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\title{Math 210A\\Homework 6}
\author{Brett Hemenway}

\begin{document}
	\maketitle
	\begin{itemize}
		\item [1.]
			%This is the "Frattini Argument".  See Rotman Ex 5.21.
			Let $G$ be finite group with $K \triangleleft G$.  Let $P$ be a Sylow $p$-subgroup of $K$.  For any $g \in G$ we have that $gPg^{-1} \subset K$ because $K$ is normal in $G$.  This means that $gPg^{-1}$ is a
			Sylow $p$-subgroup of $K$.  Since all Sylow $p$-subgroups are conjugate, we have that $kPk^{-1} = gPg^{-1}$ for some $k \in K$, so $P = (k^{-1}g)P(k^{-1}g)^{-1}$.  In other words, for any $g \in G$ there exists a
			$k \in K$ such that $k^{-1}g \in N_G(P)$.  So $g = k(k^{-1}g) \in KN_G(P)$.
		\item [2.]
			%See Rotman groups book theorem 5.39
			%Or Rotman prop 5.39 (coincidence?)
			Before we begin, recall that if $H,K$ are subgroups of $G$ and $\gcd(|H|,|K|) = 1$ then $H \cap K = \{e\}$, because the order of any element in $H \cap K$ must divide the orders of $H$ and $K$.\\
			Let $G$ be a finite group, with $|G| = p_1^{e_1}\ldots p_n^{e_n}$.  Suppose, no proper subgroup of $G$ is its own normaliser.  We have already seen that if $P$ is a Sylow subgroup of $G$, then $N(N(P)) = N(P)$, 
			so this means that $N(P) = G$, i.e. $P$ is normal in $G$.  Thus every Sylow subgroup of $G$ is normal.  Let $P_i$ denote the Sylow $p_i$-subgroup, there is only one because $P_i$ is normal.  
			Then $P_1P_2$ is a subgroup of $G$ and
			\[
				|P_1P_2| = \frac{|P_1||P_2|}{|P_1 \cap P_2|} = |P_1||P_2|
			\]
			because $P_1 \cap P_2 = \{e\}$, because the $\gcd( |P_1|, |P_2| ) = 1$.  Then, inductively, we have that $P_1P_2\cdots P_n$ is a subgroup with 
			\[
				|P_1\cdots P_n| = |P_1|\cdots|P_n|
			\]
			because $\gcd(|P_1\cdots P_{n-1}|,|P_n|) = 1$ we have $P_1 \cdots P_n$ is a subgroup of $G$ with $|P_1\cdots P_n| = p_1^{e_1}\cdots p_n^{e_n} = |G|$.  So we have that
			\[
				P_1 \cdots P_n = G
			\]
			Conversely, assume $G$ is the direct product of its Sylow groups, so $G = P_1 \times \ldots \times P_n$.  If $H$ is a subgroup of $G$, then there are subgroups $Q_1,\ldots,Q_n$ 
			with $Q_i \subset P_i$, and $H = Q_1 \times \ldots \times Q_n$.  If $H \subsetneq G$, then there exists an $i$, such that $Q_i \subsetneq P_i$.  Since $P_i$ is a $p$-group, we know from homework 3, problem 6, 
			that $Q_i \subsetneq N_{P_i}(Q_i)$.  Then 
			\[
				H \triangleleft Q_1 \times \ldots \times N_{P_i}(Q_i) \times \ldots \times Q_n
			\]
			So $H \subsetneq N_G(H)$.
		\item [3.]
			Let $G$ be a finite non-abelian group with $|G| \le 100$.
			\begin{itemize}
				\item 
					If $|G| = p$ then $G$ is cyclic and hence abelian.\\
					This eliminates
					\begin{align*}
						\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,\\
						43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}
					\end{align*}
				\item
					If $|G| = p^2$ then $G$ is abelian by Cauchy's theorem.\\
					This eliminates 
					\[
						\{4, 9, 25, 49\}
					\]
				\item
					%Rotman prop 5.56
					If $|G| = p^n$ then $|Z(G)| > p$, so $Z(G)$ is normal subgroup of $G$, or $G$ is abelian.\\
					This eliminates 
					\[
						\{ 8, 16, 27, 32, 64, 81 \}
					\]
				\item
					If $|G| = pq$ then with $P,Q$ the corresponding Sylow groups.  Assume $p < q$ then $[G:P] = q$ so $P$ is normal in $G$.\\
					This eliminates 
					\begin{align*}
						\{ 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55,\\
						57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95 \}
					\end{align*}
				\item
					%Herstein ex 2.12.12
					If $|G| = pqr$ with $p < q < r$ then $R \triangleleft G$.\\
					This eliminates 
					\[
						\{30,66,70,78\}
					\]
				\item
					%Rotman lemma 5.40
					If $|G| = p^em$ with $p \nmid m$, and $p^e \nmid (m-1)!$ then $G$ is not simple.  To see this
					consider the action of $G$ on the cosets $G/P$.  This gives a map from $\phi : G \rightarrow S_m$.
