Every Sylow subgroup is normal, then $G$ has a subgroup of order $m$ for every division $m$ of $|G|$












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Some trouble working out an algebra problem.



Suppose that every Sylow subgroup of a finite group $G$ is normal. Prove that $G$ has a subgroup of order $m$ for every divisor $m$ of $|G|$.






abstract-algebra group-theory sylow-theory







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    Title and question are two different things.
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    – Randall
    Dec 9 '18 at 5:00
















0












$begingroup$


Some trouble working out an algebra problem.



Suppose that every Sylow subgroup of a finite group $G$ is normal. Prove that $G$ has a subgroup of order $m$ for every divisor $m$ of $|G|$.






abstract-algebra group-theory sylow-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Title and question are two different things.
    $endgroup$
    – Randall
    Dec 9 '18 at 5:00














0












0








0


1



$begingroup$


Some trouble working out an algebra problem.



Suppose that every Sylow subgroup of a finite group $G$ is normal. Prove that $G$ has a subgroup of order $m$ for every divisor $m$ of $|G|$.






abstract-algebra group-theory sylow-theory







share|cite|improve this question











$endgroup$




Some trouble working out an algebra problem.



Suppose that every Sylow subgroup of a finite group $G$ is normal. Prove that $G$ has a subgroup of order $m$ for every divisor $m$ of $|G|$.






abstract-algebra group-theory sylow-theory




abstract-algebra group-theory sylow-theory






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edited Dec 9 '18 at 20:34









Cosmin

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1,4291527










asked Dec 9 '18 at 4:20









WaltWalt

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374114








  • 2




    $begingroup$
    Title and question are two different things.
    $endgroup$
    – Randall
    Dec 9 '18 at 5:00














  • 2




    $begingroup$
    Title and question are two different things.
    $endgroup$
    – Randall
    Dec 9 '18 at 5:00








2




2




$begingroup$
Title and question are two different things.
$endgroup$
– Randall
Dec 9 '18 at 5:00




$begingroup$
Title and question are two different things.
$endgroup$
– Randall
Dec 9 '18 at 5:00










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Let $|G| = n in mathbb{N}$ and let $n = p_1^{k_1} cdot p_2^{k_2} cdots p_r^{k_r}$ be the prime factor decomposition of $n$, where $r in mathbb{N}, k_i in mathbb{N}$ and $p_i$ is prime $forall i in {1,2,3,cdots,r }$.



We know that if for every $i in {1,2,3,cdots,r }$ the $p_i$-Sylow subgroups of $G$ are normal, then there is a unique $p_i$-Sylow subgroup $S_i$ (because the $p_i$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 times S_2 times cdots times S_r to G,$$ with $f( (x_1, x_2, cdots, x_r)) = x_1x_2cdots x_r.$



Now, we also know that every $p$-group of order $p^t, t in mathbb{N}$ has a subgroup of order $p^k$ for each $k in {0,1,2,3,cdots,t }$ (A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$).



Now let $m$ be a divisior of $n$. Then, $m = p_1^{i_1} cdot p_2^{i_2} cdots p_r^{i_r}$, where $i_s in {0,1,2,cdots, k_s }, forall s in {1,2,3,cdots,r }$.



Now, we have that the $p_1$-Sylow subgroup has a subgroup of order $p_1^{i_1}$, the $p_2$-Sylow subgroup has a subgroup of order $p_2^{i_2}$ and so on, every $p_u$-Sylow subgroup of $G, u in {1,2,3,cdots,r }$ has a subgroup of order $p_u^{i_u}$. The direct product from the isomorphism transfers this subgroup of $S_1 times S_2 times cdots times S_r$ to a subgroup of $G$.






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    $begingroup$

    Let $|G| = n in mathbb{N}$ and let $n = p_1^{k_1} cdot p_2^{k_2} cdots p_r^{k_r}$ be the prime factor decomposition of $n$, where $r in mathbb{N}, k_i in mathbb{N}$ and $p_i$ is prime $forall i in {1,2,3,cdots,r }$.



    We know that if for every $i in {1,2,3,cdots,r }$ the $p_i$-Sylow subgroups of $G$ are normal, then there is a unique $p_i$-Sylow subgroup $S_i$ (because the $p_i$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 times S_2 times cdots times S_r to G,$$ with $f( (x_1, x_2, cdots, x_r)) = x_1x_2cdots x_r.$



    Now, we also know that every $p$-group of order $p^t, t in mathbb{N}$ has a subgroup of order $p^k$ for each $k in {0,1,2,3,cdots,t }$ (A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$).



    Now let $m$ be a divisior of $n$. Then, $m = p_1^{i_1} cdot p_2^{i_2} cdots p_r^{i_r}$, where $i_s in {0,1,2,cdots, k_s }, forall s in {1,2,3,cdots,r }$.



