3
votes
Accepted
There is at most one prime $p$ such that if $P\in \operatorname{Syl}_p(G)$, then $N_G\left ( P \right )=P.$
As a hint, let $H$ be a maximal (proper) normal subgroup of $G$, so $|G:H|=p$ is prime. Then by the lemma that you stated, $p$ is the only prime for which a Sylow $p$-subgroup of $G$ can be self-...
- 78.2k
2
votes
Are Sylow $p$-subgroups in infinite groups conjugate?
In addition to what has been answered: if $P$ and $Q$ are arbitrary $p$-groups, then $P$ and $Q$ are Sylow $p$-subgroups of their free product $G=P*Q$. Hence, in an infinite group Sylow $p$ subgroups ...
- 42.7k
2
votes
Accepted
How to show that two groups are not isomorphic.
If $\mathbb{Z}_{75}$ is isomorphic to $\mathbb{Z}_5 \oplus \mathbb{Z}_{15}$, $\mathbb{Z}_5 \oplus \mathbb{Z}_{15}$ has a element of order $75$.
But for all $(x,y) \in \mathbb{Z}_5 \oplus \mathbb{Z}_{...
- 36
2
votes
Accepted
How to we explicitly get that #(G/H) = #G/#H from lagrange's theorem?
For each $j\in\{1,2,\ldots,n\}$, $\#g_jH=\#H$. So, since the union $g_1H\cup g_2H\cup\ldots\cup g_nH$ is a disjoint union,$$
\#G=n\times\#H=\#(G/H)\times\#H.
$$
- 267
1
vote
Accepted
$H\le G$. Number of orbits of the induced $H$-action of a transitive $G$-action: why assuming $H\unlhd G$?
Indeed, if $G\curvearrowright\Omega$ transitively then the (general version of the) orbit-stabilizer theorem says $\Omega\cong G/S$ as a $G$-set, where $S$ is the stabilizer of any element in $\Omega$ ...
- 14.6k
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