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Jul
15
comment G a finite group, M a maximal subgroup; M abelian implies G solvable?
@7115763: You have to apply Burnside's normal $p$-complement theorem to each $p$-Sylow of $M$ (which are also $p$-Sylows of $G$ as $M$ is a Hall subgroup).
Jun
15
comment
@Isaac: Wouldn't be the upvotes-to-given-answers ratio a better measure than the upvotes-to-rep ratio? The latter punishes people giving few but brilliant answers. Is the upvotes-to-given-answers ratio somewhere available?
Jun
10
comment Abelian subgroups of p-groups
@Rahul: Did you read the second paragraph of Derek Holt's answer?
Jun
9
comment Abelian subgroups of p-groups
You have to read "Burnside's classic theorem" differently: Given a $p$-group of order $p^n$ there exists a normal abelian subgroup of order $p^m$ with $n\le m(m-1)/2$. You wrongly read it as "Given a $p$-group of order $p^n$ there exists for all $m$ with $n\le m(m-1)/2$ a normal abelian subgroup of order $p^m$.", which is clearly absurd as you could choose $m>n$.
Jun
4
comment is every subgroup of a semi-direct product of groups a semi-direct product of subgroups?
For case i) do you really want $G = \mathrm{GL}(n, K)$ or did you intend to write $Q = \mathrm{GL}(n, K)$?
May
23
comment The cycle structure of the permutation $a \mapsto ma \bmod{n}$
Did you try to solve your problem in the case $n$ is a prime? Then look at the case $n$ a prime power, and finally apply the Chinese Remainder Theorem.
May
6
comment RSA: Creating a key of desired length
Choose the highest two bits of both numbers to be 11.
Apr
14
comment Conjugacy Classes of subgroups in GL(n,p)
@Alex: I'm very curious about this. How does the order help considering that there is also a cyclic group of the same order?
Apr
14
comment A proof of Sylow theorem
Sylow 3 (taking the numbering of the wikipedia) is a corollary of Sylow 1+2 by taking the action of the $G$ by conjugation on the set of $p$-Sylow subgroups.
Apr
14
comment Conjugacy Classes of subgroups in GL(n,p)
@Alex: Should (s)he compute the order of $\mathrm{GL}(n, p)$ just to get to know the group, or does the order of the group help in any way for the question?
Apr
14
comment Conjugacy Classes of subgroups in GL(n,p)
Take a look at the possible subspaces of $\mathbb{F}_p^n$ fixed by an element of order $p$ in $\mathrm{GL}(n, p)$. Do they all have the same dimension?
Mar
20
comment Non-Standard Deviation
Yes, there is something better. Yuval's suggestion wasn't just about avoiding the absolute value.
Feb
16
comment Permutation group proofs
For (2), as all elements of $S_n$ have finite order, $H_n$ is a subgroup of $S_n$ if and only if it is (non-empty and) closed under multiplication. Did you already try multiplying some elements of odd order to see if their product has odd order again? Just play around multiplying various $3$-cycles of $S_5$ as a start.
Jan
26
comment Showing $H$ is a normal subgroup of $G$
Does $H$ contain a unique $A$?
Jan
26
comment Showing $H$ is a normal subgroup of $G$
Should the map satisfy $\varphi(0) = A$? Maybe another equality helps you more...
Jan
26
comment Showing $H$ is a normal subgroup of $G$
Given a continuous map $\varphi:[0,1] \to G$, what do you know about the map $\varphi^A : x \mapsto A\cdot x\cdot A^{-1}$?
Jan
26
comment Showing $H$ is a normal subgroup of $G$
Given $A_i \in H$ and $\varphi_i$ like in the definition of $H$ for $i=1, 2$, what do you know about $\varphi = \varphi_1 \cdot \varphi_2$ where $"\cdot"$ denotes pointwise multiplication of matrices?
Jan
3
comment Dividing in $\mathbb{Z}_m[x]/p(x)$
As you're calculating modulo $3$, you have $2 = -1$.
Dec
30
comment Simplicity of $A_n$
Could you maybe elaborate more how you would prove the simplicity of $A_n$ using your ideas? I don't see how to do it, but would be very interested in it.
Dec
20
comment A finite length module is the direct sum of the image and kernel of a projection-like endomorphism
@Green Iden: You are welcome. After the deadline for your homework assignment, you could post your solution as answer.