    Now, we have that the $p_1$-Sylow subgroup has a subgroup of order $p_1^{i_1}$, the $p_2$-Sylow subgroup has a subgroup of order $p_2^{i_2}$ and so on, every $p_u$-Sylow subgroup of $G, u in {1,2,3,cdots,r }$ has a subgroup of order $p_u^{i_u}$. The direct product from the isomorphism transfers this subgroup of $S_1 times S_2 times cdots times S_r$ to a subgroup of $G$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $|G| = n in mathbb{N}$ and let $n = p_1^{k_1} cdot p_2^{k_2} cdots p_r^{k_r}$ be the prime factor decomposition of $n$, where $r in mathbb{N}, k_i in mathbb{N}$ and $p_i$ is prime $forall i in {1,2,3,cdots,r }$.



      We know that if for every $i in {1,2,3,cdots,r }$ the $p_i$-Sylow subgroups of $G$ are normal, then there is a unique $p_i$-Sylow subgroup $S_i$ (because the $p_i$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 times S_2 times cdots times S_r to G,$$ with $f( (x_1, x_2, cdots, x_r)) = x_1x_2cdots x_r.$



      Now, we also know that every $p$-group of order $p^t, t in mathbb{N}$ has a subgroup of order $p^k$ for each $k in {0,1,2,3,cdots,t }$ (A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$).



      Now let $m$ be a divisior of $n$. Then, $m = p_1^{i_1} cdot p_2^{i_2} cdots p_r^{i_r}$, where $i_s in {0,1,2,cdots, k_s }, forall s in {1,2,3,cdots,r }$.



      Now, we have that the $p_1$-Sylow subgroup has a subgroup of order $p_1^{i_1}$, the $p_2$-Sylow subgroup has a subgroup of order $p_2^{i_2}$ and so on, every $p_u$-Sylow subgroup of $G, u in {1,2,3,cdots,r }$ has a subgroup of order $p_u^{i_u}$. The direct product from the isomorphism transfers this subgroup of $S_1 times S_2 times cdots times S_r$ to a subgroup of $G$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $|G| = n in mathbb{N}$ and let $n = p_1^{k_1} cdot p_2^{k_2} cdots p_r^{k_r}$ be the prime factor decomposition of $n$, where $r in mathbb{N}, k_i in mathbb{N}$ and $p_i$ is prime $forall i in {1,2,3,cdots,r }$.



        We know that if for every $i in {1,2,3,cdots,r }$ the $p_i$-Sylow subgroups of $G$ are normal, then there is a unique $p_i$-Sylow subgroup $S_i$ (because the $p_i$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 times S_2 times cdots times S_r to G,$$ with $f( (x_1, x_2, cdots, x_r)) = x_1x_2cdots x_r.$



        Now, we also know that every $p$-group of order $p^t, t in mathbb{N}$ has a subgroup of order $p^k$ for each $k in {0,1,2,3,cdots,t }$ (A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$).



        Now let $m$ be a divisior of $n$. Then, $m = p_1^{i_1} cdot p_2^{i_2} cdots p_r^{i_r}$, where $i_s in {0,1,2,cdots, k_s }, forall s in {1,2,3,cdots,r }$.



        Now, we have that the $p_1$-Sylow subgroup has a subgroup of order $p_1^{i_1}$, the $p_2$-Sylow subgroup has a subgroup of order $p_2^{i_2}$ and so on, every $p_u$-Sylow subgroup of $G, u in {1,2,3,cdots,r }$ has a subgroup of order $p_u^{i_u}$. The direct product from the isomorphism transfers this subgroup of $S_1 times S_2 times cdots times S_r$ to a subgroup of $G$.






        share|cite|improve this answer











        $endgroup$



        Let $|G| = n in mathbb{N}$ and let $n = p_1^{k_1} cdot p_2^{k_2} cdots p_r^{k_r}$ be the prime factor decomposition of $n$, where $r in mathbb{N}, k_i in mathbb{N}$ and $p_i$ is prime $forall i in {1,2,3,cdots,r }$.



        We know that if for every $i in {1,2,3,cdots,r }$ the $p_i$-Sylow subgroups of $G$ are normal, then there is a unique $p_i$-Sylow subgroup $S_i$ (because the $p_i$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 times S_2 times cdots times S_r to G,$$ with $f( (x_1, x_2, cdots, x_r)) = x_1x_2cdots x_r.$



        Now, we also know that every $p$-group of order $p^t, t in mathbb{N}$ has a subgroup of order $p^k$ for each $k in {0,1,2,3,cdots,t }$ (A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$).



        Now let $m$ be a divisior of $n$. Then, $m = p_1^{i_1} cdot p_2^{i_2} cdots p_r^{i_r}$, where $i_s in {0,1,2,cdots, k_s }, forall s in {1,2,3,cdots,r }$.



        Now, we have that the $p_1$-Sylow subgroup has a subgroup of order $p_1^{i_1}$, the $p_2$-Sylow subgroup has a subgroup of order $p_2^{i_2}$ and so on, every $p_u$-Sylow subgroup of $G, u in {1,2,3,cdots,r }$ has a subgroup of order $p_u^{i_u}$. The direct product from the isomorphism transfers this subgroup of $S_1 times S_2 times cdots times S_r$ to a subgroup of $G$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 19:46

























        answered Dec 9 '18 at 19:39









        CosminCosmin

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        1,4291527






























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