Note that if either $p=1$ or $q=1$, $G$ is a $p{\hbox{-}}$group, which is a nontrivial center that is always normal. So assume $p\neq 1$ and $q\neq 1$.

We want to show that $G$ has a non-trivial normal subgroup. Noting that ${\sharp}G = p^2 q$, we will proceed by showing that either $n_p$ or $n_q$ must be 1.

We immediately note that \begin{align*} \begin{align*} n_p \equiv 1 \operatorname{mod}p &\quad& n_q \equiv 1 \operatorname{mod}q \\ n_p \divides q &\quad& n_q \divides p^2 ,\end{align*} \end{align*}

which forces \begin{align*} n_p \in \left\{{1, q}\right\}, \quad n_1 \in \left\{{1, p, p^2}\right\}. \end{align*}

If either $n_p =1$ or $n_q = 1$, we are done, so suppose $n_p \neq 1$ and $n_1 \neq 1$. This forces $n_p = q$, and we proceed by cases:

Case 1: $p = q$.

Then ${\sharp}G = p^3$ and $G$ is a $p{\hbox{-}}$group. But every $p{\hbox{-}}$group has a non-trivial center $Z(G) \leq G$, and the center is always a normal subgroup.

Case 2: $p > q$.

Here, since $n_p \divides q$, we must have $n_p < q$. But if $n_p < q < p$ and $n_p = 1 \operatorname{mod}p$, then $n_p = 1$.

Case 3: $q > p$.

Since $n_p \neq 1$ by assumption, we must have $n_p = q$. Now consider sub-cases for $n_q$:

$n_q = p$: If $n_q = p = 1 \operatorname{mod}q$ and $p < q$, this forces $p=1$.
$n_q = p^2$: We will reach a contradiction by showing that this forces \begin{align*} {\left\lvert {P \coloneqq\cup_{S_p \in \mathrm{Syl}(p, G)} S_p\setminus\left\{{e}\right\}} \right\rvert} + {\left\lvert { Q \coloneqq\cup_{S_q \in \mathrm{Syl}(q, G)} S_q\setminus\left\{{e}\right\}} \right\rvert} + {\left\lvert {\left\{{e}\right\}} \right\rvert} > {\left\lvert {G} \right\rvert}. \end{align*} We have \begin{align*} \begin{align*} {\left\lvert {P} \right\rvert} + {\left\lvert {Q} \right\rvert} + {\left\lvert {\left\{{e}\right\}} \right\rvert} &= n_p(q-1) + n_q(p^2 - 1) + 1 \\ &= p^2(q-1) + q(p^2 - 1) + 1 \\ &= p^2(q-1) + 1(p^2 - 1) + (q-1)(p^2-1) + 1 \quad\quad \text{(since $q > 1$) } \\ &= (p^2q - p^2) + (p^2 - 1) + (q-1)(p^2-1) + 1\\ &= p^2q + (q-1)(p^2-1) \\ &\geq p^2 q + (2-1)(2^2-1) \quad\quad\text{(since $p, q \geq 2$)} \\ &= p^2 q + 3 \\ &> p^2q = {\left\lvert {G} \right\rvert} ,\end{align*} \end{align*} which is a contradiction. $\hfill\blacksquare$

Problem 2

We’ll use the fact that $H {~\trianglelefteq~}N(H)$ for any subgroup $H$ (following directly from the closure axioms for a subgroup), and thus \begin{align*} P {~\trianglelefteq~}N(P) \quad \text{and}\quad N(P) {~\trianglelefteq~}N^2(P). \end{align*} Since it is then clear that $N(P) \subseteq N^2(P)$, it remains to show that $N^2(P) \subseteq N(P)$.

So if we let $x \in N^2(P)$, so $x$ normalizes $N(P)$, we need to show that $x$ normalizes $P$ as well, i.e. $xPx^{-1}= P$.

However, supposing that ${\left\lvert {G} \right\rvert} = p^k m$ where $(p, m) = 1$, we have \begin{align*} P \leq N(P) \leq G ~\implies ~p^k \divides {\left\lvert {N(P)} \right\rvert} \divides p^km ,\end{align*}

so in fact $P \in \mathrm{Syl}(p, N(P))$ since it is a maximal $p{\hbox{-}}$subgroup.

