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location | FI | |
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visits | member for | 4 years, 10 months |
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student.
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awarded | Necromancer |
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Lucido's three prime lemma
Also, it is possible that a nonsolvable group has the property in question. For example $\operatorname{PSL}(3,3)$ has order $2^4 \cdot 3^3 \cdot 13$ and contains an element of order $6$. |
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awarded | Self-Learner |
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Applications of Frobenius theorem and conjecture
There is also a nice direct proof by Brauer (in the paper "On A Theorem of Frobenius" in AMM), which does not require induction at all. Basically if $n = p^\alpha m$, $p$ prime, $(m,p) = 1$, then using a subgroup of order $p^\alpha$, we can define an equivalence relation $\sim$ on $G$ such that each $\sim$-class has order $p^\alpha$, and the set of solutions to $x^n = 1$ is an union of $\sim$-classes. This proof also gives you more general statements about the number of solutions in a fixed double coset $HyH$, for example. |
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awarded | Yearling |
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Index of agemo subgroups in $p$-groups
@GerryMyerson: Yes, the name comes from the common notation $\mho_1(G) = G^p$. |
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awarded | Popular Question |
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awarded | Teacher |
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answered | Applications of Frobenius theorem and conjecture |
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awarded | Enthusiast |
Feb
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answered | Lower bounds on the number of elements in Sylow subgroups |
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Lower bounds on the number of elements in Sylow subgroups
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Lower bounds on the number of elements in Sylow subgroups
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awarded | Nice Question |
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Lower bounds on the number of elements in Sylow subgroups
@Nick: Ok, I'll edit the post. What I mean by the construction is this. Fix a prime $p$ and integer $n \geq 1$. The construction shows that we can find a group $G$ with Sylow $p$-subgroups of order $p^n$ (not just some arbitrary power of $p$ like in your comment) such that $f_p(G)$ is arbitrarily large. |
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Lower bounds on the number of elements in Sylow subgroups
Perhaps one way to start with this is that given $n$, $p$ and $k$, find the smallest possible value for $f_p(G)$. I don't know if the bound $2p^{n+1}−p^n$ is sharp for the case $k=2$. It is sharp for all primes $p$ such that $2p+1$ is prime, which can be seen by the construction in my question. |
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awarded | Scholar |
Jan
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Lower bounds on the number of elements in Sylow subgroups
@Nick Gill: Yes, the lower bound will of course depend on $p$ and $n$, just like the lower bound given by Miller's thm does. Basically given $p$ and $n$, we're looking at functions $g$ such that $f_p(G) \geq g(k)$ for any finite group $G$ with Sylow subgroups of order $p^n$ and $n_p(G) = kp + 1$. Is this what you meant? |