Exercise on Maximal Group
Well, this one is actually an old random stuff I did when I was undergraduate, I found it on my old harddrive and would like to test what happen if I post it. This is actually an exercise in Martin Isaac Book, and I always forget how to do this problem properly (I mean really solve it concretely). In fact, I post it here again because I always forget how to solve it. This is a really detailed proof.
Without using Cauchy Theorem or Sylow Theorem prove that If $G$ is a finite group with unique maximal subgroup then $|G|=p^{\alpha}$, for some prime $p$.
We begin with some lemmas, and all the following lemmas assume that $G$ is finite.
Lemma 1
Let $A$ be a nonempty subset of $G$ such that $\langle A \rangle \neq G$ and for any $x\in G \backslash A$ we have $\langle A \cup \{x\} \rangle=G$ then $\langle A \rangle$ is a maximal subgroup.
Proof:
Let $H$ be a subgroup of $G$ such that $\langle A \rangle \subseteq H \subseteq G$, suppose for contrary that $H \neq \langle A \rangle $ and $H \neq G$, since $\langle A\rangle \neq G$ there exists $x \in H$ with $x \not \in \langle A \rangle $, therefore $x\not \in A$ thus $A \cup \{x\} \subseteq \langle A \rangle \cup \{x\} \subseteq H$ and hence $\langle A \cup \{x\} \rangle \subseteq H$, but by the hypothesis this means $G \subseteq H$, and we have $H=G$, a contradiction. Therefore either $H=\langle A \rangle$ or $H=G$, which implies $\langle A \rangle$ is maximal subgroup.
Lemma 2
Let $A$ and $B$ are subgroup of $G$. If $G = A \cup B$ then $A=G$ or $B=G$.
Proof:
Let $|G|=n$, $|A|=d_1$, $|B|=d_2$, and $|A\cap B| = d_3$. First we assume that $d_1 \geq d_2$, by Langrange theorem we have $d_1 \lvert n$ and $d_2 \lvert n$, also since $e \in A\cap B$ we have $d_3 \geq 1$. Observe that
\[n=|G|=|A\cup B| = |A| + |B| - |A \cap B| \leq d_1+d_1 -1 = 2d_1-1.\]
If $d_1= n$, then $|A|=|G|$, but since $A \subseteq G$, we have $A=G$, the assertion is proved. Now for $d_1 < n$ since by Langrange $d_1 | n$, we have $n=k d_1$ for a positive integer $k$, but $n \leq 2d_1-1$, thus $k \leq 2- \frac{1}{d_1}$, and we have $k \leq 1$, this leads to $n=d_1$ again. Similarly If we instead assume $d_2 \geq d_1$ then $B=G$.
Lemma 3
If $G$ has a unique maximal subgroup then $G$ is a cyclic group.
Proof:
Let $M$ be the unique maximal subgroup of $G$, let $C(M)=\langle G\backslash M \rangle$, then we have $G = M \cup C(M)$ thus by Lemma 2 we have $M=G$ or $C(M)=G$, but $M\neq G$ therefore we must have $C(M)=G$.
Define
\[\begin{eqnarray*}&\mathcal{T}_i := \{A \, \lvert A \subseteq G\backslash M \mbox{ and } \lvert A \rvert=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{L} := \{i \, \lvert \, \langle A \rangle = G \mbox{ for all } A \in \mathcal{T}_i\}
\end{eqnarray*}\]
Let $\lvert G \backslash M \rvert = \lambda$, we have \[\mathcal{T}_{\lambda} = \{A \, \lvert A \subseteq G \backslash M \mbox{ and } |A|=\lambda=\lvert G \backslash M \rvert\}=\{ G \backslash M\}\]
Since $\langle G \backslash M \rangle = G$, this means that for every $A \in \mathcal{T}_{\lambda}=\{G \backslash M\}$ we have $\langle A \rangle= \langle G \backslash M \rangle =G$, therefore $\lambda \in \mathfrak{L}$, in particular $\mathfrak{L} \neq \emptyset$.