Then $P' \coloneqq xPx^{-1}\in \mathrm{Syl}(p, N(P))$ as well, since all conjugates of Sylow $p{\hbox{-}}$subgroups are also Sylow $p{\hbox{-}}$subgroups.

But since $P {~\trianglelefteq~}N(P)$, there is only one Sylow $p{\hbox{-}}$ subgroup of $N(P)$, namely $P$. This forces $P = P'$, i.e. $P = xPx^{-1}$, which says that $x \in N(P)$ as desired. $\hfill\blacksquare$

Problem 3

By definition, $G$ is simple iff it has no non-trivial subgroups, so we will show that if ${\left\lvert {G} \right\rvert} = 148$ then it must contain a normal subgroup.

Noting that $248 = p^2 q$ where $p=2, q = 37$, we find that (for example) $n_2 \divides 37$ but $n \equiv 1 \operatorname{mod}2$; but the only odd divisor of 7 is 1, forcing $n_2 = 1$. So $G$ has a normal Sylow $2{\hbox{-}}$subgroup and we are done.

Problem 4

Let $\tau \coloneqq(t_1, t_2)$ denote the transposition and $\sigma = (s_1, s_2 \cdots, s_p)$ denote the $p{\hbox{-}}$cycle, and let $S = \left\langle{\sigma, \tau}\right\rangle$. We would like to show that $S = S_p$, and since $S \subseteq S_p$ is clear, we just need to show that $S_p \subseteq S$.

We first note that because $p$ is prime, $\sigma^k$ is a $p{\hbox{-}}$cycle for every $1\leq k \leq p$, and $\left\langle{\sigma}\right\rangle = \left\langle{\sigma^k}\right\rangle$ for any such $k$.

Then note that $t_1=s_i$ for some $i$ and $t_2=s_j$ for some $j$, so we can take $k=j-i$ to get a cycle $\sigma^k$ that sends $t_1$ to $t_2$. So without loss of generality, we can replace $\sigma$ with \begin{align*} \sigma = (t_1, t_2, \cdots ) \end{align*}

But now, we can relabel all of the elements of $S_p$ simultaneously (i.e. replace $\left\langle{\sigma, \tau}\right\rangle$ with another subgroup in the same conjugacy class) in such a way that $t_1$ becomes 1 and $t_2$ becomes 2. We can then assume wlog that \begin{align*} \tau = (1,2),\quad \sigma=(1,2,\cdots,p) \end{align*}

We can then get all adjacent transpositions: noting that \begin{align*} \begin{align*} \sigma^{-1}\tau \sigma &= (2, 3) \\ \sigma^{-2} \tau \sigma^2 &= (3, 4) \\ &\cdots \\ \sigma^{-k} \tau \sigma^k &= (k+1 \operatorname{mod}p,~~k+2\operatorname{mod}p) \quad \forall 1\leq k \leq p ,\end{align*} \end{align*}

where we use the fact that for any $\gamma\in S_p$, we have $\gamma\tau\gamma = (\gamma(1),~\gamma(2))$.

But this also gives us all transpositions of the form $(1, j)$ for each $2\leq j \leq p$: \begin{align*} \begin{align*} (2, 3)^{-1}(1, 2)(2, 3) &= (1, 3) \\ (3, 4)^{-1}(1, 3) (3, 4) &= (1, 4) \\ &\cdots \\ (j-1, j)^{-1}(1, j-1) (j-1, j) &= (1,j) \quad \forall 1\leq j \leq p .\end{align*} \end{align*}

Thus we have $J \coloneqq\left\langle{\{(1, j) \mathrel{\Big|}2\leq j \leq p\}}\right\rangle \subseteq S$.

But now if $\gamma = (g_1, g_2, \cdots, g_k) \in S_p$ is an arbitrary cycle, we can write \begin{align*} \gamma = (g_1, g_2, \cdots, g_k) = (1, g_1)( 1, g_2), \cdots (1, g_k), \end{align*}

so $\gamma \in J$. Then writing any arbitrary permutation as a product of disjoint cycles, we find that $S_p \subseteq J \subseteq S$, and so $S_p \subseteq S$ as desired. $\hfill\blacksquare$

Problem 5

Since $G$ is a $p{\hbox{-}}$group, it has a nontrivial center. Since $p$ is prime and $Z(G)$ is a subgroup, this forces ${\sharp}Z(G) \in \left\{{p, p^2}\right\}$, where $p^3$ is ruled out because this would make $G$ abelian.

