## Tag Archives: subgroup index

### The order of conductor f in the ring of quadratic integers is a subring and has index f

Let $D$ be a squarefree integer, and let $O$ be the ring of integers in the quadratic field $\mathbb{Q}(\sqrt{D})$. (I.e. $O = \mathbb{Z}[\omega]$.) For any positive integer $f$ prove that the set $O_f = \mathbb{Z}[f\omega] = \{ a + bf\omega \ |\ a,b \in \mathbb{Z} \}$ is a subring of $O$ containing the identity. Prove that $[O : O_f] = f$. (The index as an abelian group.) Prove conversely that a subring of $O$ containing the identity and having (as a subgroup) finite index $f$ is equal to $O_f$. (The ring $O_f$ is called the order of conductor $f$ in the field $\mathbb{Q}(\sqrt{D})$. The ring of integers $O$ is called the maximal order in $\mathbb{Q}(\sqrt{D})$.)

First we show that $O$ is a subgroup of $\mathbb{Z}[\omega]$. To see this, note that $0 + 0f\omega \in O$. Now let $x = a_1 + a_2 f \omega$ and $y = b_1 + b_2 f \omega$ be in $O$. Then $x-y = (a_1 - b_1) + (a_2 - b_2)f \omega \in O$, so that by the subgroup criterion, $O$ is an additive subgroup of $\mathbb{Z}[\omega]$.

Now it is easy to see that $\omega^2 = D$ or $\omega + (D-1)/4$, depending on whether $D$ is not or is 1 mod 4. Moreover, if $D$ is 1 mod 4, then $(D-1)/4$ is an integer. Thus we have $xy = (a_1b_1 + a_2b_2f^2D) + (a_1b_2 + a_2b_1)f \omega$ or $(a_1b_1 + a_2b_2f^2(D-1)/4) + (a_1b_2 + a_2b_1 + a_2b_2)f \omega$, depending on whether $D$ is not or is 1 mod 4. In either case, $xy \in \mathbb{Z}[f \omega]$. Thus $\mathbb{Z}[f \omega] \subseteq \mathbb{Z}[\omega]$ is a subring.

Now we will show that $[\mathbb{Z}[\omega] : \mathbb{Z}[f \omega]] = f$. Define a mapping $\Phi : \mathbb{Z}[\omega] \rightarrow \mathbb{Z}$ by $\Phi(a + b\omega) = b$. We claim that $\Phi$ is an additive group homomorphism. To see this, note that $\Phi((a_1 + a_2 \omega)(b_1 + b_2\omega)) = \Phi((a_1 + b_1) + (a_2 + b_2)\omega)$ $= a_2 + b_2$ $= \Phi(a_1 + a_2 \omega) + \Phi(b_1 + b_2 \omega)$. Moreover, $\Phi$ is surjective since for all $n \in \mathbb{Z}$, $\Phi(0 + n \omega) = n$. Consider now $\pi \circ \Phi : \mathbb{Z}[\omega] \rightarrow \mathbb{Z}/(f)$, where $\pi$ denotes the natural projection. We claim that $\mathsf{ker}(\pi \circ \Phi) = \mathbb{Z}[f\omega]$, as we show. $(\subseteq)$ If $\Phi(a+b\omega) = b \equiv 0$ mod $f$, then $b = b^\prime f$ for some integer $b^\prime$. Thus $a + b\omega = a + b^\prime f \omega \in \mathbb{Z}[f\omega]$. $(\supseteq)$ Clearly $\Phi(a+bf\omega) = bf \equiv 0$ mod $f$. By the First Isomorphism Theorem for groups, then, $\mathbb{Z}[\omega]/\mathbb{Z}[f\omega] \cong \mathbb{Z}/(f)$, hence $[\mathbb{Z}[\omega] : \mathbb{Z}[f\omega]] = f$.

Suppose now that $S \subseteq \mathbb{Z}[\omega]$ is a subring containing 1 and having finite index $f$. If $f = 1$, then trivially, $\mathbb{Z}[\omega] = \mathbb{Z}[1\omega]$. If $f \geq 2$, then since $1 \in S$, $\mathbb{Z} \subseteq S$, and in particular, $(f-1)a \in S$ for all integers $a$.