Since $\mathfrak{L}$ is nonempty, by the Well-Ordering Principle there exists a smallest elemet $t \in \mathfrak{L}$. If $t=1$, since $1=t \in \mathfrak{L}$, for every $A \in \mathcal{T}_1$ we have $\langle A \rangle =G$ , in particular if we take $A=\{g\}$ where $g\in G\backslash M$, we see that $A \subset G\backslash M$ and $\lvert A \rvert = 1=t$. This means $A\in \mathcal{T}_1$, so that $\langle g\rangle = \langle A \rangle =G$, thus $G$ is cyclic, and the lemma is proved whenever $t=1$.
Next we will prove that $t > 1$ would eventually end with a contradiction. If $t>1$, then $t-1$ is the largest positive integers which fails to be included on $\mathfrak{L}$, equivalently there exists $S \in \mathcal{T}_{t-1}$ with $\langle S \rangle \neq G$, notice that $|S|=t-1 >0$ thus $S\neq \emptyset$. Define
\[\begin{eqnarray*}&\mathcal{U}_i := \{ B \, \lvert B \subseteq M \mbox{ and } |B|=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{P} := \{i \, \lvert \, \langle S \cup B \rangle = G \mbox{ for all } B \in \mathcal{U}_i\}
\end{eqnarray*}\]
Observe that $M \subset S \cup M \subseteq \langle S \cup M \rangle \subset G$, since $M$ is maximal we have either $M=\langle S \cup M \rangle$ or $\langle S \cup M \rangle = G$, but $M=\langle S \cup M \rangle$ would implies $S \subset \langle S \cup M \rangle = M$ this is impossible since $S\in \mathcal{T}_1$ which means $S \subset G \backslash M$, thus the only possibility is $\langle S \cup M \rangle=G$. Let $\lvert M \rvert=\mu$, we have
\[\mathcal{U}_{\mu} = \{B \, \lvert B \subseteq M \mbox{ and } \lvert B \rvert = \mu = \lvert M \rvert\} = \{ M\}\]
Since $\langle S \cup M \rangle=G$, this means that for every $ B \in \mathcal{U}_{\mu}=\{M\}$ we have $\langle S \cup B \rangle= \langle S \cup M \rangle =G$, therefore $\mu \in \mathfrak{P}$, in particular $\mathfrak{P} \neq \emptyset$. Since $\mathfrak{P}$ is nonempty, by Well-Ordering Principle there exists a smallest element $l \in \mathfrak{P}$.
Before we proceed, we will recall what we have so far :
\[S \subset G \backslash M \qquad S \in \mathcal{T}_{t-1} \qquad \langle S \rangle \neq G\]
Also $t\in \mathfrak{L}$ and satisfies
\[\text{for all $A \in \mathcal{T}_t$ we have $\langle A \rangle = G$}\]
and $l\in \mathfrak{P}$ and satisfies
\[\text{for all $B \in \mathcal{U}_l$ we have $\langle S \cup B \rangle = G$}\]
Now we continue the proof, if $l=1$, since $1=l \in \mathfrak{P}$ we have $\langle S \cup B \rangle =G$ for every $ B \in \mathcal{U}_1$, we will prove that this would implies $\langle S \rangle $ is maximal subgroup. Indeed, take any $x\in G \backslash S$, which implies $x\not \in S$, we divide into two cases:
i) If $x\in M$ then $\{x\} \subset M$, thus $ \{x\} \in \mathcal{U}_1$ and hence $\langle S \cup \{x\}\rangle =G$.
ii) If $x \not \in M$ then $\{x\}\subset G \backslash M$, thus $S \cup \{x\} \subset G \backslash M$, also since $x\not \in S$ we have $S \cap \{x\} = \emptyset$, together with $S \in \mathcal{T}_{t-1}$ we have $\lvert S \cup \{x\}\rvert = \lvert S \rvert + 1 = t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$. Since $t\in \mathfrak{L}$ every $A \in \mathcal{T}_t$ satisfies $\langle A \rangle =G$ thus in particular $S \cup \{x\} \in \mathcal{T}_t$ implies $\langle S \cup \{x\} \rangle = G$.