Supposing that ${\sharp}Z(G) = p^2$,we would have $[G: Z(G)] = p$, and since $Z(G) {~\trianglelefteq~}G$, we can take the quotient and ${\sharp}\left(G/Z(G)\right) = p$. But this means $G/Z(G)$ is cyclic, which implies that $G$ is abelian, a contradiction.

So we must have ${\sharp}Z(G) = p$, and ${\sharp}\left(G/Z(G)\right) = p^2$.

But any group of $p^2$ is abelian, and we can characterize $G' \coloneqq[G, G]$ in the following way:

$G' \leq G$ is the unique subgroup of $G$ such that if $N {~\trianglelefteq~}G$ and $G/N$ is abelian, then $N \leq G'$.

We can thus conclude that $G' \leq Z(G)$. It can not be the case that $G' = \left\{{e}\right\}$, since this would make $G$ abelian. This forces $G' = Z(G)$ as desired. $\hfill\blacksquare$

Problem 6

Writing $f(x) = x^3 - 3x - 3 = \sum a_i x_i \in {\mathbf{Q}}[x]$, we can conclude that $f$ is irreducible over ${\mathbf{Q}}$ by Eisenstein with the prime $p=3$, since $p\divides a_0 = -3, a_1 = 3, a_2 = 0$, but $p^2 \nmid a_3 = 1$.

We can check that $f(0) < 0$ and $f(10) > 0$, so $f$ has at least one real root. By the 1st derivative test, we can find that $f$ is increasing on $(-\infty, -1)$ and less than zero, decreasing on $(-1, 1)$ and less than zero, and increasing on $(1, \infty)$, where it it attains its root. This root has multiplicity one, since $\gcd(f, f') = 1$, which means that $f$ has exactly one real root $r_0$, and thus a complex conjugate pair of roots $r_1, \overline r_1$ as well.

This means that complex conjugation is a nontrivial element $\tau$ of the Galois group $G \leq S_3$, and thus $G$ contains a 2-cycle.

The Galois group must be a transitive subgroup of $S_3$, which restricts the possibilities to $S_3, A_3$.

Since $A_3$ only contains 3-cycles, this possibility is ruled out. Thus the Galois group must be $S_3$. $\hfill\blacksquare$

Problem 7

Definition: A field $F$ is perfect if every irreducible polynomial $f(x) \in F[x]$ is separable in $\overline{F}[x]$.

Note that since $F$ is a finite field, $p$ must be a prime.

$\implies:$

Suppose all irreducible polynomials in $F[x]$ are separable. Then let $a\in K$ be arbitrary, we will show that there exists some $\beta \in K$ such that $\beta^p = a$.

Given such an $a$, define the polynomial \begin{align*} f(x) = x^p - a \in F[x]. \end{align*}

Note that $f$ is not separable, since $f'(x) = px^{p-1} = 0$ since $\mathrm{char}(F) = p$, which means (by assumption) that $f$ must be reducible.

Thus we can write $f(x) = g(x)h(x)$ where $g \in F[x]$ is some irreducible factor that divides $f$.

Noting that if $\beta \in \overline{F}$ is a any root of $f$, then \begin{align*} f(\beta) = 0 \implies \beta^p = a \implies f(x) = x^p - a = x^p - \beta^p = (x-\beta)^p, \end{align*}

and so $\beta$ is necessarily a multiple root.

Moreover, since $g\divides f$, we must have $g(x) = (x-\beta)^\ell$ for some $1 \leq \ell \leq p$.

But then we can expand $g$ using the binomial formula: \begin{align*} g(x) = (x - \beta)^\ell = \sum_{k=1}^\ell {\ell \choose k}x^{\ell-k}(-\beta)^k = x^\ell + \cdots + (-\beta)^\ell \in F[x]. \end{align*}

But since every coefficient must be in $F$, we must have $\beta^\ell \in F$. We know that $\beta^p = a \in F$ as well, but since $p$ is prime, $\gcd(p, \ell) = 1$.

We can thus find $s, t \in {\mathbf{Z}}$ such that $ps + t\ell = 1$. But then

\begin{align*} \beta = \beta^1 = \beta^{ps + t\ell} = \beta^{st} \beta^{t\ell} = (\beta^\ell)^s (\beta^p)^t, \end{align*}

where since $\beta^\ell, \beta^p \in F$, the entire RHS is in $F$, and thus the LHS $\beta\in F$ as well.