Let $a,b \in \mathbb{Z}$. Since $S$ has index $f$, $fxS = S$ for all $x \in \mathbb{Z}[\omega]$. (Here, $fx$ denotes the $f$-fold sum of $x$.) Specifically, $f(a+b\omega)S = (fa + bf \omega)S$ $= (f-1)aS + (a + bf\omega)S$ $= (a+bf \omega)S$; thus $a + bf \omega \in S$ for all $a,b \in \mathbb{Z}$, and we have $\mathbb{Z}[f\omega] \leq S$.

Now $f = [\mathbb{Z}[\omega] : \mathbb{Z}[f\omega]]$ $= [\mathbb{Z}[\omega] : S] \cdot [S : \mathbb{Z}[f\omega]]$ $= f \cdot [S : \mathbb{Z}[f\omega]]$. Thus $[S : \mathbb{Z}[f\omega]] = 1$, and we have $S = \mathbb{Z}[f\omega]$.

### A sufficient condition for Sylow intersections of bounded index

Prove that if $n_p \not\equiv 1$ mod $p^k$, then there are distinct Sylow $p$-subgroups $P$ and $Q$ in $G$ such that $P \cap Q$ has index at most $p^{k-1}$ in both $P$ and $Q$.

(We will follow the strategy of this previous exercise.)

Let $P \leq G$ be a Sylow $p$-subgroup. Now $P$ acts on $\mathsf{Syl}_p(G)$ by conjugation; note that if $Q \in \mathsf{Syl}_p(G)$ is distinct from $P$, then using the Orbit-Stabilizer theorem and Lemma 4.19, we have $|P \cdot Q| = [P : N_P(Q)]$ $= [P : P \cap N_G(Q)]$ $= [P : Q \cap P]$. Since the orbit containing $P$ itself has order 1, then $n_p(G) = 1 + \sum [P : Q_i \cap P]$, where $Q_i$ ranges over a set of orbit representatives. If every $[P : Q_i \cap P]$ is divisible by $p^k$, then $n_p(G) \equiv 1$ mod $p^k$, a contradiction. Thus some $[P : Q_i \cap P]$ is a power of $p$ and is at most $p^{k-1}$.

We can see then that $Q_i \cap P$ has index at most $p^{k-1}$ in both $P$ and $Q_i$.

### No simple groups of order 2205, 4125, 5103, 6545, or 6435 exist

Prove that there are no simple groups of order 2205, 4125, 5103, 6545, or 6435.

We begin with a lemma.

Lemma 1: Let $G$ be a finite simple group. If $|G|$ does not divide $k!$, then for no subgroup $H \leq G$ do we have $[G:H] \leq k$. Proof: Suppose to the contrary that some $H$ exists, with $[G:H] \leq k$. This affords a left permutation representation $G \rightarrow S_{G/H}$, and the kernel of this representation is contained in $H$. Since the kernel is also normal in $G$, the kernel is trivial and the representation is injective; thus $G \leq S_{G/H} \leq S_k$. But by Lagrange this means that $|G|$ divides $k!$, a contradiction. $\square$

