# Tag Info

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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