But then $\alpha = \beta^p$ where $\beta \in F$, which is exactly what we wanted to show.

$\impliedby$:

Suppose every element in $F$ admits a $p$th root in $F$, and suppose $f \in F[x]$ is an irreducible polynomial which is not separable, so it has a repeated root in $\overline F$.

Supposing that $\gcd(f, f') = g(x)$ for any polynomial $g(x)$, this would imply that $g\divides f$. But $f$ was assumed irreducible, so the only possibility is that in fact $g = f$.

But if $\gcd(f, f') = f$, since $\deg f' < f$, we can not have $f \divides f'$ unless $f'$ is identically zero.

If we thus write \begin{align*} \begin{align*} f(x) &= \sum_{k=0}^n c_k x^k, \\ f'(x) &= \sum_{k=1}^n k c_k x^{k-1} \\ &\equiv 0 ,\end{align*} \end{align*}

then for each $k$ we must have $c_k = 0$ or $k = 0$ in $F$, i.e. $c_k = 0$ or $p \divides k$.

Thus the only possible nonzero terms in $f$ must come from coefficients of $x^{kp}$ for each $k$ such that $1 \leq kp \leq n$, i.e. \begin{align*} f(x) = c_0 + c_p x^p + c_{2p} x^{2p} + \cdots \end{align*}

But this says we can write $f(x) \coloneqq g(x^p)$, where \begin{align*} g(x) = c_0 + c_p x + c_{2p} x^2 + \cdots \end{align*}

and furthermore, we can now use the assumption that $F$ is perfect to write $c_i = b_i^p$ for each $i$, yielding

\begin{align*} \begin{align*} g(x) &= b_0^p + b_p^p x^2 + b_{2p}^p x^{2} + \cdots \\ .\end{align*} \end{align*} and thus \begin{align*} \begin{align*} f(x) &= g(x^p) \\ &= b_0^p + b_p^p x^{p} + b_{2p}^p x^{2p} + \cdots \\ &= (b_0 + b_p x + b_{2p} x^2)^p \\ &\coloneqq\left( j(x) \right)^p ,\end{align*} \end{align*}

from which it follows that $j \divides f$ in $F[x]$. But since $f$ was irreducible, this is a contradiction, and so $f$ could not have had a repeated root. Thus every irreducible polynomial is separable, which is what we wanted to show. $\hfill\blacksquare$

Problem 8

Let $f(x) \in F[x]$ be irreducible, then since $p(x) \coloneqq\gcd(f, f')$ must divide $f$ and $f$ is irreducible, the only possibilities are $p(x) = 1$ or $p(x) = f(x)$.

If $p(x) = 1$, then $f$ is separable, so every root is distinct and $f$ itself is of the form $f(x^{p^e})$ where each $e=0$.

Otherwise, $p(x) = f(x)$, which forces $f'(x) = 0$ in $K[x]$. If we write \begin{align*} \begin{align*} f(x) &= \sum_{k=0}^n a_k a^k \\ f'(x) &= \sum_{k=1}^n k a_k a^{k-1} \\ ,\end{align*} \end{align*} then $f'(x) \equiv 0$ forces either $a_k = 0$, or $k = 0$ in $F$ (so $p \divides k$).

We can thus rewrite $f$ by leaving out all terms where $a_k = 0$ to obtain \begin{align*} f(x) = a_p x^p + a_{2p} x^{2p} + \cdots \end{align*} and we thus define \begin{align*} g(x) \coloneqq a_p x + a_{2p}x^{2} + \cdots \end{align*}

and we recover $f(x) = g(x^p)$. Moreover, $g$ is irreducible; otherwise if $h(x) \divides g(x)$ then $h(x^p) \divides g(x^p) = f$, where $f$ was assumed irreducible. If $g$ is separable we are done; otherwise $g$ fulfills the same hypotheses of that applied to $f$, so we can inductively continue this process to write $g(x) = g_1(x^p)$, and thus $f(x) = g(x^p) = g_1(x^{p^2})$, and so on.

