# If $p$ is a prime number and $\ \ p^{\alpha}|o(G)$, then $G$ has a subgroup of order $p^{\alpha}$

I know the proof of this result, but i have doubt in the proof of "I.N Herstein".

Let $G$ be a group of order n = $p^{\alpha}$m, where $p$ is a prime number and if $p^r$ | m but $p^{r+1} \nmid$ m

Let $\mathcal M$ be the set of all subset of $G$ which have $p^{\alpha}$ elements . Then $\mathcal M$ has elements $\left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right)$. Given $M_1,M_2$ $\in \mathcal M$, define $M_1 \thicksim M_2$ if there exist g $\in G$ such that $M_1 = M_2$g. This is an equivalence relation. We claim that there is at least one equivalence class of elements in $\mathcal M$ such that number of elements in this class is not multiple of $p^{r+1}$, for if $p^{r+1}$ is a divisor of the size of each equivalence class , then $p^{r+1}$ would be a divisor of of the number of elements in $\mathcal M$. Since $\mathcal M$ has $\left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right)$. and $\ \ p^{r+1} \nmid \left( \begin{array}{c} m p^{\alpha}\\ p^{\alpha} \end{array} \right)$. , this cannot be the case. Let $\{ M_1,M_2,...........,M_n\}$ be such an equivalence class in $\mathcal M$. If g$\in G$ , for each i = 1,2,.......,n, $M_ig = M_j$ for some j, $1\leq j \leq n$. We define $H = \{g \in G \ \ | \ \ M_1g = M_1 \}$. Clearly H is a subgroup of G.

I have doubt in this argument

Author claim that $\ \ no(H) = o(G) = p^{\alpha}m$. since $p^{r+1} \nmid n$ and $p^{\alpha +r}| p^{\alpha}m$ = no($H$), it must follow that $p^{\alpha}$ | o($H$) and so o($H) \geq p^{\alpha}$ . However , if $m_1 \in M_1$, then for all h $\in H$, $m_1 h \in M_1$. Thus $M_1$ has at least o($H$) diatinct elements. However $M_1$ was a subset of $G$ containing $p^{\alpha}$ elements.

Thank you.

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Can you finish writing the first sentence? – blue Mar 8 '14 at 3:10
Better to pick one of $p$ and $P$ - using both is confusing. – Thomas Andrews Mar 8 '14 at 3:15
@Thomas Andrew :,sorry for this, I have ediditin my problrm – user120386 Mar 8 '14 at 4:00

ADD I think you want to show that if $|G|=p^kw,(w,p)=1$ then $G$ contains a subgroup of order $p^j$ for $j=1,2,\ldots,j$. Then argument below needs to be fixed but slightly when one shows $\mathcal O(M_1)$ has cardinality not divisible by $p$. Suppose $|G|=p^\alpha p^rm,(m,p)=1$. Here you use the valuation of $p$ in $\displaystyle\binom{p^{\alpha +r}m}{p^\alpha}$ is $r$ to show there is an orbit $\mathcal O(M_1)$ with size not divisible by $p^{r+1}$, and continue mutatis mutandi.

Recall that a group $G$ acts on a set $S$ if there a map $G\times S\to S$ that sends a pair $(g,s)$ to a "product" $g\cdot s$ such that $1\cdot s=s$ and $g\cdot (h\cdot s)=(gh)\cdot s$. Alternatively, there is a group homomorphism $G\to {\rm Sym}(S)$.

The orbit of an element $s\in S$ is the set of all $G$-multiples of $s$, that is $\mathcal O(s)=\{g\cdot s:g\in G\}$. It can be checked that $s\sim s'\iff g\cdot s=s'$ for some $g\in G$ is an equivalence relation, so that $\mathcal O(s)$ is indeed the equivalence class of $s$.

The stabilizer of an element $s\in S$ is the set of all elements in $g$ that fix $s$, that is, $g\cdot s=s$, so ${\rm stab}(s)=\{g\in G:g\cdot s=s\}$. One can see that this is indeed a subgroup of $G$ for any set $S$ and any action.

One can see the more elements that fix $s$; the smaller the orbit of $s$ will be, and reciprocally. This is put neatly in the orbit-stabilizer theorem, which says that for any $s\in S$, the product of the cardinalities $|\mathcal O(s)||{\rm stab}(s)|$ is always constant and equal to $|G|$.

