Let G be a finite group of order $p^{r}m$ where $p \in \mathbb{P}, r>0$ and $p$ does not divide $m$.

One of Sylow's theorem is this:

If $n$ is the number of Sylow p-subgroups then $n$ divides $m$ and $n \equiv 1 \pmod{p}$.

I made an observation and wonder if they are true.

Since p-group/ Sylow p-subgroups are contained in G, G contains at least one p-group/Sylow p-subgroup. In general, there are m p-group/Sylow p-subgroup. Is this correct?

Secondly, by the above theorem I have stated, how does the fact that n divides m denote the index of the p-group/ Sylow p-subgroup in G?

The index of a subgroup in a group G is the distinct number of the subgroup in G. This is given by the fact that the order of the subgroup divides the order of a group G.

Greatly appreciate if anyone would illuminate my doubts.

Thanks in advance.

  • 2
    $\begingroup$ It is not the case that in general $m$ is the number of Sylow $p$-subgroups. Alas, I cannot understand your remaining questions. $\endgroup$ – Lord Shark the Unknown Jul 13 '17 at 6:46
  • $\begingroup$ However, $m$ is indeed always the index of a $p$-Sylow subgroup of $G$. This is simply because a $p$-Sylow subgroup has order $p^r$ by definition. $\endgroup$ – Sebastian Schoennenbeck Jul 13 '17 at 7:58

As to your first question, as noted in a comment, consider for instance $S_{3}$, and take $p = 3$, so $m = 2$, but $n = 1$.

As to

The index of a subgroup in a group $G$ is the distinct number of the subgroup in $G$.

you probably mean

The index of a subgroup $H$ in a group $G$ is the number of distinct left, say, cosets of the subgroup $H$ in $G$.

Note that this number is denoted by $\lvert G : H \rvert$, and it is called the index of $H$ in $G$.

The reason $n \mid m$ (which appears to be what you are asking) is that $n = \lvert G : N_{G}(P) \rvert$, where $P$ is a Sylow $p$-subgroup, and $N_{G}(P)$ is its normalizer in $G$. Now since $P \le N_{G}(P)$ we have by Lagrange $$ m = \lvert G : P \rvert = \lvert G : N_{G}(P) \rvert \cdot \lvert N_{G}(P) : P \rvert = n \cdot \lvert N_{G}(P) : P \rvert, $$ so that $n \mid m$.

  • $\begingroup$ Thank you very much. Something for me to digest. $\endgroup$ – Mathematicing Jul 13 '17 at 8:15

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