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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.
Apr
22
comment Does A5 have a subgroup of order 6?
If you are able to calculate the normalizer of $\langle(123)\rangle$, then you have just to check if it contains an element of order $2$.
Apr
22
comment Matrix Derivations-Research
I guess the first sentence makes quite some people stop reading your question (the length is also intimidating). "$DT_3(R)$ is the upper triangular matrix with the diagonal being the same element" cannot be correct, as "the" indicates that there is only one. It should be "a". But then reading on, one gets to know that $DT_3(R)$ is meant to be the set of all $3\times3$ matrices whose diagonal elements are all the same. By the way, is $R$ any ring or the ring/field $\mathbb R$ of the real numbers?
Apr
21
comment subsets of $\mathbb{Z}_2^{p}$ up to permutation equivalence
@ThomasAndrews: I read "fixed permutation of the places" as permutation matrices, not all permutations.
Apr
20
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
Hint for (x): Assuming that Jacobson calls "effective action" what otherwise is known as "faithful action", you can try to show that the kernel $K$ of this action is a proper subgroup of $N_G(V)$, that is centralized by an element of order $5$ (look at $Aut(K))$, so cannot contain elements of order $2$. Then lead $K=Z_3$ normal in $G$ to a contradiction.
Apr
20
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
For (vi) you can use $N_G(P)/C_G(P)\le Aut(P) = Z_2$.