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May
28
suggested approved edit on Matrix and field extension
May
27
comment Classifying groups of order 18
@CensiLI: In one case the involution fixes an element of $C_3^2$, in the other case not. So once the center of $G$ is $C_3$ and once trivial.
May
27
comment How can you tell if a normal subgroup induces a semidirect product?
Sometimes you can find the complement as the point stabilizer for a suitable group action. Look for example at the proofs of the theorems of Schur-Zassenhaus and of Gaschuetz in the book "The Theory of Finite Groups" by Kurzweil and Stellmacher. Another example is the Frattini argument: If a group $G$ acts on a set $\Omega$ such that a normal subgroup $N$ acts transitively, then $G = G_\omega N$ for each $\omega\in\Omega$. If $N_\omega=1$, then $G_\omega$ is a complement of $N$ in $G$.
May
25
comment $PSL(2,7)$ as a subgroup of $A_7$
After establishing that there are eight 7-Sylows, you could shorten your proof by observing that they are self-centralizing in $S_7$. As Aut($C_7$)=$C_6$ you get that $|G|=8 |N_G(C_7)|$ divides $8×7×6$. With mentioning that $ab$ has order 3, you are done.
May
24
revised Is $SL(2, 3) $ a subgroup of $SL(2, p)$ for $ p>3$?
added full proof
May
23
comment For any finite group $G$ and for any natural number $n$, does there exist a group $H$ such that $|H|=n|G|$ and $G$ is a normal subgroup of $H$?
-1 Your last three questions had really similar answers. How about thinking a sec before posting?
May
20
comment How many non-isomorphic groups of order 122 are there?
You showed that there is a normal subgroup of order 61 (a prime) and another subgroup of order 2. So $G$ is a semidirect product $Z_{61}\rtimes Z_2$. The possible semidirect products depend on the chosen homomorphisms $Z_2\to Aut(Z_{61})$. How many such homomorphisms do you know?
May
12
comment A finite group which is isomorphic to $PSL(2,p)$
Is its order and the fact that the $2$-, $p$- and $q$-Sylow subgroups are not normal the only information you have about the group? It looks hard to prove the existence of $p$ involutions as a first step, since multiplying the order by $2$ one would get such groups with a unique element of order $2$.
May
12
comment if o(a) is equal to exponent of finite abelian group G then $G=<a>\times K$
You could choose $h\in G\setminus \langle a\rangle$ of minimal order (with respect to not being power of $a$), show that it has prime order and use induction by looking at $G/\langle h\rangle$.
May
5
comment $|G|=p_1p_2p_3$ distinct primes with $p_i \nmid p_j-1$ then $G$ is cyclic
If you start using Sylow for the smallest of the three primes, say $p_1$, you either get a normal $p_1$-Sylow subgroup (and finish off by factoring it out) or so many (how many?) elements of order $p_1$ that there aren't enough for more than one $p_2$- or $p_3$-Sylow group (and finish off by showing that elements of order $p_1$ have to centralize normal $p_2$- or $p_3$-Sylow groups).
May
4
comment Galois group of the extension $E:= \mathbb{Q}(i, \sqrt{2}, \sqrt{3}, \sqrt[4]{2})$
I'd guess that most people would look at the quotients $Gal(K/\mathbb{Q})$ instead of looking at $Gal(E/K)$ like you. I doubt that this would make the calculation easier, but it fits better to your goal of understanding the projective limit.
May
4
comment Galois group of the extension $E:= \mathbb{Q}(i, \sqrt{2}, \sqrt{3}, \sqrt[4]{2})$
"E is a galois extension of $\mathbb{Q}$ of dimension 16, it being a galois extension of the splitting field L" is not quite correct, as being a galois extension is not transitive. You need $3\in\mathbb{Q}$ here.
Apr
29
comment Find the conjugacy classes of $A_5$
The size of a conjugacy class equals the index of the centralizer of any element of this class. For the centralizers you get that the centralizer in $A_5$ is just the centralizer in $S_5$ intersected with $A_5$. If the centralizer was already contained in $A_5$, its index in $A_5$ is half of its index in $S_5$, so the conjugacy class splits in two (this happens for example for $(12345)$ because its centralizer is $\langle(12345)\rangle$). Otherwise, i.e. if an element of $S_5\setminus A_5$ is contained in the centralizer, the conjugacy class stays one in $A_5$.
Apr
27
comment Sylow p-subgroups and set X not divisible by p
Do you know about the transfer homomorphism?
Apr
24
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
Clearer hint for (vi): If $|N_G(P)| = 36$ take $H=N_G(P)$, otherwise take $H=C_G(P)$ which has index $2$ in $N_G(P)$ as $N_G(P)/C_G(P) = Z_2$ ($C_1$ contains a nontrivial automorphism of $P$ and by my first comment there are no others).
Apr
24
comment Proving a group is $PSL(2,q)$ with $q>3$ odd.
What's the intended connection between both theorems? Rough idea would be enough (and appreciated a lot).
Apr
23
comment Groups with “few” subgroups
To be pedantic, you will have to replace the induction on $m$ by a double induction on $n$ and $m$...
Apr
23
comment Groups with “few” subgroups
@DerekHolt: As your proof if phrased in the moment, I have problems seeing why all $SP$ have to be different (but I think you can fix that, for example by working in $N/P$): Take $G=S_4, p^r = 8, k=3$. For $i=2$ I pick $P=V_4$, so $N=G$ and $h=3$. As the $s$ (=two) different subgroups of order $3$ I take two different $3$-Sylow subgroups. But then both give the same $SP = A_4$.
Apr
23
comment Groups with “few” subgroups
@coffeemath: Yes, you are right. It should have been "at most".
Apr
23
comment Groups with “few” subgroups
A finite group $G$ of order $n$ such that the number of elements $x$ with $x^d=1$ is less than $d$ for all divisors $d$ of $n$ has to be cyclic. I wonder, if a variation of this topic can lead to a proof that there are at least as many subgroups as divisors.