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I am a former assistant-professor of mathematics at the Univ. of Amsterdam, with specialization in group theory and representation theory of groups. I work for IBM, though not in research.


2d
comment Question on order of elements in groups (subgroups)
In 2. your $H$ is not a subgroup, it should be for example $H=\{e,a,b, ab\}$, since it must be closed under group multiplication. With $ab=ba$. Then $H$ has order 4 and you are back in business.
Oct
20
comment Part of simple proof of nontrivial center in p-group
@Math83 - no problem, you learned something I guess!
Oct
20
comment Part of simple proof of nontrivial center in p-group
The [index=number of cosets] of a subgroup is a divisor of the order of the whole group!
Oct
20
comment Covering finite groups by unions of proper subgroups
Nice reference of Garonzi!
Oct
20
answered Covering finite groups by unions of proper subgroups
Oct
18
awarded  Popular Question
Oct
17
comment Let $a$ be an element of order $n$ in a group $G$. If $a^m$ has order $n$, then $m$ and $n$ are relatively prime.
BTW In general $ord(a^m)=\frac {ord(a)}{gcd(m,ord(a))}$
Oct
17
comment Let $a$ be an element of order $n$ in a group $G$. If $a^m$ has order $n$, then $m$ and $n$ are relatively prime.
Group elements can only have powers of an integer, not a rational number in general. If $g \in G$, then $g^\frac 34$ makes no sense. There is a class of groups, called divisible groups where one tries to mimic something like this.
Oct
17
comment Let $a$ be an element of order $n$ in a group $G$. If $a^m$ has order $n$, then $m$ and $n$ are relatively prime.
Almost correct - you cannot caluclate $(a^{mn})^\frac 1q$. Better write $(a^m)^{n'}=(a^m)^\frac nq= (a^n)^\frac mq= e^\frac mq=e$.
Oct
17
comment Cosets of $H$ in the group $S_4$
Maybe you should start with writing out the cosets $(14)H$ and $H(14)$ and see what happens.
Oct
17
answered If $G=G_0\geq \ldots\geq G_n=\{1\}$ then $\displaystyle|G|=\prod_{i=0}^{n-1} |G_i/G_{i+1}|$?
Oct
15
answered Combinatorial group theory books
Oct
14
answered Product of finite order elements in a group
Oct
14
comment Order of groups and elements
$H$ is actually a subgroup?
Oct
13
comment Definition of “one element normalizes another element”
Yes, this is not the right formulation on the Wiki page - it should be $f$ normalizes $\lambda[G]$, so the image of $G$ in $\text{Sym}(G)$.
Oct
13
comment Definition of “one element normalizes another element”
Yes, commutes would be appropriate - if $a$ and $b$ commute, then $a$ normalizes, and even centralizes $\langle b \rangle$.
Oct
10
revised Isomorphism of factor groups
edited body
Oct
10
answered Isomorphism of factor groups
Oct
8
revised $G$ infinite abelian group with $[G:H]$ finite for every non trivial subgroup $H$ , to prove $G$ is cyclic
edited body
Oct
8
comment Prove that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$
No problem, you are welcome!