bio | website | |
---|---|---|
location | FI | |
age | ||
visits | member for | 4 years |
seen | 13 hours ago | |
stats | profile views | 284 |
student.
Sep 24 |
awarded | Autobiographer |
Jun 9 |
awarded | Self-Learner |
May 12 |
comment |
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. |
May 7 |
awarded | Yearling |
Feb 2 |
comment |
Index of agemo subgroups in $p$-groups
@GerryMyerson: Yes, the name comes from the common notation $\mho_1(G) = G^p$. |
Jul 3 |
awarded | Popular Question |
Mar 10 |
awarded | Teacher |
Feb 16 |
answered | Applications of Frobenius theorem and conjecture |
Feb 14 |
awarded | Enthusiast |
Feb 5 |
answered | Lower bounds on the number of elements in Sylow subgroups |
Jan 23 |
revised |
Lower bounds on the number of elements in Sylow subgroups
added 4 characters in body |
Jan 23 |
awarded | Editor |
Jan 23 |
revised |
Lower bounds on the number of elements in Sylow subgroups
deleted 22 characters in body; added 129 characters in body |
Jan 23 |
awarded | Nice Question |
Jan 23 |
comment |
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. |
Jan 21 |
comment |
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. |
Jan 21 |
awarded | Scholar |
Jan 21 |
comment |
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? |
Jan 21 |
accepted | Applications of Frobenius theorem and conjecture |
Jan 20 |
asked | Lower bounds on the number of elements in Sylow subgroups |