because there may exist $a,b \in G - P$ such that $a.b \in P$. – Lord KK Jun 15 at 7:02
• H is not necessarily a subgroup even if G is abelian group when P will obviously in the centre of group. Take $G= Z_3 \times Z_4$. – Lord KK Jun 15 at 7:10
• Ah yeah i'm wrong, my intuition here was thinking about the generators of the groups, but I translated that wrong into the proof. The proper thing to say is that if $a_1, \dots , a_n$ generate $G$ then take the generators of $P$ and then $H$ is the subgroup generated by their compliment. – Juan Sebastian Lozano Jun 15 at 7:17