					If $G$ were simple, then $\ker(\phi) = \{e\}$, so $G \simeq \phi(G) \subset S_m$.  So we have
					$p^em | m!$, which means that $p^e | (m-1)!$.\\
					This eliminates 
					\[
						\{36,72,100\}
					\]
				\item
					If $|G| = p^aq$ then $G$ is not simple by problem 7.\\
					This eliminates 
					\[
						\{ 12, 18, 20, 24, 28, 40, 44, 45, 48, 50, 52, 56, 63, 75, 80, 88, 96, 98, 99 \}
					\]
				\item
					If $|G| = 84 =2^2 \cdot 3 \cdot 7$.  Then by the third Sylow theorem, there are $n_7 \equiv 1 \mod 7$ Sylow $7$-subgroups.  
					But $n_7 | 2^2\cdot 3$, so $n_7 = 1$.  Thus the Sylow $7$-subgroup is normal, and $G$ is not simple.
				\item
					If $|G| = 90 = 2\cdot 3^2 \cdot 5$.  Then by the third Sylow theorem, there are $n_3 \equiv 1 \mod 3$ Sylow $3$-subgroups.
					But $n_3 | 2\cdot 5$, so $n_3 = 1$.  Thus the Sylow $3$-subgroup is normal, and $G$ is not simple.
				\item
					%See Rotman prop 5.67
					This leaves only $|G| = 60 = 2^2 \cdot 3 \cdot 5$.  So we must show that if $G$ is simple $G \simeq A_5$.  So we suppose $G$ is simple with $|G| = 60$.
					By the third Sylow theorem there are $n_2 \equiv 1 \mod 2$ Sylow $2$-subgroups of $G$.  Since $n_2 | 60$, we have $n_2 \in \{3,5,15\}$.  If $n_2 = 3$, the
					action of $G$ on $G/P_2$ gives us a nontrivial map from $G$ to $S_3$.  The kernel of this map would be a proper normal subgroup of $G$, so we conclude $n_2 \ne 3$.\\
					If $n_2 = 5$, then the action of $G$ on the cosets of the $G/P_2$ gives us a map from $G$ to $S_5$, as $G$ is simple, this map is injective, so $G \subset S_5$.  Since $|G| = 60$, we conclude
					$G = A_5$.  Suppose $n_2 = 15$, then let $P,Q$ be distinct Sylow $2$-subgroups of $G$.  If $P \cap Q \ne \{e\}$, then let $e \ne x \in P \cap Q$.  We have $P \cup Q \subset C_G(x)$, since 
					$P$ and $Q$ are abelian.  We also have that $4 | C_G(x)$ since $P$ is a subgroup of $C_G(x)$, and $|C_G(x)| | 60$.  This gives us that $|C_G(x)| \in \{12,20,60\}$.  If $|C_G(x)| = 60$, then 
					$x \in Z(G)$, in which case $Z(G)$ is normal in $G$.  If $|C_G(X)| = 20$, then the action of $G$ on the cosets of $C_G(x)$ gives us a nontrivial map from $G$ to $S_3$, and the kernel of this map 
					would be a a normal subgroup of $G$.  If $|C_G(x)| = 12$, then the action of $G$ on the cosets $G/C_G(x)$ gives us a map from $G$ to $S_5$, which, as we saw above, gives $G = A_5$.
					Now we consider the case when $P \cap Q = \{e\}$, so all Sylow $2$-subgroups have trivial intersection.  By problem 6, since $n_2 = 15 \equiv 3 \mod 4$, we can find $P,Q$ with $[P:P\cap Q] < 4$, which 
					means $P \cap Q \ne \{e\}$, so we have a contradiction.  Thus $G \simeq A_5$.
			\end{itemize}
		\item [4.]
