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Aug
4
comment How many $p$-elements does $G$ have?
@JackSchmidt: I'm interested! I spent a lot of time with this and other things related to solutions of $x^n = 1$.. are you going to post an answer or should I send you an email?
Aug
4
comment How many $p$-elements does $G$ have?
I have also asked a related question here
Aug
4
answered How many $p$-elements does $G$ have?
Aug
4
comment References about the exponent of automorphism groups of finite groups.
For general $G$ one result which might be useful is the fact that any automorphism of $G$ has order at most $|G| - 1$. A proof can be found in "Finite Group Theory" by Isaacs.
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