In summary for every $x\in G \backslash S$, we have $\langle S \cup \{x\} \rangle = G$, by Lemma 1 this means $\langle S\rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S\rangle=M$, but then $S \subset M $ also we have $S\subset G \backslash M$, thus $S=\emptyset$, contradiction,
, hence $l \neq1$.
If $l>1$, then $l-1$ is the largest positive integers which fails to be included on $\mathfrak{P}$, equivalently there exists $V \in \mathcal{U}_{l-1}$ with $\langle S \cup V \rangle \neq G$. We will prove that $\langle S \cup V \rangle$ is maximal subgroup. Indeed, take any $x \in G \backslash \left(S \cup V\right) = G\backslash S \cap G \backslash V$, which implies $x\not \in S$ and $s \not \in V$
we divide into two cases:
i) If $x\in M$, note that $V \in \mathcal{U}_{l-1}$ means that $V \subset M$ and $\lvert V \rvert = l-1$, therefore $V \cup \{x\} \subset M$. And since $x \not \in V$ we have $V \cap \{x\}=\emptyset$, thus $\lvert V \cup \{x\}\rvert = \lvert V \rvert + 1 = l-1+1=l$, therefore $V \cup \{x\} \in \mathcal{U}_{l}$, since $l \in \mathfrak{P}$ we have $\langle S \cup V \cup \{x\}\rangle= G$.
ii) If $x\not \in M$ then $x \in G \backslash M$, note that $S \in \mathcal{T}_{t-1}$ means that $S \subset G\backslash M$ and $\lvert S \rvert = t-1$, therefore $S \cup \{x\} \subset G\backslash M$. And since $x \not \in S$ we have $S \cap \{x\}=\emptyset$, thus $\lvert S\cup \{x\} \rvert=\lvert S \rvert + 1=t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$, since $t\in \mathfrak{P}$ we have $\langle S \cup \{x\} \rangle =G$. Furthermore we have $G=\langle S \cup \{x\} \rangle \subset \langle S \cup V \cup \{x\} \rangle$, which implies $\langle S \cup V \cup \{x\}\rangle= G$.
In summary, for every $x \in G \backslash \left(S \cup V\right)$, we have $\langle S \cup V \cup \{x\}\rangle= G$, by Lemma 1 this means $\langle S \cup V \rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S \cup V \rangle=M$, but then $S \subset \langle S \cup V \rangle = M$, by the fact that $S\subset G \backslash M$, we would have $S=\emptyset$, contradiction. And the proof is completed.
Now we have $G=\langle g \rangle$ by Lemma 3, let $|G|=n=p_1^{\alpha_1} \cdots p_j^{\alpha_j}$, where $p_1,\cdots,p_j$ are all prime numbers. Consider
\[H_{p_k} = \langle g^{p_k} \rangle\]
We will prove that $H_{p_k}$ is maximal subgroup for every $k$. Indeed, if S is a subgroup with $S \neq G$ and $S \neq H_{p_k}$ satisfies $H_{p_k} \subseteq S \subseteq G$, then $d_k=|H_{p_k}|$ divides $|S| \leq |G|$, but $|H_{p_k}|=\frac{n}{p_k}$, therefore
\[\frac{n}{p_k} \leq |S| \leq n \qquad \text{and} \qquad \text{$\frac{n}{p_k}$ divides $\lvert S\rvert$ divides $n$}\]
So that there exist $u, v\in \mathbb{N}$ such that $n= v |S|$ and $|S|=u \left( \frac{n}{p_k}\right)$, thus $n= (u)\, (v) \left(\frac{n}{p_k}\right)$, which implies $p_k = u v$, since $p_k$ is prime we should have $u=1$ or $v=1$, but then $|H_{p_k}|=|S|$ or $|S|=|G|$ which together with $H_{p_k} \subseteq S \subseteq G$ would result to $H_{p_k}=S$ or $S=G$, thus $H_{p_k}$ is a maximal subgroup for every $k$.
Since $G$ has only one maximal subgroup that is $M$, we should have $H_{p_k} =M= H_{p_l}$ for all $k$ and $l$, therefore $\frac{n}{p_k}=|H_{p_k}|=|H_{p_l}|=\frac{n}{p_l}$ so that $p_k=p_l$ for all $k$ and $l$, thus $|G|$ actually has only one prime factor and we conclude that $|G|=p^{\alpha}$.