			Let $G$ be a simple group with $100 < |G| \le 200$, and $|G| \ne 168$.  Using the arguments from the previous problem we are left to consider the cases
			$|G| \in \{108,120,126,132,144,150,180,200\}$.  But $p^e \nmid (m-1)!$ for $\{108,150,156,200\}$, so we are left with
			\[
				|G| \in \{120,126,132,144,180\}
			\]
			At this point we begin looking at cases individually.  Notice that if $H$ is a subgroup of $G$ with index $n$, if $n! < |G|$, then $G$ is not simple because the action of
			$G$ on the cosets $G/H$ induces a map from $G$ to $S_n$ which must have nontrivial kernel.  We will use this repeatedly in the following arguments.  We will use $n_p$ to 
			denote the number of Sylow $p$-subgroups of $G$.
			\begin{itemize}
				\item 
					If $G = 120 = 2^3 \cdot 3 \cdot 5$, by the third Sylow theorem, $n_5 \in \{1,6\}$.  If $n_5 = 6$, then the action of $G$ on the cosets of a Sylow 5-subgroup gives a map from $G$ to $S_6$.  If
					this map is not injective, then its kernel is normal in $G$, so we may assume $G \subset S_6$.  Then $G \cap A_6$ is normal in $G$, so we conclude that $G \cap A_6 = G$, because $G \cap A_6$ 
					cannot be trivial since $H \cap A_6$ is nontrivial for every subgroup $H$ of $S_6$ of order greater than $2$.  So $G \subset A_6$.  The action of $G$ on the cosets $A_6/G$ gives a map from 
					$G$ to $S_3$ because $|A_6| = 360$.  This map is nontrivial, so its kernel is a normal subgroup of $G$, and consequently $G$ is not simple.
				\item
					If $G = 126 = 2 \cdot 3^2 \cdot 7$, then by the third Sylow theorem there are $n_7 \equiv 1 \mod 7$ Sylow $7$-subgroups of $G$.  We also have that 
					$n_7 | 2\cdot 3^2$, so $n_7 = 1$, and the Sylow $7$-subgroup is normal by the second Sylow theorem.
				\item
					If $G = 132 = 2^2 \cdot 3 \cdot 11$.  By the third Sylow theorem $n_3 \in \{1,4,22\}$, $n_{11} \in \{1,12\}$.  If $n_3 = 4$, then $G$ is simple because $|G| > 4!$.  So we conclude that $n_3 = 22$.  
					The intersection of two Sylow 3-subgroups must be the identity because groups of order 3 have no proper subgroups.  Similarly for the Sylow 11-subgroups.  Now let us count the number of elements.  
					There are $22 \cdot (3-1) = 44$ elements of order $3$, similarly there are $12(11-1) = 120$ elements of order $11$.  But $44 + 120 > 132$, so we conclude that we cannot have $n_3 = 22$ and $n_{11} = 12$, 
					so one of them must be smaller, hence $G$ is not simple.
				\item
					If $G = 144 = 2^4 \cdot 3^2$.  By the third Sylow theorem $n_2 \in \{1,3,9\}$ and $n_3 \in \{1,4,16\}$.  If $n_2 \in \{1,3\}$ or $n_3 \in \{1,4\}$ we conclude that $G$ is not simple, so suppose 
					$n_2 = 9$ and $n_3 = 16$.  $n_3 \not \equiv 1 \mod 3^2$, so by problem 6, there are Sylow 3-groups $P_{3,1},P_{3,2}$ with $|P_{3,1} \cap P_{3,2}| = 3$.  Let $P = P_{3,1} \cap P_{3,2}$.  
					Then $P \triangleleft P_{3,1}$ and $P \triangleleft P_{3,2}$ by Cauchy's theorem.  So $P_{3,1}$ and $P_{3,2}$ are contained in $N_G(P)$.  Thus $3^2 + 3^2 - 3 = 15 \le |N_G(P)|$.  But $9$ divides 
					$|N_G(P)|$, so $|N_G(P)| \ge 18$.  We know that $|N_G(P)| = 3^2k$, but $N_G(P)$ contains at least two Sylow 3-subgroups, so by the third Sylow theorem, $k \ge 3+1$, so $|N_G(P)| \ge 36$.
					This means that $[G:N_G(P)] \le 4$, but $144 > 4!$, so we conclude that $G$ is not simple.
