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I am trying to grasp some moments of First Sylow's Theorem from Herstein's book:

Theorem (Sylow): If $p$ is a prime number and $p^{\alpha}\mid o(G)$, then $G$ has a subgroup of order $p^{\alpha}$.

Firstly he proves the following combinatorial fact: If $n=p^{\alpha}m$ and $\nu_p(m)=r$ i.e., $p^r\mid m$ but $p^{r+1}\nmid m$. Then $\nu_p(\binom{p^{\alpha} m }{p^{\alpha}})=r$.

Proof (by Wielandt): Let $\mathcal{M}$ be the set of all subsets of $G$ which have $p^{\alpha}$ elements. Thus $\mathcal{M}$ has $\binom{p^{\alpha} m }{p^{\alpha}}$ elements. Given $M_1, M_2\in \mathcal{M}$ define $M_1\sim M_2$ if there exists an element $g\in G$ such that $M_1=M_2g$. It is easy to verify that this defines an equivalence relation on $\mathcal{M}$. It's easy to see that there is at least one equivalence class of elements in $\mathcal{M}$ such that the number of elements in this class is not a multiple of $p^{r+1}$. Let $\{M_1, \dots, M_n\}$ be such an equivalence class in $\mathcal{M}$ where $p^{r+1}\nmid n.$ By our very definition of equivalence in $\mathcal{M}$, for any $g\in G$, for each $i=1,\dots, n$ exists $j=1,\dots, n$ such that $M_ig=M_j$.

I have understood almost everything except the last sentence. How to show that if $g\in G$ and $i\in \{1,\dots,n\}$ then $\exists j\in \{1,\dots,n\}$ such that $M_ig=M_j$.

I have tried by contradiction but it does not give result.

Would be very grateful for help.

RFZ
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2 Answers2

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How to show that if $g\in G$ and $i\in \{1,\dots,n\}$ then $\exists j\in \{1,\dots,n\}$ such that $M_ig=M_j$.

That is simply by definition of an equivalence class:

  • All elements of $\{M_1,\ldots , M_n\}$ are related to each other;
  • No elements outside of $\{M_1,\ldots , M_n\}$ are related to elements in $\{M_1,\ldots , M_n\}$.

If you pick $M_i$ in that set, and $g\in G$, by definition of $\sim$ you have $M_i\sim M_ig$. Therefore $M_ig$ has to be one of the $M_j$.

Arnaud Mortier
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  • Let me ask you the question: We take an arbitrary $g\in G$ and some $M_i$. Let's consider $M_ig$. How to show that $M_ig$ is one of elements of ${M_1, \dots, M_n}$? However, it does not look so simple to me. The element $g$ is arbitrary :( – RFZ Feb 22 '18 at 11:45
  • It is by the very definition of an equivalence class. It is like saying, let ${\ldots, -2,0,4,6,\ldots}$ be the set of all even numbers; take one of them, say $e_i$, and an arbitrary $g\in \Bbb Z$. How to prove that $e_ig$ is one of the elements of ${\ldots, -2,0,4,6,\ldots}$? -well I'm not very proud of that example but it is this simple. – Arnaud Mortier Feb 22 '18 at 11:53
  • Another way to put it, if you could find $g$ and $M_i$ such that $M_ig$ was not one of the $M_j$, then the set ${M_1,\ldots M_n}$ would not be an equivalence class. – Arnaud Mortier Feb 22 '18 at 12:26
  • After some pondering during the walking I have the following approach: Let $[M]={M_1,\dots, M_n}$. Let $g\in G$ and $M_i\in [M]$. It's easy to show that $[M_i]=[M]$. Since $M_ig\sim M_i$ then $M_ig\in [M_i]$ and also $M_ig\in [M_ig]$ so $[M_i]=[M_ig]=[M]$ then $[M_ig]=[M]$ then $M_ig\in [M]$ so $M_ig=M_j$ for some $j$. – RFZ Feb 22 '18 at 15:34
  • That's all correct! Now that you've got that, you can learn a shorter way to say it :-) – Arnaud Mortier Feb 22 '18 at 15:49
  • Hi, dear Arnaud! Could you take a look at this page, please? https://math.stackexchange.com/questions/2684164/non-abelian-groups-of-order-pq-with-p-mid-q-1/2684185#2684185 Could you provide some answer? – RFZ Mar 09 '18 at 19:08
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Suppose $\;\{M_1,..,M_n\}=[M]\;$ , meaning: for any $\;i=1,2..,n\;$ there exists $\;g_i\in G\;$ s.t $\;M_i=Mg_i\;$ , so for $\;j=1,2,..,n\;$ we get:

$$\begin{cases}M_i=Mg_i\\{}\\ M_j=Mg_j\end{cases}\;\;\implies\;\;M_i=Mg_i=\left(M_jg_j^{-1}\right)g_i=M_j\left(g_j^{-1}g_i\right)$$

Conclusion: for any $\;g\in G\;$ and for any $\;M_i\in[M]\;$, we have that also $\;M_ig\in[M]\;$, because $\;[M]=\{M_1,...,M_n\}\;$ is one (complete) equivalence class.

Arnaud Mortier
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DonAntonio
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  • I guess it should be: $M_i=Mg_i$ instead of $M_i=Mg_1$ and $g_i$ instead of $g_I$. I cannot get your conclusion. It is unclear for me ;( How it follows from your above reasoning? Could you clarify it, please? – RFZ Feb 22 '18 at 10:16