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seen Sep 24 at 10:54

Sep
16
comment Automorphisms inducing automorphisms of quotient groups
@Swlabr: If $N$ contains a non-central element $n \in N\setminus \operatorname{Z}(G)$, then the non-trivial inner automorphism induced by $n$ maps to the trivial automorphism on $G/N$. Hence your map is not one-to-one. Maybe it's more natural to look at the map $\operatorname{Aut}(G) \to \operatorname{Aut}(G/N)\times\operatorname{Aut}(N)$ defined in the obvious way instead?
Sep
15
comment Automorphisms inducing automorphisms of quotient groups
@Swlabr: (about your case of a semi-direct product) Do you know any groups except $1$ and $Z_2$ with trivial automorphism group?
Sep
15
comment Maximal Subgroups and order of a group
@D B Lim: (his comment got deleted and replaced during I posted my comment) Every proper subgroup $X$ of $G$ is contained in a maximal subgroup. $H$ is the only maximal subgroup, so $X \le H$. Now take $X = \langle x \rangle$.
Sep
15
comment Maximal Subgroups and order of a group
@D B Lim: Your problem (reducing to the cyclic case).
Sep
15
comment Maximal Subgroups and order of a group
The solution is embarrassing simple.
Aug
13
comment Why does the number of orbits of the stabilizer of an element in a transitive group not depend on the element?
$G$ acts component-wise on $E×E$. As $G$ is transitive, the diagonal $\Delta$ is an orbit. The number of orbits of $G$ on $E×E$ equals the number of orbits of $S_x$ on $E$. I think that as the same is obviously true for $y$, the authors of the books considered the equality you are looking for to be trivial.
Aug
9
comment Group of order $3^a\cdot5\cdot11$ has a normal Sylow 3-group
@Steve: Instead of counting you could repeat the first part of your proof with $N$ and $11$.
Aug
4
comment If $|H|$ and $[G:K]$ are relatively prime, then $H \leq K$
@Geoff: You could also add a hint and a spoiler warning at the beginning of your (nice) answer and undelete it.
Aug
4
comment If $|H|$ and $[G:K]$ are relatively prime, then $H \leq K$
dls wrote "A hint would be perfect.". This is slightly more than a hint...
Aug
4
comment If $|H|$ and $[G:K]$ are relatively prime, then $H \leq K$
For a proof without using Sylow's theorem: take the order information from the isomorphism $KH/H \cong K/(K\cap H)$.
Jul
28
comment Does this class of cipher have a name? What weaknesses does it have?
I wouldn't feel comfortable encrypting two plaintexts with the same $g$.
Jul
25
comment Teach me a simple, efficient division algorithm
Take the long division algorithm as given in Knuth: The Art of Computer Programming, Vol. 2. In the 3rd edition this is Algorithm D in section 4.3.1 on page 272. Googling for the terms knuth "algorithm d" site:books.google.com I got as top link "Hacker's delight" where a C-version of that algorithm is given.
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?