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comment Group of order 30 can't be simple
How do you get in (2) that 5-Sylow subgroups are normal if $s=6$?
2d
comment Group of order 30 can't be simple
Part (3) of your postscript surprises me a bit. Since when is 6 a prime number?
Dec
22
comment Group of order 30 can't be simple
@Nishant: I wanted to point out to you that Thompson's transfer lemma is another possible generalization of this line of argument, but then the first hit on google was this. You get the conclusion of your comment by taking $k=1$ in your question from July (looks a bit odd, but that's probably the reason you didn't make your observation back then).
Dec
20
comment Group of order $p^2$ is commutative with prime $p$
Try to use first Lemma 2 and then Lemma 1.
Dec
20
comment Group of order $p^2$ is commutative with prime $p$
Sylow's theorems are more commonly used for groups that are not $p$-groups.
Dec
20
comment Why do we negate the imaginary part when conjugating?
If you work with i and j as described by you, a and b wouldn't be unique anymore.
Dec
19
awarded  Constituent
Dec
19
comment Density of odd vs even group orders that are not forced to be simple by Sylow's Theorem
How does the statistic look for multiples of 3 versus non-multiples?
Dec
19
comment Density of odd vs even group orders that are not forced to be simple by Sylow's Theorem
For odd primes $p$ the number $p+1$ is even. On one hand the same holds for $3p+1, 5p+1, \dots$, so you loose many possible $n_p$'s. On the other hand you have for small $p$ a good chance that $p+1$ is twice a prime or more generally a small multiple of a prime.
Dec
12
awarded  Caucus
Dec
8
answered Odd order matrix in $GL_n(\mathbb F_2)$ that doesn't commute with any order $2$ matrix?
Dec
8
comment Odd order matrix in $GL_n(\mathbb F_2)$ that doesn't commute with any order $2$ matrix?
Try multiplication with a generator of $\mathbb{F}_{2^n}^\times$.
Dec
5
comment Equivalence of two relations in Braid groups
In your definition of $A_{i,j}$ there is a typo: $\sigma_{i+1}^2$ should be $\sigma_{i+1}^{-1}$. I tried to correct it, but "Edits must be at least 6 characters long".
Dec
4
comment group without involution is 2-divisible
@rmznyzgyr You don't have to prove that it is a direct sum, just a set-theoretic union.
Dec
4
comment group without involution is 2-divisible
Can you solve the problem for $G = Z_n$ with $n$ odd?
Dec
4
comment Writing this $G$-set explicitly as union of orbits
To substantiate my pessimism: It's quite common to misinterpret the classification of finite abelian groups. One tends to forget that the factors are not uniquely defined as subgroups of $G$ (only the $p$-Sylow subgroups are). $G=Z_2\times Z_4$ for example, has four elements of order 4, a unique element of order 2 that is twice any element of order 4 and two other elements of order 2. Any of the two elements of order 2, and any element of order 4 can be used to define the subgroups to write $G$ as inner product $Z_2\times Z_4$ (4 possibilities in total as $Z_4$ has 2 elements of order 4).
Dec
4
comment Maps to all finite cyclic groups factor implies map to integers factors
How about $H = \mathbb{Q}/\mathbb{Z}$?
Dec
3
comment Writing this $G$-set explicitly as union of orbits
Unfortunately you cannot hope to find nice representatives in general, just because you found them for some special case. What you can do here, is to choose representatives $x_i$ for $G/(A+B)$ (I use additive notation as $G$ is abelian), i.e., $G = \stackrel{.}{\cup}_i x_i + (A+B)$, and get then $G/A\times G/B = \stackrel{.}{\cup}_i G\cdot (x_i+A/A, B/B)$ (where $\cdot$ denotes the operation of $G$). I wouldn't hope for nice choices of the $x_i$ for general $G$ and $A+B$.
Dec
2
awarded  Revival
Dec
2
comment Is there a group which has precisely all finite groups as subgroups?
@PavelC: I just thought about Jeremy's claim that the group is quasisimple: A proper normal subgroup being finite would have a centralizer of finite index, hence be central.