Continue Reading
Without using Cauchy Theorem or Sylow Theorem prove that If $G$ is a finite group with unique maximal subgroup then $|G|=p^{\alpha}$, for some prime $p$.
We begin with some lemmas, and all the following lemmas assume that $G$ is finite.
Lemma 1
Let $A$ be a nonempty subset of $G$ such that $\langle A \rangle \neq G$ and for any $x\in G \backslash A$ we have $\langle A \cup \{x\} \rangle=G$ then $\langle A \rangle$ is a maximal subgroup.
Proof:
Let $H$ be a subgroup of $G$ such that $\langle A \rangle \subseteq H \subseteq G$, suppose for contrary that $H \neq \langle A \rangle $ and $H \neq G$, since $\langle A\rangle \neq G$ there exists $x \in H$ with $x \not \in \langle A \rangle $, therefore $x\not \in A$ thus $A \cup \{x\} \subseteq \langle A \rangle \cup \{x\} \subseteq H$ and hence $\langle A \cup \{x\} \rangle \subseteq H$, but by the hypothesis this means $G \subseteq H$, and we have $H=G$, a contradiction. Therefore either $H=\langle A \rangle$ or $H=G$, which implies $\langle A \rangle$ is maximal subgroup.
Lemma 2
Let $A$ and $B$ are subgroup of $G$. If $G = A \cup B$ then $A=G$ or $B=G$.
Proof:
Let $|G|=n$, $|A|=d_1$, $|B|=d_2$, and $|A\cap B| = d_3$. First we assume that $d_1 \geq d_2$, by Langrange theorem we have $d_1 \lvert n$ and $d_2 \lvert n$, also since $e \in A\cap B$ we have $d_3 \geq 1$. Observe that
\[n=|G|=|A\cup B| = |A| + |B| - |A \cap B| \leq d_1+d_1 -1 = 2d_1-1.\]
If $d_1= n$, then $|A|=|G|$, but since $A \subseteq G$, we have $A=G$, the assertion is proved. Now for $d_1 < n$ since by Langrange $d_1 | n$, we have $n=k d_1$ for a positive integer $k$, but $n \leq 2d_1-1$, thus $k \leq 2- \frac{1}{d_1}$, and we have $k \leq 1$, this leads to $n=d_1$ again. Similarly If we instead assume $d_2 \geq d_1$ then $B=G$.
Lemma 3
If $G$ has a unique maximal subgroup then $G$ is a cyclic group.
Proof:
Let $M$ be the unique maximal subgroup of $G$, let $C(M)=\langle G\backslash M \rangle$, then we have $G = M \cup C(M)$ thus by Lemma 2 we have $M=G$ or $C(M)=G$, but $M\neq G$ therefore we must have $C(M)=G$.
Define
\[\begin{eqnarray*}&\mathcal{T}_i := \{A \, \lvert A \subseteq G\backslash M \mbox{ and } \lvert A \rvert=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{L} := \{i \, \lvert \, \langle A \rangle = G \mbox{ for all } A \in \mathcal{T}_i\}
\end{eqnarray*}\]
Let $\lvert G \backslash M \rvert = \lambda$, we have \[\mathcal{T}_{\lambda} = \{A \, \lvert A \subseteq G \backslash M \mbox{ and } |A|=\lambda=\lvert G \backslash M \rvert\}=\{ G \backslash M\}\]
Since $\langle G \backslash M \rangle = G$, this means that for every $A \in \mathcal{T}_{\lambda}=\{G \backslash M\}$ we have $\langle A \rangle= \langle G \backslash M \rangle =G$, therefore $\lambda \in \mathfrak{L}$, in particular $\mathfrak{L} \neq \emptyset$.
