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2d
comment Why lower central series imply $N$ series?
@Zuriel: Sorry, yeah that didn't make sense. We could instead argue by induction on $i$
2d
revised Why lower central series imply $N$ series?
deleted 16 characters in body
2d
answered Why lower central series imply $N$ series?
Oct
19
awarded  Nice Answer
Oct
10
awarded  Yearling
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30
awarded  Explainer
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24
awarded  Autobiographer
Sep
3
comment Proving that a group of order $112$ is not simple
@ErginSuer: Let $H \leq S_n$. If $H$ has order $> 2$, then you can see that $H$ must contain an even permutation $\neq 1$. Thus $H \cap A_n$ is a nontrivial normal subgroup of $H$. In the case where $H$ is simple this implies $H \cap A_n = H$, ie. $H \leq A_n$.
Aug
17
revised Let $G$ be a group and $K$ be subgroup of order $p^a$. I am trying to show that $|\{H \leq G : K \subset H, |H|=p^{a+b}\}| \equiv 1 \pmod{p}$
edited body
Aug
17
comment Let $G$ be a group and $K$ be subgroup of order $p^a$. I am trying to show that $|\{H \leq G : K \subset H, |H|=p^{a+b}\}| \equiv 1 \pmod{p}$
First reduce (by induction) to the case where $b = 1$. Then in that case, the number of subgroups of order $p^{a+1}$ that contain $K$ are in bijection with the subgroups of order $p$ in $N_G(K)/K$.
Aug
7
awarded  Enlightened
Aug
7
awarded  Nice Answer
Aug
4
awarded  Pundit
Aug
4
comment Sequence of primes by concatenating digits in a given base.
@DarthGeek: At least in base 10, there is no example of such an infinite sequence. See mathworld.wolfram.com/TruncatablePrime.html
Aug
4
comment Sequence of primes by concatenating digits in a given base.
$333333331 = 17 \cdot 19607843$
Aug
2
comment Is every Banach space isometrically isomorphic to the dual of a normed space?
math.stackexchange.com/questions/242034/…
Jul
30
comment Characterization of nilpotent groups
@JoelMoreira: It might be a good idea to look up the results about Engel groups, perhaps there is something there that would give you a proof or a counterexample for the bounded class question
Jul
30
comment Characterization of nilpotent groups
@DerekHolt: The survey here mentions at page 4 that Golod has constructed a group that is a finitely generated, non-nilpotent Engel group such that every subgroup generated by two elements is finite. Since finite Engel groups are nilpotent, that should give you an example.
Jul
30
revised Groups of order $n^2$ that have no subgroup of order $n$
better title
Jul
30
comment Groups of order $n^2$ that have no subgroup of order $n$
@DerekHolt: Thanks for saving me the trouble!