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4

$|G|$ refers to the order of $G$ (abelian or not). Action by left multiplication is always transitive (consequence of the so-called 'sudoku property' of groups). The conjugation action however, is not transitive for any non-trivial group $G$.


4

There is no equivalence (or 'bi-implication'), only the following implication: If $G$ is a group acting on a non-empty set $A$, then the relation $\sim$ on $A$ defined by $$a\sim b\qquad\Leftrightarrow\qquad (\exists g\in G)(a=gb),$$ is an equivalence relation. The 'iff' in your question only serves to define the relation on $A$.


3

Let $G$ acting on a set $X$ by $(g,a)\mapsto g.a$, for each $a\in X$ the stabilizer of $a$ under ( or relatively to) this action is the maximal sub-group $N$ of $G$, s.t $g.a=a, \forall g\in H$; the standard notation of this stabilizer is $St(a)$. When $G$ acts on the set of all its subgroup by conjugation the stabilizer $St(H)$ of a subgroup $H$ ...


3

A finite group of order $p$ or $p^2$ for some prime $p$ is abelian. Then the proper subgroups of groups of order $1$ up to $11$ are always abelian. For $12$, we have $S_3\times C_2$ which has a non-abelian subgroup ($S_3$).


3

Hint Let $V=\{1,(12)(34),(13)(24),(14)(23)\}$ a subgroup of $\mathfrak S_4$. This group has 4 element, and is normal in $\mathfrak S_4$. If you can prove that $$\mathfrak S_4/V\cong \mathfrak S_3,$$ then, if $$\varphi:\mathfrak S_4/V\longrightarrow \mathfrak S_3$$ is such an isomorphism, then, $\varphi\circ \pi$ is the researched homomorphism where $$\pi:\...


3

$\mathbb Z_2$ is abelian. Thus conjugation can't change the first member of an element of $\mathbb Z_2\oplus S_3$.


3

The action is non-trivial because for a fixed $p$, all Sylow-$p$ subgroups of a group $G$ are conjugate to one another. This is the second Sylow theorem. So the kernel of $G\to S^3$ is proper, since there are elements of $G$ that does a non-trivial permutation of the three Sylow-$2$ subgroups. The kernel is also non-trivial, since it's a homomorphism from a ...


3

As Derek says in a comment, it should be $|N_2|\le |G|/8$, which just barely is enough to still make the $2/3$ bound. The transfer theorem tells us that the Sylow subgroup is not in the center of its normalizer. If the order is $4$, that means the normalizer has order at least $12$. Since the number of Sylow subgroups is the index of the normalizer, and the ...


2

Because if $a\in B$, $a\ne1_Β$ and $H=\langle a\rangle=\{a^n\mid n\in\mathbb{Z}\}$, then from Lagrange's Theorem we have that $|H| \mid |B|$ so $a$ has order $p$ and so $H=G$.


2

Okay, I have found $2p+17$ groups of order $2pq^2$ when $p,q$ are primes and $2p \mid q-1$. There is just one case where I haven't been able to enumerate the number of groups. Let $G$ be a group of order $2pq^2$ as above. First, the third Sylow theorem tells us that there is a unique $q$-Sylow group of $G$, call it $Q$. Next, the Schur-Zassenhaus theorem ...


2

SMALL number theory person here. First of all, the problem doesn't really depend on you taking the finite subsets. We may as well just call $A = \{ \sum_{n\in\mathbb{N}} \delta_n a_n : \mbox{$\delta_n\in\{0,1\}$ and all but finitely many $\delta_n$ are zero}\}$ and call $R$ the set of remainders of the elements of $A$ when divided by $q$. The problem is ...


1

For the proof of the theorem, consider the action in the setting $$G\times\wp(G)\to\wp(G),$$ where $\wp(G)$ is the $G$'s power-set, defined, of course, as $g\cdot A=gAg^{-1}$, then, by the bijection ${\rm Orb}\leftrightarrow{\rm G/St}$, we have $\#{\rm Orb}(A)=[G:N_G(A)]$, since $$N_G(A)=\{g\in G\mid gAg^{-1}=A\}={\rm St}(A).$$


1

Yes, (P1) implies (P2). You can find Hall subgroups of all orders by taking intersections of the largest Hall subgroups, which (P1) assumes exist. For example, if $|G|=p^nm=q^rs$ with $p$ and $q$ distinct primes, and $H$ and $K$ are subgroups of orders $m$ and $s$, then since $|HK| \le |G|$, we get $|H \cap K| \ge m/q^r = s/p^n$. But $|H \cap K|$ is not ...


1

All subgroups of $B$ must be of an order that divides $B.$ If $B$ is of prime order, then its only subgroups are the trivial subgroups. Suppose $b$ is in $B$ and $b$ is not the identity. If B has no non-trivial subgroups, it must be the case that everything in B can be expressed as $b^k.$


1

The second Sylow theorem says that all of the Sylow $2$-subgroups are conjugate to one another. If they are $H_1,H_2$, and $H_3$, there are $g_2,g_3\in G$ such that $H_2=H_1^{g_2}$ and $H_3=H_1^{g_3}$, and of course then $H_3=H_2^{g_2^{-1}g_3}$. Thus, the action is clearly not trivial. The kernel is proper, since the action doesn’t collapse the group, and ...



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