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Subgroups of a p-group

  1. Aug 8, 2007 #1
    Problem
    Suppose for all subgroups [tex]H,K[/tex] of a finite group [tex]G[/tex], either [tex]H \subset K[/tex] or [tex]K \subset H[/tex]. Show that [tex]G[/tex] is cyclic and its order is the power of a prime.

    Attempt
    I think I get the intuition: if [tex]H[/tex] and [tex]K[/tex] are not the same, then one of them must be the trivial subgroup and the other must be [tex]G[/tex] itself. So if [tex]g \in G[/tex] but [tex]g \notin H[/tex], then [tex]\left\langle g \right \rangle[/tex] is a subgroup containing [tex]g[/tex], so by hypothesis, [tex]H \subset \left\langle g \right \rangle[/tex]. From here I want to show that [tex]H[/tex] is actually the trivial subgroup. No idea yet about the power of a prime thing. Can anyone provide a hint? Thanks!
     
  2. jcsd
  3. Aug 8, 2007 #2

    Dick

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    Homework Helper

    Your intuition is not quite correct. Suppose H is a proper subgroup of G. Consider the subgroups H_g=gHg^(-1) for g in G. What's the containment relation between H and H_g? What kind of a subgroup is H? This should be enough of a hint to get you started. For the prime thing, if p and q are prime factors of G, then we know there are subgroups Hp and Hq of order p and q respectively. What is their containment relation?
     
    Last edited: Aug 8, 2007
  4. Aug 8, 2007 #3
    There is a minor mistake. You should have said, [tex]G[/tex] is a non-trivial group.
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    My Own Question: If the same conditions apply for an INFINITE group G, does the it mean it is cyclic? (Possibly Zorn's Lemma).
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    Here is the answer to your question.
    Say [tex]p\not = q[/tex] are two primes dividing [tex]G[/tex]. By Cauchy's theorem there are subgroups [tex]H \mbox{ and }K[/tex] respectively so that [tex]|H|=p \mbox{ and }|K|=q[/tex]. But how can (without lose of generality) [tex]H\subseteq K[/tex] since all its non-identity elements are of order [tex]p[/tex] while all the non-identity elements of [tex]K[/tex] are of order [tex]q[/tex]. A contradiction. Hence two distint primes do not divide [tex]G[/tex]. And so [tex]|G|=p^n[/tex]. Now we show that [tex]G\simeq \mathbb{Z}_{p^n}[/tex]. Assume that [tex]G[/tex] is not cyclic. Let [tex]a\not = e[/tex] be any element, then [tex]\left< a \right>[/tex] does not exhaust [tex]G[/tex] by assumption. So choose [tex]b\in G \mbox{ with }b\not \in \left< a \right> [/tex] and construct [tex]\left< b \right>[/tex]. But hypothesis [tex]\left< a \right> \subseteq \left< b\right> \mbox{ or }\left< b \right> \subseteq \left< a\right>[/tex]. But that is a contradiction. Q.E.D.
     
  5. Aug 8, 2007 #4
    I made a mistake in my other post when I said [tex]\left< a \right> \subseteq \left< b\right> \mbox{ or }\left< b \right> \subseteq \left< a\right>[/tex] is a contradiction. It is not. It however means that [tex]\left< a\right> \subseteq \left< b \right>[/tex] since [tex]b\not \in \left< a \right> \implies \left< b\right> \not \subseteq \left< a\right>[/tex]. Thus, [tex]\left< a\right> \subset \left< b \right> [/tex]. So we can choose [tex]c\in G \not \in \left< b \right>[/tex] to get [tex]\left< a\right> \subset \left< b \right> \subset \left< c\right>[/tex]. Thus, continuing this we can an ascending chain condition of subgroups properly contained in another. Since [tex]G[/tex] is finite it means this chain must terminate and hence there is an element which generates the full group.
    Q.E.D.
     
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