meta user
network profile
| bio | website | math.sunysb.edu/~sdalton |
|---|---|---|
| location | Stony Brook, NY | |
| age | 29 | |
| visits | member for | 2 years, 9 months |
| seen | 27 mins ago | |
| stats | profile views | 1,206 |
Grad Student!
Please help push Pets.SE into the next stage.
|
May 18 |
awarded | Constituent |
|
May 14 |
comment |
Ring such that $x^4=x$ for all $x$ @Jonathan: While I agree this seems somewhat opaque, this is a natural way of beginning to prove the much stronger version of this theorem (Jacobson's theorem): look at idempotents, and the differences between them. |
|
May 6 |
awarded | Caucus |
|
May 1 |
comment |
Solving $|Aut(H)| = |Aut(G)|$ Even without including the trivial subgroup, any cyclic subgroup of order $p$ has automorphism group of size $p-1$. So if $|G|$ is divisible by two primes, you're out of luck. If $G$ is a p-group, looking at $G/\Phi(G)$ shows it would have to be cyclic. A quick count then shows $G$ would have to be cyclic of prime order. |
|
May 1 |
comment |
Prove that if $A - A^2 = I$ then $A$ has no real eigenvalues No work or motivation shown - vote to close. |
|
Apr 30 |
comment |
circle reflections in hyperbolic geometry No, reflect the unit circle in the other circle (to get another circle). |
|
Apr 30 |
comment |
Is the group $\langle a, b \ | \ a^3, b^3, b^2a, ba^2, a^2b, ab^2\rangle$ normal in $\langle a, b\rangle$? And the answer is yes. |
|
Apr 30 |
comment |
Is the group $\langle a, b \ | \ a^3, b^3, b^2a, ba^2, a^2b, ab^2\rangle$ normal in $\langle a, b\rangle$? I think the question is "given the free group on generators $a$ and $b$, is the subgroup $\langle a^3, b^3, b^2a, ba^2, a^2b, ab^2\rangle$ a normal subgroup?" |
|
Apr 29 |
comment |
Non-central subgroup which is maximal among abelian normal subgroups is self-centralizing? $S_3\times A_5$? |
|
Apr 29 |
comment |
A group of order 2 in a group of order 60. $G/H$ is a group of order $30$, which has a normal Sylow 5-subgroup, so its preimage is normal... |
|
Apr 28 |
comment |
Prove the intersection of a Sylow $p$-subgroup and a subgroup is the unique Sylow $p$-subgroup I have voted to close your question. Not because I think it is a bad question, but because you have shown no work or effort of your own. |
|
Apr 27 |
comment |
geometrically finite hyperbolic surface of infinite volume You can also look at the twice-punctured complex plane. |
|
Apr 27 |
comment |
Why is every answer of $5^k - 2^k$ divisible by 3? -1: OP tried nothing on their own. |
|
Apr 26 |
awarded | Quorum |
|
Apr 24 |
comment |
Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$ Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$. |
|
Apr 24 |
comment |
Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$ While this is not exactly what you asked, it is pretty straighforward that the tetrahedral group is $A_4\cong PSL(2,3)$. And of course the binary tetrahedral group is the "double cover" of that group. So this is closely related to why $A_4$ is isomorphic to $PSL(2,3)$, which I believe has been discussed on this site before. |
|
Apr 23 |
comment |
Groups with 20 Sylow subgroups I wonder if you can use the conjugation action of $G$ on the Sylows to, WLOG, embed $G$ into $S_{20}$ as a doubly transitive group. The $760$ case shows there must be some elements of order $3$. Basically, the odd order subgroup of $N_G(P)$ in $S_{20}$ "should" give the subgroup of order $19\cdot9$ in $PSL(2,19)$. I have no idea how to prove $G$ has a subgroup of order $20$. |
|
Apr 23 |
comment |
Groups with 20 Sylow subgroups You can do this without caring about the Sylow 5- or 19-subgroups. If there are 95 Sylow 2-subgroups, there is a subgroup of index 5 or 19, which then contains a normal Sylow 19- or 5-subgroup. Sylow's theorems show this is normal in all of G. Other counts for Sylow 2-subgroups reduce once again to this case. |
|
Apr 23 |
comment |
A generalization of Galois connections what is this nonsense? |
|
Apr 23 |
revised |
Divergence of the series $\sum\limits_{n=1}^\infty\frac{(n!)^n}{n^{4n}}$ deleted 4 characters in body |
