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Jan
2
comment Is $x \mapsto \text{Stabilizer}(x)$ an equivariant map?
The action on the conjugates of $H$ is conjugation, not left multiplication. So the claim is that $f(gx) = gf(x)g^{-1}$.
Dec
29
answered Characteristic subgroups of a direct product of groups
Dec
29
comment Maximal solvable subgroup not Borel
One example is the normalizer of a maximal torus in $\operatorname{SL}_2(\mathbb{C})$.
Dec
21
answered What does “exponent 2 nilpotency class 2” mean?
Dec
21
revised If $G/Z(G)$ is finite, then $|G'| < \infty$
added 3 characters in body
Dec
21
comment If $G/Z(G)$ is finite, then $|G'| < \infty$
@Myself: You are probably right! This answer was a bit hasty, we should take the coproduct here.
Dec
19
answered If $G/Z(G)$ is finite, then $|G'| < \infty$
Dec
16
comment Groups such that inclusion on collection of all its subgroup is a total order
All elements in such a group clearly have order power of $p$ for some fixed prime $p$. If $G$ is infinite, then you can build a chain of subgroups $1 < \langle x_1 \rangle < \langle x_2 \rangle < \cdots$ such that each $x_i$ has order $p^i$ and then $G$ must equal to the union of this chain. I think that implies that $G$ must be the Prüfer $p$-group
Dec
8
awarded  Caucus
Dec
6
comment When does $(ab)^n = a^n b^n$ imply a group is abelian?
Actually you get all the examples you need from the Heisenberg group (second point in your answer), see my answer.
Dec
6
answered When does $(ab)^n = a^n b^n$ imply a group is abelian?
Dec
6
comment When does $(ab)^n = a^n b^n$ imply a group is abelian?
@HenningMakholm: No, because it is false. Take $G$ to be a nonabelian group of exponent $3$. Then for any $x \in G$ we have $x^{-2} = x$, so $(xy)^{-2} = xy = x^{-2}y^{-2}$.
Nov
30
answered Existing of such automorphism subgroup
Nov
28
answered Suppose G is a group, p is prime , Then the number of elements of G of order p is multiple of (p-1)
Nov
26
comment If all sylow subgroups are cyclic, prove that G is solvable
Any textbook on group theory that introduces the transfer homomorphism should have a proof of this fact. There is also an alternative proof using Frobenius theorem (in a finite group $G$, the number of solutions to $x^n = 1$ is a multiple of $\gcd(n, |G|)$. Why do you want to avoid using Burnside's theorem? I think that no "elementary" proof is known.
Nov
24
answered How is number of conjugacy class related to the order of a group?
Nov
21
comment Groups of Order $n$
No such formula is known. There are some special cases where you have a formula (eg. for the number of groups of squarefree order). But currently there is no reason to expect that it will ever be easy to compute the number of groups of order $n$ (for arbitrary $n$). For example, nobody knows how many groups of order $2048$ exist.
Nov
17
comment Is this subset of group, a subgroup of it?
This question was asked before, I gave an answer here. Where did you find this problem by the way? I am interested because I've only seen it here on Math.SE.
Nov
14
answered How many $A_5$ are there inside $A_6$?
Nov
14
comment $G$ is product of its center and commutator
Any reductive linear algebraic group (over an algebraically closed field) has this property.