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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 $i$.
Jun
3
comment Subgroups of a cyclic group and their order.
Yes it can. Any cyclic group of order $> 2$ has more than one generator.
Jun
3
answered Subgroups of a cyclic group and their order.
May
29
comment find a group with the property :
This subgroup is the direct sum of the $\mathbb{Z}_n$, and is often denoted by $\mathbb{Z}_1 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \ldots$
May
27
awarded  Popular Question
May
21
comment Schur's Lemma in Group Theory
Also, $G = \operatorname{SL}_2(F)$ is an example for any finite field $F$ other than $\mathbb{F}_2$ and $\mathbb{F}_3$. The reason is the same, the only nontrivial proper subgroup of $G$ is $Z(G) = \{I, -I\}$.
May
19
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
Just like before, $H = \oplus_{a \in A} \mathbb{Z}$, the direct sum of $|A|$ copies of $\mathbb{Z}$, ie. the free abelian group with basis $A$.
May
15
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
@AlexB.: Let $H = \oplus_{a \in A} \mathbb{Z}$, ie. the free abelian group with basis $A$. Then there exists a surjective homomorphism $H \rightarrow A$. Also, we can choose $D = \oplus_{a \in A} \mathbb{Q}$.
May
9
answered Quickest route to the structure theorem for finitely generated modules over a PID