Since $\mathfrak{L}$ is nonempty, by the Well-Ordering Principle there exists a smallest elemet $t \in \mathfrak{L}$. If $t=1$, since $1=t \in \mathfrak{L}$, for every $A \in \mathcal{T}_1$ we have $\langle A \rangle =G$ , in particular if we take $A=\{g\}$ where $g\in G\backslash M$, we see that $A \subset G\backslash M$ and $\lvert A \rvert = 1=t$. This means $A\in \mathcal{T}_1$, so that $\langle g\rangle = \langle A \rangle =G$, thus $G$ is cyclic, and the lemma is proved whenever $t=1$.
Next we will prove that $t > 1$ would eventually end with a contradiction. If $t>1$, then $t-1$ is the largest positive integers which fails to be included on $\mathfrak{L}$, equivalently there exists $S \in \mathcal{T}_{t-1}$ with $\langle S \rangle \neq G$, notice that $|S|=t-1 >0$ thus $S\neq \emptyset$. Define
\[\begin{eqnarray*}&\mathcal{U}_i := \{ B \, \lvert B \subseteq M \mbox{ and } |B|=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{P} := \{i \, \lvert \, \langle S \cup B \rangle = G \mbox{ for all } B \in \mathcal{U}_i\}
\end{eqnarray*}\]
Observe that $M \subset S \cup M \subseteq \langle S \cup M \rangle \subset G$, since $M$ is maximal we have either $M=\langle S \cup M \rangle$ or $\langle S \cup M \rangle = G$, but $M=\langle S \cup M \rangle$ would implies $S \subset \langle S \cup M \rangle = M$ this is impossible since $S\in \mathcal{T}_1$ which means $S \subset G \backslash M$, thus the only possibility is $\langle S \cup M \rangle=G$. Let $\lvert M \rvert=\mu$, we have
\[\mathcal{U}_{\mu} = \{B \, \lvert B \subseteq M \mbox{ and } \lvert B \rvert = \mu = \lvert M \rvert\} = \{ M\}\]
Since $\langle S \cup M \rangle=G$, this means that for every $ B \in \mathcal{U}_{\mu}=\{M\}$ we have $\langle S \cup B \rangle= \langle S \cup M \rangle =G$, therefore $\mu \in \mathfrak{P}$, in particular $\mathfrak{P} \neq \emptyset$. Since $\mathfrak{P}$ is nonempty, by Well-Ordering Principle there exists a smallest element $l \in \mathfrak{P}$.
Before we proceed, we will recall what we have so far :
\[S \subset G \backslash M \qquad S \in \mathcal{T}_{t-1} \qquad \langle S \rangle \neq G\]
Also $t\in \mathfrak{L}$ and satisfies
\[\text{for all $A \in \mathcal{T}_t$ we have $\langle A \rangle = G$}\]
and $l\in \mathfrak{P}$ and satisfies
\[\text{for all $B \in \mathcal{U}_l$ we have $\langle S \cup B \rangle = G$}\]
Now we continue the proof, if $l=1$, since $1=l \in \mathfrak{P}$ we have $\langle S \cup B \rangle =G$ for every $ B \in \mathcal{U}_1$, we will prove that this would implies $\langle S \rangle $ is maximal subgroup. Indeed, take any $x\in G \backslash S$, which implies $x\not \in S$, we divide into two cases:
i) If $x\in M$ then $\{x\} \subset M$, thus $ \{x\} \in \mathcal{U}_1$ and hence $\langle S \cup \{x\}\rangle =G$.
ii) If $x \not \in M$ then $\{x\}\subset G \backslash M$, thus $S \cup \{x\} \subset G \backslash M$, also since $x\not \in S$ we have $S \cap \{x\} = \emptyset$, together with $S \in \mathcal{T}_{t-1}$ we have $\lvert S \cup \{x\}\rvert = \lvert S \rvert + 1 = t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$. Since $t\in \mathfrak{L}$ every $A \in \mathcal{T}_t$ satisfies $\langle A \rangle =G$ thus in particular $S \cup \{x\} \in \mathcal{T}_t$ implies $\langle S \cup \{x\} \rangle = G$.
In summary for every $x\in G \backslash S$, we have $\langle S \cup \{x\} \rangle = G$, by Lemma 1 this means $\langle S\rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S\rangle=M$, but then $S \subset M $ also we have $S\subset G \backslash M$, thus $S=\emptyset$, contradiction,
, hence $l \neq1$.