The orbit stabilizer theorem admits a rather simple proof: define a map from the orbit $\mathcal O(s)$ to the set of cosets of ${\rm stab}(s)$ in $G$ by sending $g\cdot s\mapsto g\,{\rm stab}(s)$. One must check this map is well defined, for we might have $g\cdot s=g'\cdot s$ for two possibly distinct $g,g'$. But if $g\cdot s=g'\cdot s$ we have that $g^{-1}g'$ stabilizes $s$, thus $g^{-1}g'\in {\rm stab}(s)$ and we know this means $g\,{\rm stab}(s) =g'\,{\rm stab}(s)$. It is clear the map is surjective for $g$ ranges through all elements of $G$ in $\{g\cdot s:g\in G\}=\mathcal O(s)$. Finally, this map is injective (we kinda already did): if $g\,{\rm stab}(s)=g'\,{\rm stab}(s)$ then $g^{-1}g'\cdot s=s$ so $g'\cdot s=g\cdot s$.

Since the cardinality of the coset space $G/{\rm stab}(s)$ is $|G|/|{\rm stab}(s)|$, the claim is proven. The orbit-stabilizer theorem carries a bit more information indeed: it says $\mathcal O(s)$ and the coset space $G/{\rm stab}(s)$ are equivalent $G$-sets. More generally, if $G$ acts transitively on a set $S$; then $S$ and $G/{\rm stab}(s)$ are equivalent $G$-sets for any $s\in S$.

The orbit-stabilizer theorem then has a very important corollary: since -- as we said before -- the orbits are equivalence classes of an equivalence relation on the set $S$, we can write $$\tag 1 |S|=\sum_{i=1}^n|\mathcal O(s_i)|$$ for some set or representatives $s_1,\ldots,s_n$. Using the orbit-stabilizer theorem we can write $$|S|=\sum_{i=1}^n|G:{\rm stab}(s_i)|$$

The proof you're presented with is due to Wielandt. If $\mathcal M$ is the family of all subsets of $G$ of size $p^\alpha$, then $G$ acts on $\mathcal M$ by left multiplication, it sends $M\in\mathcal M$ to $gM$ which is also a subset of size $p^\alpha$ for $x\mapsto gx$ is a bijection. Since this is actually multiplication in $G$, it is almost trivial to check this is an action.

The cardinality of $\mathcal M$ is the binomial coefficient in question $\binom{p^\alpha m}{p^\alpha}$, which one can prove is not divisible by $p$. Since we know that $|\mathcal M|$ is equal to the sum of all orbits by $(1)$ of $\mathcal M$ under $G$, there must exist an orbit, that is, an element $M_1\in\mathcal M$ and $\mathcal O(M_1)=\{gM_1:g\in G\}$ whose cardinality is not divisible by $p$ either -- if all orbits had size divisible by $p$, $(1)$ shows $|\mathcal M|$ would have size divisible by $p$; which is not the case.

The orbit-stabilizer theorem says then that $|G|=|{\rm stab}\, M_1|\,|\mathcal O(M_1)|$, so that ${\rm stab }\, M_1=\{g\in G:gM_1=M_1\}$ is a subgroup of $G$ with cardinality $p^\alpha m$ for some $m\geqslant 1$ -- since $p\nmid |\mathcal O(M_1)|$, all of $p$ in $|G|$ must go into $|{\rm stab}\, M_1|$.

The final claim is that ${\rm stab}\, M_1$ injects into $M_1$, so that indeed $m=1$. Fix $g_0\in M_1$, and consider the map ${\rm stab}\, M_1\to M_1$ that sends $h\to hg_0\in M_1$, because by definition of ${\rm stab}\, M_1$ indeed $hg_0\in M_1$. If $g,h\in {\rm stab}\, M_1$ are such that $gb_0=hb_0$ then cancelling $b_0$ gives $g=h$; so this is an injection.

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@ Pedroff Tamaroff : My internet is not working properly, so I could not not reply your answer, sorry for this. I have understood your point , but I want to ask : there exist an equivalence class of elements in $\mathcal M$ such that number of elements in this class is not multiple of $p^{r+1}$, for if $p^{r+1}$ is a divisor of the size of each equivalence class .Why Author take number of element in this equivalence class is n which is also the order of G. – user120386 Mar 12 '14 at 4:33