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}.
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}.
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