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Aug
4
awarded  Revival
Aug
2
awarded  group-theory
Aug
1
answered Normal products and radicals in finite groups
Aug
1
comment No group of order 400 is simple
@Prism: Here $\langle P, Q \rangle$ contains the set $PQ$ which has order $5^3 = 125$. So $\langle P, Q \rangle$ would be too small in those cases.
Jul
31
answered What is the standard notation to represent the set of primes?
Jul
30
comment A group $G$ with $G'$ abelian and every abelian normal subgroup finite
@MarshalKurosh: In this case $G_p$ itself is an abelian, normal, infinite subgroup of $G_p$. So the assumptions of the statement are not satisfied.
Jul
29
comment A group $G$ with $G'$ abelian and every abelian normal subgroup finite
I don't understand why you need the stuff with $Z(G)$. To me it seems that your proof works when you just let $A$ be any abelian normal subgroup of $G$ containing $G'$.
Jul
28
comment A set which satisfies all conditions for a Group except associativity
A loop is not required to have two-sided inverses.
Jul
28
comment A set which satisfies all conditions for a Group except associativity
Yet $0 - x = x$ if and only if $x = 0$
Jul
28
comment A set which satisfies all conditions for a Group except associativity
But what is the identity?
Jul
28
comment Subgroups of a cyclic group and their order.
@user1729: Suppose that $H$ is a subgroup of order $d$, where $d \mid n$. Now $H = \langle a^k \rangle$ for some $k \mid n$. Then $a^k$ has order $n/k$, but on the other hand $a^k$ must also have order $d$. Hence $k = n/d$.
Jul
26
awarded  Guru
Jul
25
awarded  Good Answer
Jul
25
awarded  Mortarboard
Jul
25
awarded  Nice Answer
Jul
25
answered Prove that $\sqrt 2 + \sqrt 3$ is irrational
Jul
23
answered Is $G$ a semidirect product of $Z(G)$ and $\operatorname{Inn}(G)$?
Jul
19
answered Normal subgroup of direct product of two groups
Jul
16
comment Is a finite group determined by the family of all its 2-generated subgroups?
For the $Sub_1$ case, consider two distinct $p$-groups of same order, of exponent $p$ (with $p$ odd). For example, let $G_1$ be the elementary abelian group of order $p^3$ and $G_2$ the Heisenberg group of order $p^3$. So if $Sub_1(G_1)$ and $Sub_1(G_2)$ are isomorphic, it does not necessarily follow that $G_1$ and $G_2$ are isomorphic.
Jul
11
answered If $G$ is a finite group of order $n$, why is it isomorphic to its centralizer in $S_n$?