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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
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
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 $G/A\times G/B$ 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 $G/A\times G/B$ 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
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.
Dec
2
comment Is there a group which has precisely all finite groups as subgroups?
@PavelC: The center $Z$ is finite and intersects subgroups of $G$ isomorphic to $S_n$ (n>2) trivially. If you factor out the center, therefore $G/Z$ still has your property. If it has a non-trivial center, repeat. This has to stop after finitely many steps, as otherwise you would get an infinite subgroup not containing any simple subgroup of $G$.
Dec
2
comment Questions concerning $\mathbb Z_3[x]/(x^3+2x-1)$
What's your $\mathbb Z_3$? The field with three elements or the 3-adic numbers?
Dec
2
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
@user123733: The general proof strategy "$G$ has no element of order 6 implies $G$ is isomorphic $A_4$" is clear to you?
Dec
2
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
Which direction do you have problems with?
Dec
2
comment Generating a group by its $q$-elements.
How about $G=S_4$ and $q=3$?
Dec
2
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
@user123733: Please tell me which step you have problems with.
Dec
1
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
Assuming that $G$ has no element of order 6 is equivalent to the centralizer of a 3-Sylow having order 3. The normalizer of a 3-Sylow has order either 3 or 12 (Sylow). As $Z_3$ has only one nontrivial automorphism, we know that $G$ has four 3-Sylows and hence four elements not of order 3. These are the elements of the (therefore unique and hence normal) 2-Sylow subgroup $V$. As any 3-Sylow acts non-trivially on $V$, we can exclude $V=Z_4$ and get $V=Z_2\times Z_2$ with the elements of order 3 permuting the elements of order $2$. That's $A_4$.
Nov
29
comment Permutation group of a set
@DerekHolt: Ups, sorry. You're right, of course.