1. Note that $2205 = 3^2 \cdot 5 \cdot 7^2$. Let $G$ be a simple group of order 2205. Note that $|G|$ does not divide $13!$ since the largest power of 7 dividing $13!$ is $7^1$; by the lemma, $G$ has no subgroups of index less than or equal to 13. By Sylow’s Theorem, we have $n_7(G) = 15$. Note that $15 \not\equiv 1$ mod 49, so that by Lemma 13, there exist $P_7, Q_7 \in \mathsf{Syl}_7(G)$ such that $P_7 \cap Q_7 \neq 1$. Then $P_7 \cap Q_7 = P_0 \cong Z_7$. Moreover, $P_0$ is normal in $P_7$ and $Q_7$ since it is maximal in both; thus $P_7, Q_7 \leq N_G(P_0)$. Note that both $P_7$ and $Q_7$ are Sylow 7-subgroups of $N_G(P_0)$, so that $|N_G(P_0)| = 7^2 \cdot k$ where $k \geq 8$ (using Sylow). By Lagrange and because $G$ is simple, we have $k \in \{9,15\}$. But if $k=9$, then $N_G(P_0)$ has index 5, and if $k=15$, then $N_G(P_0)$ has index 3, both cases yielding a contradiction.
2. Note that $4125 = 3 \cdot 5^3 \cdot 11$. Let $G$ be a simple group of order 4125. By Sylow’s Theorem, we have $n_5(G) \in \{1,11\}$. Note that $|G|$ does not divide $14!$, since the largest power of 5 that divides $14!$ is $5^2$. By the lemma, no subgroup of $G$ has index less than 15. However, we have $n_5 = 11$, so that $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$; this is a contradiction.
3. Note that $5103 = 3^6 \cdot 7$. Let $G$ be a simple group of order 5103. By Sylow’s Theorem we have $n_3(G) \in \{1,7\}$. Note that $|G|$ does not divide $8!$, since the highest power of 3 which divides $8!$ is $3^2$. By the lemma, no subgroup of $G$ has index 7, a contradiction since $[G:N_G(P_3)] = 7$ for each Sylow 3-subgroup $P_3$.
4. Note that $6545 = 5 \cdot 7 \cdot 11 \cdot 17$. Let $G$ be a simple group of order 6545. By Sylow’s Theorem, $n_5(G) \in \{1,11\}$. Now $|G|$ does not divide $16!$, so that by the lemma $G$ contains no subgroups of index 11. This is a contradiction since $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$.
5. Note that $6435 = 3^2 \cdot 5 \cdot 11 \cdot 13$. Let $G$ be a simple group of order 6435. By Sylow’s Theorem, we have $n_5(G) \in \{1,11\}$. Now $|G|$ does not divide $12!$, so that by the lemma no subgroup has index 11. This is a contradiction since $[G:N_G(P_5)] = 11$ for each Sylow 5-subgroup $P_5$.

### The center of a subgroup of index 2 in a 2-group with cyclic center has rank at most 2

Use the preceding exercise to prove that if $P$ is a 2-group which has a cyclic center and $M$ is a subgroup of index 2 in $P$, then $Z(M)$ has rank at most 2.

Note that $M \leq P$ is normal and that $Z(M) \leq M$ is characteristic; thus $Z(M)$ is normal in $P$. Moreover, letting $Z(M)_2 = \{ z \in Z(M) \ |\ z^2 = 1 \}$, we know that $Z(M)_2 \leq Z(M)$ is characteristic, so that $Z(M)_2 \leq P$ is normal. Moreover, $Z(M)_2$ is elementary abelian by construction. Note that, by §5.2 #7, $Z(M)$ and $Z(M)_2$ have the same rank.

If $w \in P$, then conjugation by $w$ is an automorphism of $Z(M)_2$. If we choose $w \notin M$, and let $z$ be fixed by $w$– that is, $wz = zw$, then we have $z \in C_P(M)$ (since $z \in Z(M)$) and $z \in C_P(\langle w \rangle)$. Since $P = M \langle w \rangle$, in fact $w \in Z(P)$. Thus the subgroup of $Z(M)_2$ which is fixed under conjugation by $w$ is contained in $Z(P)$, and thus has rank 1.

By this previous exercise, $Z(M)_2$ has rank at most 2, so that $Z(M)$ also has rank at most 2.

### Every maximal subgroup of a finite solvable group has prime power index

Prove that every maximal subgroup of a finite solvable group has prime power index.

Let $G$ be a finite solvable group. We will proceed by induction on the breadth $k$ of $G$; that is, the number of prime factors in the factorization of $|G|$, including multiplicity.

For the base case, if $k = 1$, we have $G \cong Z_p$ for some prime $p$. Clearly then the (unique) maximal subgroup of $G$, namely 1, has prime power index.

For the inductive step, suppose that every finite solvable group of breadth at most $k$ satisfies the conclusion. Let $G$ be a finite solvable group of breadth $k+1$.

