Reputation
652
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
9 15
Newest
 Yearling
Impact
~5k people reached

Jun
4
comment Classification of all finite elementary $p$-groups.
I wouldn't expect the groups to be uncomplicated just because they are semidirect products. Every extension embeds into a wreath product, which is a (special) semidirect product.
Jun
4
comment Classification of all finite elementary $p$-groups.
Any element of $G\setminus M$ generates a complement of $M$, as its order is $p$ by assumption.
Jun
3
comment Do I or don't I, have to verify the existence of the identity in a subgroup?
As closure normally refers only to "closed under multiplication", not to inverses, I guess you meant to write "inverses exist" instead of "identities exist" in the first line of your answer. But then you still miss to mention that $H$ has to be non-empty. The empty set is closed under products and taking inverse, but is not a subgroup.
Jun
1
comment Show that $D$ is a normal subgroup.
Conjugation is a group homomorphism. Try the 2nd approach.
May
30
comment Extend isomorphism of subgroups to homomorphism of groups
Even for abelian groups $G_1=G_2$ and $H_1=H_2$ the homomorphism does not have to exist: Take the direct product $C_2\times C_4$ of two cyclic groups of orders $2$ and $4$ and its subgroup isomorphic to $C_2\times C_2$ of elements whose squares are $1$. An automorphism of $C_2\times C_2$ exchanging the element of order $2$ in $C_4$ with any other element of order $2$ can not be extended.
May
28
revised Matrix and field extension
replaced the tag group-theory by the tag linear-algebra
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
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).