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I am trying to understand the following theorem:

Let $G$ be a finite group, $p$ a prime number, and let's suppose $|G|=p^ns$ s.t. $p$ doesn't divide $s$

Let $n_p$ be the number p-sylow subgroups of G then we have

$$\begin{cases}n_p |s\\n_p \equiv 1 \mod p\end{cases}$$

Now my problem is with $n_p$ I don't understand why $n_p$ can be anything different from 1. a p-sylow subgroup is a p-subgroup which is maximal, but I don't understand why there can be more than one maximal subgroup of a group G

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  • $\begingroup$ There can be more than one. Perhaps the best is, to consider an example. Take matrices, i.e., all Sylow $p$-subgroups of $GL_2(\mathbb{F}_p)$. We have $p+1$ Sylow-$p$ subgroups. Or write down all Sylow $2$-subgroups of $S_3$. There are $3$ of them. $\endgroup$ – Dietrich Burde Jun 16 '17 at 19:02
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    $\begingroup$ You're confusing "maximal" with "maximum". $\endgroup$ – rschwieb Jun 16 '17 at 19:09
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Maximal means "not contained in anything else" not "everything else is contained in it."

Look at $S_3$ where you have three Sylow-$2$ subgroups, $\{(1), (ij)\}$ for any $1\le i\ne j\le 3$.

Or if you prefer a simpler example of where "maximal" means this, look at a case with some sets: consider $\{\varnothing, \{1\},\{2\}\}$ with respect to inclusion both $\{1\}$ and $\{2\}$ are maximal because they are not properly contained in anything larger, even though neither contains the other.

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Consider the set of all subgroups of a given group $G$, with the relation 'being a subgroup' it is a partial order (a lattice actually) and 'directly below' G you could have several subgroups. Now consider the same for finite dimensional vector spaces, take for example $\mathbb{R}^n$, how many different subspaces does it have with dimension $(n-1)$?. In the set of all subspaces, those subspaces would be the maximal ones, then your $p$-Sylow subgroups would be those subgroups 'directly below' $G$ when considering the set of all $p$-subgroups.

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Let $G=A_5$. Then $G$ is simple.

$|G|=60=2^2.3.5$

Here $n_2 \neq 1$.For, if we had exactly one Sylow-$2$ then it would be normal.

In fact, in any simple group the Sylow-$p$ would have to be necessarily greater than $1$.

I think your problem is in understanding of what "maximal subgroup" means. "A maximal subgroup H of a group G is a proper subgroup, such that no proper subgroup K contains H strictly. In other words H is a maximal element of the partially ordered set of proper subgroups of G."

For instance in $A_5$ , $H_1 =\{(),(1 2)(34),(13)(24),(14)(23)\}$ and $H_2=\{(),(12)(35),(13)(25),(15)(23) \}$ are two distinct Sylow-$2$ subgroups.

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