If $G$ is simple, then $G^\prime = 1$ (since $G$ is solvable). Thus $G$ is an abelian simple group, and we have $G \cong Z_p$ for some prime $p$; this is a contradiction.

Suppose now that $G$ is not simple, so that nontrivial normal subgroups exist. $H \leq G$ be a maximal subgroup and let $M \leq G$ be a minimal nontrivial normal subgroup. Now $G/M$ is a finite solvable group of breadth at most $k$. There are two cases:

If $M \leq H$, then $H/M \leq G/M$ is maximal by the Lattice Isomorphism Theorem. By the Third Isomorphism Theorem and the induction hypothesis, $[G:H] = [G/M : H/M]$ is a prime power.

If $M \not\leq H$, then $G = MH$. Now $[G:H] = [HM:H]$ $= |HM|/|H|$ $= (|H| \cdot |M|/|H \cap M|)/|H|$ $= |M|/|H \cap M|$ is a prime power, since $M$ is a $p$-group.

### Maximal subgroups of a finite nilpotent group have prime index

Prove that a maximal subgroup of a finite nilpotent group has prime index.

Let $G$ be a finite nilpotent group and $M \leq G$ a maximal subgroup. By Theorem 3, $M < N_G(M)$ is a proper subgroup, so that (since $M$ is maximal) $N_G(M) = G$. Thus $M \leq G$ is normal. Now $G/M$ is a finite nilpotent group.

Suppose $[G:M]$ is composite, with a proper prime divisor $q$. By this previous exercise, there is a normal subgroup $\overline{H} \leq G/M$. By the Lattice Isomorphism Theorem, $\overline{H} = H/M$ for some normal subgroup $H \leq G$ with $M \leq H$. Since $M$ is maximal, either $H = M$ or $H = G$, so that $q$ is not a proper divisor of $[G:M]$– a contradiction.

Thus $[G:M]$ is prime.

### Properties of the image and kernel of the p-power map on a finite abelian group

Let $A$ be a finite abelian group (written multiplicatively) and let $p$ be a prime. Let $A^p = \{ a^p \ |\ a \in A \}$ and $A_p = \{ x \in A \ |\ x^p = 1 \}$. (I.e. $A^p$ and $A_p$ are the image and kernel of the $p$-power map, respectively.

1. Prove that $A/A^p \cong A_p$. (Show that they are both elementary abelian and have the same order.)
2. Prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of index $p$. (Reduce to the case where $A$ is an elementary abelian $p$-group.)

1. We begin with some lemmas.

Lemma 1: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then there exists an isomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$. Proof: Let $\pi$ denote the natural projection from $Z$ to $Z/Z^p$ (for $Z \in \{A,B\}$). If $x \in A^p$, then $x = y^p$ for some $y \in A$. Then $(\pi \circ \theta)(x) = (\pi \circ \theta)(y^p)$ $= \pi(\theta(y)^p)$ $= B^p$, so that $A^p \leq \mathsf{ker}\ (\pi \circ \theta)$. By the remarks on page 100 in D&F, there exists a unique group homomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$ such that $\theta^\prime \circ \pi = \pi \circ \theta$. ($\theta^\prime$ is injective) Suppose $xA^p \in \mathsf{ker}\ \theta^\prime$. Then $\theta^\prime(\pi(x)) = 1$, thus $(\theta^\prime \circ \pi)(x) = 1$, so that $(\pi \circ \theta)(x) = 1$, hence $\theta(x) \in B^p$. Now since $\theta$ is surjective, $\theta(x) = \theta(y)^p$ for some $y \in A$. Thus $x = y^p$, so that $x \in A^p$. Thus $\mathsf{ker}\ \theta^\prime = 1$, so that $\theta^\prime$ is injective. ($\theta^\prime$ is surjective) Let $xB^p \in B/B^p$. Now $xB^p = (\pi \circ \theta)(y)$ for some $y \in A$, so that $xB^p = \theta^\prime(\pi(y))$. Hence $\theta^\prime$ is surjective. $\square$