If $l>1$, then $l-1$ is the largest positive integers which fails to be included on $\mathfrak{P}$, equivalently there exists $V \in \mathcal{U}_{l-1}$ with $\langle S \cup V \rangle \neq G$. We will prove that $\langle S \cup V \rangle$ is maximal subgroup. Indeed, take any $x \in G \backslash \left(S \cup V\right) = G\backslash S \cap G \backslash V$, which implies $x\not \in S$ and $s \not \in V$
we divide into two cases:
i) If $x\in M$, note that $V \in \mathcal{U}_{l-1}$ means that $V \subset M$ and $\lvert V \rvert = l-1$, therefore $V \cup \{x\} \subset M$. And since $x \not \in V$ we have $V \cap \{x\}=\emptyset$, thus $\lvert V \cup \{x\}\rvert = \lvert V \rvert + 1 = l-1+1=l$, therefore $V \cup \{x\} \in \mathcal{U}_{l}$, since $l \in \mathfrak{P}$ we have $\langle S \cup V \cup \{x\}\rangle= G$.
ii) If $x\not \in M$ then $x \in G \backslash M$, note that $S \in \mathcal{T}_{t-1}$ means that $S \subset G\backslash M$ and $\lvert S \rvert = t-1$, therefore $S \cup \{x\} \subset G\backslash M$. And since $x \not \in S$ we have $S \cap \{x\}=\emptyset$, thus $\lvert S\cup \{x\} \rvert=\lvert S \rvert + 1=t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$, since $t\in \mathfrak{P}$ we have $\langle S \cup \{x\} \rangle =G$. Furthermore we have $G=\langle S \cup \{x\} \rangle \subset \langle S \cup V \cup \{x\} \rangle$, which implies $\langle S \cup V \cup \{x\}\rangle= G$.
In summary, for every $x \in G \backslash \left(S \cup V\right)$, we have $\langle S \cup V \cup \{x\}\rangle= G$, by Lemma 1 this means $\langle S \cup V \rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S \cup V \rangle=M$, but then $S \subset \langle S \cup V \rangle = M$, by the fact that $S\subset G \backslash M$, we would have $S=\emptyset$, contradiction. And the proof is completed.
Now we have $G=\langle g \rangle$ by Lemma 3, let $|G|=n=p_1^{\alpha_1} \cdots p_j^{\alpha_j}$, where $p_1,\cdots,p_j$ are all prime numbers. Consider
\[H_{p_k} = \langle g^{p_k} \rangle\]
We will prove that $H_{p_k}$ is maximal subgroup for every $k$. Indeed, if S is a subgroup with $S \neq G$ and $S \neq H_{p_k}$ satisfies $H_{p_k} \subseteq S \subseteq G$, then $d_k=|H_{p_k}|$ divides $|S| \leq |G|$, but $|H_{p_k}|=\frac{n}{p_k}$, therefore
\[\frac{n}{p_k} \leq |S| \leq n \qquad \text{and} \qquad \text{$\frac{n}{p_k}$ divides $\lvert S\rvert$ divides $n$}\]
So that there exist $u, v\in \mathbb{N}$ such that $n= v |S|$ and $|S|=u \left( \frac{n}{p_k}\right)$, thus $n= (u)\, (v) \left(\frac{n}{p_k}\right)$, which implies $p_k = u v$, since $p_k$ is prime we should have $u=1$ or $v=1$, but then $|H_{p_k}|=|S|$ or $|S|=|G|$ which together with $H_{p_k} \subseteq S \subseteq G$ would result to $H_{p_k}=S$ or $S=G$, thus $H_{p_k}$ is a maximal subgroup for every $k$.
Since $G$ has only one maximal subgroup that is $M$, we should have $H_{p_k} =M= H_{p_l}$ for all $k$ and $l$, therefore $\frac{n}{p_k}=|H_{p_k}|=|H_{p_l}|=\frac{n}{p_l}$ so that $p_k=p_l$ for all $k$ and $l$, thus $|G|$ actually has only one prime factor and we conclude that $|G|=p^{\alpha}$.