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Jun
12
comment What do you need to perform Karatsuba multiplication?
The formula holds in any (commutative) ring. You can always choose $a=0$, but the choice of $a$ and $b$ given $x$ and $y$ reminds me of division with rest with remainder and therefore Euclidean rings.
Jun
12
comment What do you need to perform Karatsuba multiplication?
Do you want to know when this formula holds (should be in any ring) or when this formula is useful (e.g., for making multiplication faster)?
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 $A_5$.