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Feb
25
comment Does $H/N$ isomorphic to $K/N$ imply $H$ conjugates with $K$?
And $G$ is additionally abelian (again of order $8$).
Feb
13
comment $G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate
Do you assume that there exists a maximal subgroup?
Jan
23
comment Show that over finite fields of characteristic two the finite group $C_3$ has no nontrivial representation
Do you know how the multiplicative group of the field with four elements looks like?
Nov
24
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
@Amin: Yes, for odd $n$ the group $C_n$ anyway doesn't contain many possible images for an element of order $2$.
Nov
23
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
@Amin: For $n=3$ one has $D_6 = S_3$ according to Rescy_'s convention.
Nov
20
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
@Amin: You seem to assume the convention that $D_{2n}$ is the symmetry group of the regular $(2n)$-gon, whereas Rescy_ uses the other convention that $D_{2n}$ has order $2n$ (see the definition of $r$ in 2.).
Nov
20
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
For all homomorphisms $\phi : D_{2n}\to C_n$ you know that $\phi(r^2) = 1$. Homomorphisms are uniquely determined by the images of a generating set. In your case $r$ and $s$ generate $D_{2n}$, so you have to find their possible images in $C_n$ and check which possible combinations $(\phi(r), \phi(s))$ can come from (=can be extended to) a group homomorphism. Observe that $\phi(r^2) = 1$ implies that $\phi(r)$ has order $1$ or $2$. (What do you know about $\langle r^2\rangle$ if $r$ has odd order?)
Nov
19
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
The commutator subgroup of $G$ (usually denoted $G'$ like in Amin's answer) is the subgroup $\langle x^{-1}y^{-1}xy | x, y\in G\rangle$ generated by all commutators $[x, y] := x^{-1}y^{-1}xy$ in $G$. Your equation $\phi(ab) = \dots = \phi(ba)$ can be rewritten as $\phi(a^{-1}b^{-1}ab) = 1$, so you were on the right track. Next you should try to find all commutators in $D_{2n}$.
Nov
18
comment Number of homomorphisms from $D_{2n}$ to $C_n$?
You always have the trivial homomorphism between groups (mapping everything to the neutral element). For homomorphisms to an abelian group, it is helpful to know the "commutator subgroup" of the other group. Do you know what a commutator is?
Nov
2
comment If $K$ is a normal subgroup of $G$, is it true that $G$ is isomorphic to the direct product $K \times (G / K)$?
Is it true for $G=Z_4$ and $K=Z_2$? ($Z_n$ = cyclic group with $n$ elements)
Nov
2
comment About centralizers of involutions in finite simple groups
The nice thing about involutions is that any two of them generate a dihedral subgroup. If its order is $2n$ with $n$ odd, then both involutions are conjugated to each other (in the dihedral subgroup, hence also in $G$). If $n$ is even, the center of the dihedral subgroup is $Z_2$, therefore it contains an involution whose centralizer contains both involutions. In both cases you got some nontrivial information for free. (For an application for infinite groups see my answer to Is there a group which has precisely all finite groups as subgroups?.)
Oct
15
comment Prove that a proper subgroup never equals its normalizer.
Do you know that $|G/H| = 0 \bmod p$? (reason: $H$ is a proper subgroup of a $p$-group.) What can you conclude then from knowing $|F| = 0\bmod p$ and $|F|\ge 1$?
Oct
13
comment Prove that a proper subgroup never equals its normalizer.
You need to show that $|F| > |\{H\}| = 1$, not $|F| > |H|$, as $F$ is a set of cosets ($H$ being one of the cosets).
Oct
6
comment If $G/Z(G)$ is cyclic then $G$ is abelian – what is the point?
An application of your application is mathoverflow.net/questions/211159/…
Sep
29
comment Special case of Schur-Zassenhaus theorem
@MikkoKorhonen: If you look at the elegant proofs of Schur-Zassenhaus for an abelian normal subgroup and of the basic properties of the transfer homomorphism given in the book "The Theory of Finite Groups" by Kurzweil and Stellmacher, you'll see how closely related both are (as Derek already commented).
Sep
21
comment Bruhat Decomposition and Iwasawa Decomposition for Finite Groups?
For the Bruhat decomposition you need a BN-pair, which in turn gives you a Tits building. Their classification for rank greater 2 show that you are dealing with a group of Lie type, and not a general group. If you are interested in generalizing theory from Lie (type) groups, take a look at "Subgroup complexes" by Peter Webb in Proceedings of Symposia in Pure Mathematics 47, pp 349-365, from 1987.
Sep
14
comment If $G=G_1\times G_2$ does it follow that $H\leq G \implies H=H_1\times H_2$ where $H_1\leq G_1$ and $H_2\leq G_2$
With different letters the same question was asked here.
Sep
11
comment Intuition of Prime decomposition in Galois extensions of subgroups of a cyclotomic polynomial
Let $\omega = \frac{1+\sqrt{3}}{2}$ be a 3rd root of unity, so $\mathbb{Q}(\omega)$ will be one of your subfields. Over $\mathbb{Z}$ you get $\mathbb{Z}[\omega]/(p) = \mathbb{Z}[X]/(p, 1+X+X^2) = \mathbb{F}_p[X]/(1+X+X^2)$, and therefore $p$ is a prime element of $\mathbb{Z}[\omega]$ iff $1+X+X^2$ is irreducible over $\mathbb{F}_p[X]$. Is this really the case for (let's say) $p=7$?
Sep
10
comment left regular representation of $Q_8$
What do you mean with "Can any one please explain this solution?"? Where does the solution come from and which steps do you not understand?
Sep
10
comment left regular representation of $Q_8$
Two hints: (1) You have to identify the symmetric group on the set $Q_8$ with $S_8$ (which works since $Q_8$ has $8$ elements). (2) If you know two elements $x, y$ of $Q_8$ that generate $Q_8$ (i.e., $\langle x, y\rangle = Q_8$, then their images under your $f$ will generate a subgroup of $S_8$ isomorphic to $Q_8$.