Lemma 2: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then $\theta[A_p] = B_p$. Proof: $(\subseteq)$ If $x \in \theta[A_p]$, then $x = \theta(y)$ where $y \in A_p$. Now $y^p = 1$, so that $\theta(y)^p = 1$, hence $x^p = 1$. Thus $x \in B_p$. $(\supseteq)$ Let $x \in B_p$. Then x^p = 1\$. Since $\theta$ is surjective, $x = \theta(y)$ for some $y \in A$. Now $\theta(y^p) = \theta(y)^p = x^p = 1$, and since $\theta$ is injective, $y^p = 1$. Thus $y \in A_p$, hence $x \in \theta[A_p]$. $\square$

Lemma 3: Let $G$ be a finite abelian group of order $n$, let $m$ be an integer, and let $\varphi : G \rightarrow G$ denote the $m$-power homomorphism. If $\mathsf{gcd}(m,n) = 1$, then $\varphi$ is an isomorphism. Proof: Since $G$ is finite, it suffices to show that $\varphi$ is injective. Let $x \in \mathsf{ker}\ \varphi$. Then $x^m = 1$, so that $|x|$ divides $m$. By Lagrange, $|x|$ divides $n$. Thus $|x| = 1$, hence $x = 1$, and $\varphi$ is injective. $\square$

Now to the main result. By Theorem 5 in the text, $A \cong A_1 \times A_2$ where $A_1$ is a $p$-group of rank $t$ and $p$ does not divide $|A_2|$. By a lemma to the previous exercise, $\varphi = \varphi_1 \times \varphi_2$, where each map is the $p$-power map on $A_1 \times A_2$, $A_1$, and $A_2$, respectively.

We have $\mathsf{ker}\ \varphi = \mathsf{ker}(\varphi_1 \times \varphi_2)$ $= (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$ by the previous exercise.

Similarly, $(A_1 \times A_2)/\mathsf{im}\ \varphi = (A_1 \times A_2)/(\mathsf{im}\ \varphi_1 \times \mathsf{im}\ \varphi_2)$ $\cong (A_1/\mathsf{im}\ \varphi_1) \times (A_2/\mathsf{im}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$, again using the previous exercise.

Now the first conclusion follows using Lemmas 1 and 2.

2. We begin with some more lemmas.

Lemma 4: Let $E_{p^k}$ be the elementary abelian group of order $p^k$, and suppose $E_{p^k} = \langle S \rangle$ where $|S| = k$. Finally let $G$ be a group. If $\overline{\varphi} : S \rightarrow G$ is a mapping such that $|\overline{\varphi}(s)| = p$ for each $s \in S$ and $\overline{\varphi}(s) \overline{\varphi}(t) = \overline{\varphi}(t) \overline{\varphi}(s)$ for all $s,t \in S$, then $\overline{\varphi}$ extends uniquely to a group homomorphism $\varphi : E_{p^k} \rightarrow G$. Proof: Every element of $E_{p^k}$ can be written uniquely as $x = s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}$, where $s_1 \in S$ and $0 \leq a_k < p$. Define $\varphi(x) = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$. $\varphi$ is well defined since the $S$-expansion of $x$ is unique, and is a homomorphism since the $\overline{\varphi}(s)$ commute with one another. To see uniqueness, suppose $\psi : E_{p^k} \rightarrow G$ is a group homomorphism such that $\psi(s) = \varphi(s)$ for all $s \in S$. Then $\psi(x) = \psi(s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k})$ $= \psi(s_1)^{a_1} \psi(s_2)^{a_2} \cdots \psi(s_k)^{a_k}$ $= \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$ $= \varphi(x)$. Thus $\varphi$ is unique. $\square$

Note that every subgroup of $A$ of order $p$ is contained in $\mathsf{ker}\ \varphi$, so that in fact the order $p$ subgroups of $A$ and $\mathsf{ker}\ \varphi$ coincide. Now because $\mathsf{ker}\ \varphi \cong Z_{p^t}$ is elementary abelian, every nonidentity element has order $p$ and thus generates an order $p$ subgroup. Each such subgroup is generated by $p-1$ elements; thus there are $(p^t-1)/(p-1)$ order $p$ subgroups in $\mathsf{ker}\ \varphi$, thus in $A$.