				\item 
					If $G = 180 = 2^2 \cdot 3^2 \cdot 5$.  By the third Sylow theorem $n_3 \in \{1,4,10\}$.  If $n_3 = 4$, then notice $180 > 4!$, so $G$ is not simple.  So $n_3 = 10$.  By problem 6, there are two Sylow 
					3-subgroups of $G$ $P_{3,1},P_{3,2}$, with $P = P_{3,1} \cap P_{3,2}$ and $|P| = 3$.  Now consider $N_G(P)$.  Since $P_{3,1}$ and $P_{3,2}$ are abelian by Cauchy's theorem, we have 
					$|P_{3,1} \cup P_{3,2}| = 15 \le |N_G(P)|$.  Since $N_G(P)$ contains two subgroups of order $9$, its order must be divisible by $9$.  But it has at least two Sylow 3-subgroups, $P_{3,1},P_{3,2}$, 
					so by the third Sylow theorem $|N_G(P)| = 3^2k$ where $k \ge 3+1$, so $|N_G(P)| \ge 36$, so $[G:N_G(P)] \le 5$.  But $180 > 5!$, so we conclude that $G$ is not simple.
			\end{itemize}
		\item [5.]
			%see Rotman p275
			Let $F$ be a finite field with $q = p^m$ elements.  Let 
			\[
				STL_n(F) = \{ (a_{ij}) | a_{ij} = 0 \mbox{ if $i < j$ and $a_{ii} = 1$ } \}
			\]
			First we show that $STL_n(F)$ is a subgroup of $GL_n(F)$.  Notice that $A \in STL_n(F)$ iff $A = I + U$ where $I$ is the $n\times n$ identity matrix and $U$ is strictly upper triangular. 
			\\Then
			\[
				(I + U_1)(I + U_2) = I + (U_1 + U_2 + U_1U_2)
			\]
			but $U_1 + U_2 + U_1U_2$ is strictly upper triangular, so $STL_n(F)$ is closed under matrix products.  Note that $U^n = 0$ for any strictly upper triangular matrix $U$.  Now let
			\[
				B = I - U + U^2 - \ldots + (-1)^{n-1}U^{n-1}
			\]
			Then
			\begin{align*}
				(I + U)B	&=	(I + U)(I - U + U^2 - \ldots + (-1)^{n-1}U^{n-1})\\
							&=	I + (-1)^{n-1}U^n\\
							&= 	I\\
							&=	B(I + U)
			\end{align*}
			So $B = (I+U)^{-1}$, $B \in STL_n(F)$ because $-U+U^2-\ldots+(-1)^{n-1}U^{n-1}$ is strictly upper triangular.  So we have that $STL_n(F)$ is a subgroup of $GL_n(F)$.\\
			By HW 1.3 
			\[
				|STL_n(F)| = q^{\frac{n^2-n}{2}}
			\]
			and
			\begin{align*}
				|GL_n(F)| 	&= (q^n-1)(q^n-q)\ldots(q^n-q^{n-1})\\
							&= q^{\sum_{i=1}^{n-1} i}(q^n-1)(q^{n-1}-1)\ldots(q-1)\\
							&= q^{\frac{n^2-n}{2}}(q^n-1)\ldots(q-1)
			\end{align*}

			Let $UT_n(F)$ denote the set of invertible upper triangular matrices over $F$.
			Let $(a_{ij})$ and $(b_{ij})$ in $UT_n(F)$ then if $(a_{ij})(b_{ij}) = (c_{ij})$, we have 
			\[
				c_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}
			\]
			So in particular
			\[
				c_{ii} = \sum_{k=1}^{n} a_{ik}b_{ki}
			\]
			But $a_{ik} = 0$ for $i < k$ and $b_{ki} = 0$ for $k < i$, becaus $(a_{ij})$ and $(b_{ij})$ are upper triangular, so
			\[
				c_{ii} = a_{ii}b_{ii}
			\]
			This means that that map 
			\[
				\left( 
					\begin{array}{ccc} a_{11} & \ldots & a_{1n}\\ \vdots &\ddots &\vdots\\ 0 &\ldots &a_{nn} \end{array} 
				\right) 
				\mapsto 
				\left( 
					\begin{array}{ccc} a_{11} & \ldots & 0\\ \vdots &\ddots &\vdots\\ 0 &\ldots &a_{nn} \end{array} 
				\right) 
			\]
			which sets all above diagonal terms to zero, is a homomorphism from $UT_n(F)$ to the set of diagonal matrices.  The kernel of this map is clearly $STL_n(F)$, so $STL_n(F)$ is a normal subgroup of $UT_n(F)$.
			In particular $UT_n(F) \subset N_{GL_n(F)}(STL_n(F))$.\\
			I do not know how to do the reverse inclusion.  Is there a way to do it without resorting to multiplying matrices?
		\item [6.]
