Reputation
13,998
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
1 23 69
Newest
 Nice Answer
Impact
~137k people reached

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 $S_n$?
Jul
4
comment Probability that two randomly chosen permutations will generate $S_n$.
For references, see A001691 and A071605 from OEIS. To me it seems that Q2 is hard, at least if you want exact values. I think you can figure out bounds from the many papers that study the probability of generating $S_n$ with a pair of random permutations. For a recent result in this direction and some history about the problem, the paper "A. Maroti and C. M. Tamburini, Bounds for the probability of generating the symmetric and alternating groups, Arch. Math. (Basel), 96 (2011), 115-121" looked interesting.
Jun
18
comment Are Clifford groups very *non-commutative*?
Do you mean $ab = gba$ so $g = [a,b]$?
Jun
15
accepted If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$?
Jun
15
comment Intersection of the $p$-sylow and $q$-sylow subgroups of group $G$
Because $P, P_1, P_2, \ldots, P_n$ are distinct, you have $P \cap P_i = 1$. Note that $(A \times B) \cap (C \times D) = A \cap C \times B \cap D$ and similarly for cartesian products with more than two factors.
Jun
15
comment Intersection of the $p$-sylow and $q$-sylow subgroups of group $G$
Yes, whenever $p < q$ are primes and $q \equiv 1 \mod{p}$, there exists a nonabelian group of order $pq$. It can be constructed as a semidirect product, see for example this question.
Jun
15
comment If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$?
Anyway, this seems like a pretty cool theorem. So for easy examples, just take $G$ and $H$ with no common direct factor such that $Z(G) = Z(H) = 1$ and $(|G|, |H|) > 1$. For example $G = S_3$ and $H = D_{10}$ works, in this case $G \times H$ has order $60$, perhaps this could be the smallest example.
Jun
15
comment If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$?
If you look at the statement of the theorem, we need to assume that $G$ and $H$ have no common direct factor. For example, consider $G = H = S_3$. In this case $\operatorname{Aut}(G \times H)$ has order $72$ and $\operatorname{Aut}(G) \times \operatorname{Aut}(H)$ has order $36$. However, $Z(G) = Z(H) = 1$ so $|\hom(G,Z(H))| = |\hom(H,Z(G))| = 1$. Thus the formula does not hold in this case.
Jun
15
asked If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$?
Jun
15
answered Intersection of the $p$-sylow and $q$-sylow subgroups of group $G$
Jun
14
answered Some group theory interpretion problem
Jun
12
answered A subgroup containing a kernel of a group homomorphism into an abelian group is a normal subgroup.
Jun
4
comment If all Sylow subgroups are normal then the group is solvable
I think your hint is a bit misleading. It is possible that $H_i \trianglelefteq G$ and $G = H_1 H_2 \ldots H_k$ and $H_i \cap H_j = 1$ for $i \neq j$, yet $G$ is not isomorphic to $H_1 \times H_2 \times \ldots \times H_k$. For more than two normal subgroups $H_i \trianglelefteq G$ with $G = H_1 H_2 \ldots H_k$, we have that $G$ is a direct product of the $H_i$ when $H_i \cap H_1 H_2 \ldots H_{i-1} H_{i+1} \ldots H_k = 1$ for all $i$.