Let $B \leq A$ be a subgroup of index $p$. Now let $x \in A^p$, say $x = y^p$, and suppose $x \notin B$. Then $y \notin B$, so that $A = \langle y \rangle B$. Now $A/B$ is a group of order $p$, so that $(yB)^p = xB = B$; then $x \in B$, a contradiction. Thus $A^p \leq B$. Moreover, by the Third Isomorphism Theorem, $[A/A^p : B/A^p] = [A:B] = p$. By the Lattice Isomorphism Theorem, the index $p$ subgroups of $A/A^p \cong Z_p^t$ correspond precisely to the index $p$ subgroups of $A$. In particular, it suffices to count the order and index $p$ subgroups in an elementary abelian $p$-group.

Now $\mathsf{Aut}(Z_p^t)$ acts on $\mathcal{M} = \{ M \leq Z_p^t \ |\ [Z_p^t : M] = p \}$ by $\varphi \cdot M = \varphi[M]$. We claim that this action is transitive. To that end, let $M_1, M_2 \leq Z_p^t$ be index $p$ subgroups. Now $M_1 \cong M_2 \cong Z_p^{t-1}$ since $M_1$ and $M_2$ are elementary abelian, thus $M_1 = \langle S_1 \rangle$ and $M_2 = \langle S_2 \rangle$ where $|S_1| = |S_2| = t-1$. Let $\overline{\psi} : S_1 \rightarrow S_2$ be a bijection, and choose some $t_1 \in Z_p^t \setminus M_1$ and $t_2 \in Z_p^t \setminus M_2$. Now $\overline{\psi} \cup \{(t_1,t_2)\}$ extends by Lemma 4 to a homomorphism $\psi : Z_p^t \rightarrow Z_p^t$. Clearly $\psi$ is surjective, hence an isomorphism, so that $\psi \in \mathsf{Aut}(Z_p^t)$. Moreover, we see that $\psi[M_1] = M_2$. Thus this action is transitive, and we have $\mathsf{Aut}(Z_p^t) \cdot M = \mathcal{M}$ for all index $p$ subgroups $M$.

Now let $M \leq Z_p^t$ be an arbitrary index $p$ subgroup and let $\varphi$ be an automorphism of $Z_p^t$ which stabilizes $M$. As above, $M = \langle S \rangle$ for some set $S$ with $|S| = t-1$. Moreover, if $x \in Z_p^t \setminus M$ then $Z_p^t = \langle S \cup \{x\} \rangle$ since $M$ is maximal. By Lemma 4, $\varphi$ is determined uniquely by is action on $S$ and $x$. We see also that since $\varphi$ stabilizes $M$, we must have $\varphi(s) \in M$ for each $s \in S$. Since $M \cong Z_p^{t-1}$ and $\varphi|_M \in \mathsf{Aut}(M)$, there are $\prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ distinct choices we can make for $\varphi[S]$. Moreover, $\varphi(x)$ may be chosen arbitrarily so long as $\varphi(x) \notin M$; thus there are $p^t - p^{t-1} = p^{t-1}(p-1)$ choices for $\varphi(x)$. Thus in total $\mathsf{Stab}(M)$ contains $(p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ elements.

By the Orbit-Stabilizer Theorem and Lagrange's Theorem, $|\mathcal{M}| = (\prod_{i=0}^{t-1} (p^t - p^{t-1-i}))/((p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}))$ $= (\prod_{i=1}^t (p^t - p^{t-i}))/((p-1) \prod_{i=1}^{t-1}(p^{t} - p^{t-i}))$ $= (p^t-1)/(p-1)$.

Thus the theorem is proved.

### For n at least 5, Alt(n) is the only subgroup of index less than n in Sym(n)

Prove that $A_n$ is the only proper subgroup of index less than $n$ in $S_n$ for $n \geq 5$.