			Let $G$ be a finite group and $n_p = |\text{Syl}_pG| > 1$.  Suppose $n_p \not \equiv 1 \mod p^\alpha$ for some $\alpha > 1$.  Then let $P$ be a Sylow $p$-subgroup, and consider 
			the action of $P$ on $\text{Syl}_pG$ by conjugation.  Let $\mathcal{O}_1,\ldots,\mathcal{O}_s$, be the orbit decomposition.  Let $\mathcal{O}_1$ be the orbit containing $P$.  Then
			$|\mathcal{O}_1| = 1$.  Suppose $[P : P \cap Q] \ge p^\alpha$ for $i > 1$.  Then $|\mathcal{O}_i| \ge p^\alpha$ for $i > 1$ so 
			\[
				n_p	= \sum |\mathcal{O}_i| = 1 \mod p^\alpha
			\]
			But this contradicts our assumption, so we conclude that $|\mathcal{O}_i| < p^\alpha$ for some $i$.
		\item [7.]
			Let $G$ be a group with $|G| = p^aq$, with $p,q$ prime.\\
			We wish to show that $G$ is solvable.  We proceed by induction on $a$.  Recall that $G$ is solvable if $H$, both $G/H$ are solvable for a subgroup $H$.  Recall also that every $p$-group is solvable.\\
			We know that a group of order $pq$ is solvable\\
			Assume every group of order $p^bq$ is solvable for $b < a$\\
			Let $|G| = p^aq$.
			If $G$ has only one Sylow $p$-subgroup then it is normal, so we are done by our induction hypothesis.  Suppose $G$ has more than one Sylow $p$-subgroup.  Let $D$ be the intersection of two Sylow $p$-subgroups 
			such that if $D \subset P_i \cap P_j$ then $D = P_i \cap P_j$.  Now consider the map
			\begin{align*}
				\phi: P \mapsto	P \cap N_G(D)
			\end{align*}
			If $D \subset P$, and $D \subset Q$, then by the maximality of $D$, we have that $P \cap Q = D$.  If, additionally, we have $P \cap N_G(D) = Q \cap N_G(D)$.  If $P \ne Q$, then by the maximality of $D$
			we have that $N_G(D) \subset D$, so $D = N_G(D)$.  But $D$ is a $p$-group, $D \subsetneq P$, so by HW 3, problem 6, $D \subsetneq N_G(D)$.  So we conclude that $P = Q$.  So $\phi$ is injective on the set of 
			Sylow $p$-groups containing $D$.  Since any $p$-group is contained in Sylow $p$-group, there is a Sylow $p$-group of $N_G(D)$ containing $D$, then for a Sylow $p$-subgroup of $N_G(D)$, 
			there is a Sylow $p$-subgroup of $G$ containing it.  This means that there exists a Sylow $p$-group of $G$ with $D \subset P$, and $P \cap N_G(D)$ a Sylow $p$-subgroup of $G$.  Since all Sylow $p$-groups of 
			$N_G(D)$ are conjugate in $N_G(D)$, for any Sylow $p$-group $P^\prime$ of $N_G(D)$, there exists an $n \in N_G(D)$ with $P^\prime = nPn^{-1} \cap N_G(D)$, and $D \subset nPn^{-1}$ because $n \in N_G(D)$, 
			and $D \subset P$.  So we have set up a one-to-one correspondence between Sylow $p$-subgroups of $G$ containing $D$, and Sylow $p$-subgroups of $N_G(D)$.  So far we have not used that 
			$|G| = p^aq$, only that $p | |G|$.  If $N_G(D) = G$, then $D$ is normal, and $G$ is not simple.  If $|N_G(D)| = p^b$, then $N_G(D)$ has only one Sylow $p$-subgroup, so by our correspondence, 
			there is only one Sylow $p$-subgroup of $G$ containing $D$, thus there must only be one Sylow $p$-subgroup of $G$, call it $P$, so $P \triangleleft G$.  $P$ is solvable because it is a $p$ group, 
			and $G/P$ is solvable by our induction hypothesis, so $G$ is solvable.  This leaves only the case $|N_G(D)| = p^bq$ where $b < a$, in this case the Sylow $p$-group of $N_G(D)$ is unique by our 
			induction assumption, so there is only one Sylow $p$-subgroup of $G$ containing $D$, and so the Sylow $p$-subgroup, $P$, is normal in $G$.  So we have $P$ is a $p$-group so it is solvable and 
			$G/P$ is solvable by the induction hypothesis, so $G$ is solvable.
	\end{itemize}
\end{document}