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Jul
8
comment Permutation of cosets
As $\gamma$ fixes all subgroups, for every element $g$ one gets that $\gamma(g)$ generates $\langle g\rangle$.
Jul
2
comment Rotman's exercise 2.8 “$S_n$ cannot be imbedded in $A_{n+1}$”
From your comment I conclude that you realized in meanwhile that $A_4$ has an elementary abelian subgroup of order $4$, i.e., two commuting elements of order $2$.
Jul
1
comment Are there groups of order $p^4q^2$ which are not semi-direct product?
verret mentioned in a comment to a recent answer by Derek Holt to a similar question the existence of a group of order $144 = 2^4\cdot 3^2$ that cannot be written as semi-direct product of any of its proper subgroups.
Jun
30
comment Covering groups
@lattice: An easier example of a "double cover" is the quaternion group $Q_8$ with $8$ elements. It is a double cover of the elementary abelian group $V_4 = Z_2\times Z_2$ with $4$ elements. If $Q_8$ acts on something, then in general its quotient group $V_4 = Q_8/Z(Q_8)$ doesn't necessarily act on it, as $V_4$ is not a subgroup of $Q_8$.
Jun
29
comment On cyclic decomposition of element in $S_n$
The cyclic decomposition of $x$ of prime order $p$ has to be $1^{n-p}p^1$ (This follows from @whacka 's formula "order = lcm(cycle lengths)").
Jun
29
comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative
On second thought, taking a maximal abelian subgroup won't work. Instead assuming $G$ finite, I'd reduce to the case $G$ being a $p$-group. $(g, h)\mapsto [g, h]$ induces a non-degenerate skew-symmetric bilinear map $G/Z\times G/Z \to G'$. I'd induct on the order of $G/Z$ by finding a matching element for an element of $G/Z$ of maximal order.
Jun
28
comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative
A $p$-group $P$ is called special if its center $Z(P)$, derived subgroup $P'$ and Frattini subgroup $\Phi(P)$ are all equal. If this subgroup is additionally cyclic, then $P$ is called extraspecial. As there are extraspecial groups of order $p^n$ for all odd n and primes $p$, you won't be able to finish your proof with $H = G'$. Try looking at a maximal abelian subgroups $A$ of $G$ instead, and see if $H = A/Z(G)$ works.
Jun
27
comment A problem on order of a Group.
@AlexM: It also follows from $Hg = G\setminus H = gH$.
Jun
24
comment How to compute the automorphism group of split metacyclic groups?
(continuation of last comment) So $G$ is generated by $x$ and $y$, and $\phi$ centralizes $x$ and maps $y$ to a conjugate $y^z$ for some $z\in Z_p$ (as $Z_k$ is abelian). Hence $\phi$ is conjugation by $z$, an element of $Z_p$, proving the claim. To get $AGL(1, p)\le Aut(G)$ you probably know already that it follows from $G$ being normal in it with trivial centralizer.
Jun
24
comment How to compute the automorphism group of split metacyclic groups?
I think you can show that the kernel of the homomorphism $Aut(G) \to Aut(Z_p)$ (that exists since $Z_p$ is characteristic in $G$) is $Z_p$: If $\phi\in Aut(G)$ centralizes $Z_p$, i.e., $\phi(x)=x$ for a generator of $Z_p$, then take a generator $y$ of $Z_k$, which acts by multiplication on $Z_p$, let's say $x^y = x^a$ for some $a\in \mathbb{N}$. Now $x^a = \phi(x^a) = \phi(x^y) = \phi(x)^{\phi(y)} = x^{\phi(y)}$, so $\phi(y) = x^b\cdot y$ for some $b\in \mathbb{N}$. But all elements of the form $x^b\cdot y\in G$ are conjugated by an element of $Z_p$. (to be continued)
Jun
24
comment On Thompson conjecture
What do you mean with "$N(G) = N(A_n)$"? Is $G$ a group that happens to have the same sizes of conjugacy classes as $A_n$ (with or without multiplicity?)?
Jun
23
comment What is subgroup lattice of GL$(n,\mathbb F_q)$?
What do you need the subgroup lattice for?
Jun
23
comment How can you tell if a normal subgroup induces a semidirect product?
An example using the Frattini argument you can find in this answer by Derek Holt.
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.