To see that every root of $f$ has multiplicity $p^e$, note that if $f(\alpha) = 0$ then $g(\alpha^{p^e}) = 0$. But $g$ is separable, so $(x - \alpha^{p^e}) \divides g(x)$ in $K[x]$ and thus $(x^{p^e} - \alpha^{p^e}) \divides g(x^{p^e}) = f$ in $\overline{K}[x]$ where $\overline K$ is an algebraic closure of $K$. But then $x^{p^e} - \alpha^{p^e} = (x-\alpha)^{p^e} \divides f(x)$, which precisely says that $\alpha$ is a root of multiplicity $p^e$.

Problem 9

Let $x = [{\mathbf{Q}}(\zeta + \zeta^{-1}) : {\mathbf{Q}}]$.

Noting that \begin{align*} \zeta(\zeta + \zeta^{-1}) = \zeta^2 + 1, \end{align*}

if we let \begin{align*} f(x) = x^2 - (\zeta + \zeta^{-1})x + 1 \in {\mathbf{Q}}(\zeta + \zeta^{-1})[x], \end{align*} then $f(\zeta) = 0$.

Since ${\mathbf{Q}}(\zeta + \zeta^{-1}) \subset {\mathbf{R}}$, ${\mathbf{Q}}(\zeta)$ is a proper extension over this field, so if $d \coloneqq[{\mathbf{Q}}(\zeta) : {\mathbf{Q}}(\zeta + \zeta^{-1})]$ then $d > 1$. The fact that $\zeta$ is a root of $f$ shows that $d \leq 2$, so $d = 2$. We also know that $[{\mathbf{Q}}(\zeta) : {\mathbf{Q}}] = \phi(n)$.

We thus have \begin{align*} \begin{align*} [{\mathbf{Q}}(\zeta) : {\mathbf{Q}}] &= [{\mathbf{Q}}(\zeta) : {\mathbf{Q}}(\zeta + \zeta^{-1})] [{\mathbf{Q}}(\zeta + \zeta^{-1}) : {\mathbf{Q}}] \quad\implies\quad \phi(n) = 2 x ,\end{align*} \end{align*}

and so $x = \frac{\phi(n)}{2}$ as desired. $\hfill\blacksquare$

Problem 10

Suppose $K/F$ is a finite, normal, Galois extension.

Part 1

We have $F \leq E \leq K$. Suppose that

$K / F$ is cyclic, so ${ \mathsf{Gal}}(K / F)$ is a cyclic group,
$E / F$ is normal

We then want to show that

$E/F$ is cyclic, i.e. ${ \mathsf{Gal}}(E/F)$ is cyclic, and
$K/E$ is cyclic, i.e. ${ \mathsf{Gal}}(K/E)$ is cyclic.

By the fundamental theorem of Galois theory, $E/F$ is normal if and only if

${ \mathsf{Gal}}(K/E) {~\trianglelefteq~}{ \mathsf{Gal}}(K/F)$, and
${ \mathsf{Gal}}(E/F) \cong { \mathsf{Gal}}(K/F) / { \mathsf{Gal}}(K/E)$.

Since ${ \mathsf{Gal}}(K/F)$ is a cyclic group and every subgroup of a cyclic group is itself cyclic, (a) lets us conclude that (1) holds.

Similarly, since ${ \mathsf{Gal}}(K/F)$ is a cyclic group and every quotient of a cyclic group is cyclic, (b) lets us conclude (2).

Part 2

By the Galois correspondence, all intermediate fields will correspond to subgroups of ${ \mathsf{Gal}}(K/F)$. Since this group is cyclic, we are reduced to analyzing the subgroup lattice of a generic cyclic group.

But if $G = \left\langle{x \mathrel{\Big|}x^n = e}\right\rangle$ where ${\sharp}G = n$, then there is one and only one subgroup of index $d$ and order $\frac{n}{d}$ for every $d$ dividing $n$, given by $H_d \coloneqq\left\langle{x^d}\right\rangle$.

So we have $[G: H_d] = d$, so $H_d$ corresponds to a field $E_d/ F$ of degree $d$ where $F \leq E_d \leq K$. This can be done for every $d$ dividing $n$, and since $K/F$ is a Galois extension, $n = {\left\lvert {{ \mathsf{Gal}}(K/F)} \right\rvert} = [K: F]$, and this can be done for every divisor of $[K: F]$ as desired. $\hfill\blacksquare$

Midterm

Case 1: \(p = q\).

Case 2: \(p > q\).

Case 3: \(q > p\).