Large subgroups of finite groups
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From: DLH on 9 Aug 2010 01:35 On Aug 8, 6:24 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Aug 8, 4:59 pm, DLH <roddl...(a)mit.edu> wrote: > > > Call a subgroup H of a finite group G large if |H| > [G:H], or, > > equivalently, |H| > sqrt(|G|). > > Any group of order p^k, where p is a prime and k = 0, 1, or 2, > > obviously has no large proper subgroups. > > > The question is whether every other finite group has a large proper > > subgroup. A minimal counterexample must be simple: > > Suppose G is not simple. Let N be a maximal normal subgroup of G. Then > > G/N is a simple group smaller than the minimal counterexample G. > > If G/N is nonabelian, G/N has a large proper subgroup H whose preimage > > under the /N quotient map is a large proper subgroup of G. > > If G/N is cyclic of prime order p, then we must have |N| < p > > Shouldn't that be |N|<= p? Since N is not large, then |N|>[G:N]=p does > not hold, but that only gives |N|<=p, not |N|<p. > > But if |N|=p, then |G|=p^2, which you will want to exclude from the > set of "minimal counterexamples" anyway (you are only considering > counterexamples among the groups that are not of order 1, p, or p^2, > right?) Yes. Thanks for filling in the hole in my reasoning. > > > for a > > counterexample. But then G has subgroups of order p which are large > > proper subgroups. > > You are missing a case here. > > What happens if G/N is of order p^2? Then G/N has a subgroup of order > p, which pulls back to a normal subgroup of index p in G and of order > greater than p, giving a contradiction. G/N is not of order p^2 because N is a _maximal_ normal subgroup of G. ---- DLH (See the initial message of the thread for my real email address.)
From: DLH on 9 Aug 2010 01:39 On Aug 9, 1:35 am, DLH <roddl...(a)mit.edu> wrote: > On Aug 8, 6:24 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Aug 8, 4:59 pm, DLH <roddl...(a)mit.edu> wrote: > > > > Call a subgroup H of a finite group G large if |H| > [G:H], or, > > > equivalently, |H| > sqrt(|G|). > > > Any group of order p^k, where p is a prime and k = 0, 1, or 2, > > > obviously has no large proper subgroups. > > > > The question is whether every other finite group has a large proper > > > subgroup. A minimal counterexample must be simple: > > > Suppose G is not simple. Let N be a maximal normal subgroup of G. Then > > > G/N is a simple group smaller than the minimal counterexample G. > > > If G/N is nonabelian, G/N has a large proper subgroup H whose preimage > > > under the /N quotient map is a large proper subgroup of G. > > > If G/N is cyclic of prime order p, then we must have |N| < p > > > Shouldn't that be |N|<= p? Since N is not large, then |N|>[G:N]=p does > > not hold, but that only gives |N|<=p, not |N|<p. > > > But if |N|=p, then |G|=p^2, which you will want to exclude from the > > set of "minimal counterexamples" anyway (you are only considering > > counterexamples among the groups that are not of order 1, p, or p^2, > > right?) > > Yes. Thanks for filling in the hole in my reasoning. > > > > > > for a > > > counterexample. But then G has subgroups of order p which are large > > > proper subgroups. > > > You are missing a case here. > > > What happens if G/N is of order p^2? Then G/N has a subgroup of order > > p, which pulls back to a normal subgroup of index p in G and of order > > greater than p, giving a contradiction. > > G/N is not of order p^2 because N is a _maximal_ normal subgroup of G. > > ---- DLH > > (See the initial message of the thread for my real email address.) And one more correction: "it its true" should be "it's true" in my original post.
From: Arturo Magidin on 9 Aug 2010 12:25
On Aug 9, 12:35 am, DLH <roddl...(a)mit.edu> wrote: > On Aug 8, 6:24 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Aug 8, 4:59 pm, DLH <roddl...(a)mit.edu> wrote: > > > > Call a subgroup H of a finite group G large if |H| > [G:H], or, > > > equivalently, |H| > sqrt(|G|). > > > Any group of order p^k, where p is a prime and k = 0, 1, or 2, > > > obviously has no large proper subgroups. > > > > The question is whether every other finite group has a large proper > > > subgroup. A minimal counterexample must be simple: > > > Suppose G is not simple. Let N be a maximal normal subgroup of G. Then > > > G/N is a simple group smaller than the minimal counterexample G. > > > If G/N is nonabelian, G/N has a large proper subgroup H whose preimage > > > under the /N quotient map is a large proper subgroup of G. > > > If G/N is cyclic of prime order p, then we must have |N| < p > > > Shouldn't that be |N|<= p? Since N is not large, then |N|>[G:N]=p does > > not hold, but that only gives |N|<=p, not |N|<p. > > > But if |N|=p, then |G|=p^2, which you will want to exclude from the > > set of "minimal counterexamples" anyway (you are only considering > > counterexamples among the groups that are not of order 1, p, or p^2, > > right?) > > Yes. Thanks for filling in the hole in my reasoning. > > > > > > for a > > > counterexample. But then G has subgroups of order p which are large > > > proper subgroups. > > > You are missing a case here. > > > What happens if G/N is of order p^2? Then G/N has a subgroup of order > > p, which pulls back to a normal subgroup of index p in G and of order > > greater than p, giving a contradiction. > > G/N is not of order p^2 because N is a _maximal_ normal subgroup of G. Fair enough; basically, my short paragraph above shows why the index cannot be p^2 under that assumption. Six of one, half a dozen of the other... -- Arturo Magidin |