Let $n \geq 5$ and let $H \leq S_n$ be a proper subgroup. Now $S_n$ acts on the left cosets of $H$ by left multiplication. Let $\varphi : S_n \rightarrow S_{[S_n:H]}$ be the permutation representation induced by this action, and let $K$ be its kernel. Note that $K \leq H$ is proper in $S_n$ and that $K$ is normal in $S_n$. By this previous exercise, there are two possibilities.

If $K = A_n$, then since $A_n$ is maximal, $H = A_n$.

If $K = 1$, then $\varphi$ is injective and we have $S_n \leq S_{[S_n : H]}$. If $[S_n : H] < n$, this yields a contradiction.

Thus for $n \geq 5$, $S_n$ has precisely one proper subgroup of index less than $n$, namely $A_n$.

### For n at least 5, the index of a subgroup of Alt(n) is at least n

Prove that $A_n$ does not have a proper subgroup of index less than $n$ for all $n \geq 5$.

We begin with a lemma.

Lemma: If $n \geq 3$, then $n!/2 > (n-1)!$. Proof: For the base case, note that $3!/2 = 3 > 2 = 2!$. For the inductive step, suppose $n!/2 > (n-1)!$. Now $n+1 > n$, so that $(n+1)!/2 > n!$. $\square$

Let $n \geq 5$, and let $H < A_n$ be a proper subgroup. Note that $A_n$ acts on the left cosets of $H$ by left multiplication; let $\varphi : G \rightarrow S_{[A_n : H]}$ be the permutation representation induced by this action, and let $K$ be the kernel of this action. Note that $K \leq \mathsf{stab}(H) = H < A_n$. Since $A_n$ is simple and $K$ is normal, we have $K = 1$. Thus $\varphi$ is injective.

If $[A_n : H] < n$, we have $A_n \leq S_{n-1}$. By the lemma this is a contradiction. Hence no proper subgroup has order less than $n$.

### A congruence condition on the index of subgroups containing Sylow normalizers

Let $P$ be a Sylow $p$-subgroup of $G$ and let $M$ be any subgroup of $G$ which contains $N_G(P)$. Prove that $[G : M] \equiv 1 \mod p$.

We prove a slightly stronger result by induction. First a definition.

Let $G$ be a finite group and $H \leq G$ a subgroup. The distance from $H$ to $G$ is the length $d$ of the longest chain of proper subgroups $K_i$ such that $H = K_0 < K_1 < \ldots < K_d$.

Lemma: Let $G$ be a finite group and $P \leq G$ a Sylow $p$-subgroup. If $H,K \leq G$ are subgroups such that $N_G(P) \leq H \leq K$, then $[K : H] \equiv 1 \mod p$. Proof: We proceed by induction on the distance $d$ from $N_G(P)$ to $K$. For the base case, if $d = 0$, then $K = N_G(P)$ so that $H = N_G(P)$, and we have $[K : H] = 1$. For the inductive step, suppose that for some $d \geq 0$, if the distance from $N_G(P)$ to $K$ is at most $d$ then for all $H$ such that $N_G(P) \leq H \leq K$ we have $[K : H] \equiv 1 \mod p$. Now let $K \leq G$ be a subgroup containing $N_G(P)$ such that the distance from $N_G(P)$ to $K$ is $d+1$, and suppose further that $N_G(P) \leq H \leq K$. First, if $H = K$, we have $[K : H] = 1$ so that the conclusion holds. Suppose now that the inclusion $H \leq K$ is proper. Note that $[K : N_G(P)] = [K : H] \cdot [H : N_G(P)]$. By the induction hypothesis, we have $[H : N_G(P)] \equiv 1 \mod p$. Recall that $N_K(P) = K \cap N_G(P)$, so that $[K : N_G(P)] = [K : N_K(P)]$ $= n_p(K) \equiv 1 \mod p$ by Sylow's Theorem. Modulo $p$, we thus have $[K:H] \equiv 1$. $\square$

In particular, if $K = G$ and $N_G(P) \leq M \leq G$, we have $[G:M] \equiv 1 \mod p$.