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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.


2h
revised If $G/Z(G)$ is of size $qp$ and $p-1$ is not divisible by $q$ then $G/Z(G)$ is cyclic?
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2h
answered If $G/Z(G)$ is of size $qp$ and $p-1$ is not divisible by $q$ then $G/Z(G)$ is cyclic?
1d
comment Let $G = \langle a \rangle$ be a cyclic group of order $n$. Show that for every divisor $d$ of $n$, there exists a subgroup $G$ whose order is $d$.
No, no, that is not necessary. I put $e$ as the order of $a^{\frac{n}{d}}$, so $e$ is by definition the smallest natural number such that $(a^{\frac{n}{d}})^e=1$. You already showed that $(a^{\frac{n}{d}})^d=1$, and this implies $e \mid d$ (can you see why?). Conversely, $(a^{\frac{n}{d}})^e=1$, but since $n$ is the order of $a$, we get $n \mid \frac{ne}{d}$, say $\frac{ne}{d}=fn$. This implies $e=df$, whence $d \mid e$. But we already had $e \mid d$, so $d=e$ is the order of $a^{\frac{n}{d}}$. Hope this clarifies it.
1d
comment Let $G = \langle a \rangle$ be a cyclic group of order $n$. Show that for every divisor $d$ of $n$, there exists a subgroup $G$ whose order is $d$.
$e$ is here just a natural number - not 2.7182818 ... :-)). In this case the order of $a^{\frac{n}{d}}$. You have to check if $d|e$ and $e|d$, whence $e=d$. In group theory the identity element is denoted as $1$ (that is what I use mostly), $e$, $1_G$.
1d
comment Let $G = \langle a \rangle$ be a cyclic group of order $n$. Show that for every divisor $d$ of $n$, there exists a subgroup $G$ whose order is $d$.
Yes, but you also have to show that if $(a^{\frac{n}{d}})^e=1$, then $d \mid e$.
1d
answered Let $G = \langle a \rangle$ be a cyclic group of order $n$. Show that for every divisor $d$ of $n$, there exists a subgroup $G$ whose order is $d$.
Jan
26
comment Volume of Revolution Verification
That is great! You can show only once the infinitesimal delta summation approach for the Riemann integral for any function and then let them use the formula freely. A nice example is $f(x)=\sqrt{1-x^2}$, defined on $[-1,1]$ and gives you the volume of a sphere of radius $1$.
Jan
26
answered Volume of Revolution Verification
Jan
26
answered Can every group be extended to ring with idenity
Jan
23
comment Number of solutions to an equation
Working over a field different from $\mathbb{C}$ the situation could be totally different.
Jan
23
revised Number of solutions to an equation
added 4 characters in body
Jan
23
answered On special normal subgroup of a group
Jan
22
comment Why is the derivative changed during trigonometric substitution?
Please provide an example - by the way $df(x)=f'(x)dx$ and that is what you are using in substitutions.
Jan
18
answered Example where a finite group $G$ of order $n$ has no subgroup of order $m$
Jan
16
comment How to determine if the subset $K=\{ g\in S_4|2^g=2\}$ is a subgroup of $S_4$?
The identity element of course or $\begin{pmatrix}1&2&3&4\\4&2&1&3\end{pmatrix}$. And $K$ is actually a subgroup $S_4$, which is more than a set without an algebraic structure. Maybe you can show now that $K \cong S_3$.
Jan
10
comment Is a normal subgroup of G, normal in any subgroups of G containing it?
Yes, in general, if $N \unlhd G$, then $N \cap H \unlhd H$ for any subgroup $H$ of $G$. Try to prove it, it is easy. Your case would be $N \subseteq H$, so $N \cap H=N$.
Jan
10
comment Is a normal subgroup of G, normal in any subgroups of G containing it?
If $G' \subseteq K$, then automatically $G' \unlhd K$. For $G' \unlhd G$ and hence $G' \cap H \unlhd H$ for any subgroup $H$ of $G$.
Jan
10
comment Subgroup of the quotient group of the commutator G'.
OK Meitar, but each subgroup of $G/G'$ can be represented as $H/G'$, where $G' \subseteq H \leq G$.
Jan
10
answered Subgroup of the quotient group of the commutator G'.
Jan
8
comment Two groups that are the automorphism groups of each other
Wielandt [1939] proved the classical result that the automorphism tower of any centerless finite group terminates in finitely many steps. Rae and Roseblade [1970] proved that the automorphism tower of any centerless Cernikov group terminates in finitely many steps. Hulse [1970] proved that the the automorphism tower of any centerless polycyclic group terminates in countably many steps. Simon Thomas [1985] proved that the automorphism tower of any centerless group eventually terminates. Hamkins proved that every group has a terminating transfinite automorphism tower. Centerless or not.