$G_1, G_2$ finite groups, for all primes $p$, Sylow $p$-subgroups of $G_1$ and $G_2$ are isomorpic and...












0












$begingroup$


Decide whether the following staement is true of false. If true, prove it. If false, provide a counterexample



Let G1, G2 be a finite groups such as for all prime p, p-sylow subgroups of G1 isomorpic ($cong$) to p-sylow subgroups of G2 and |G1|=|G2| then G1$cong$G2.



I think this statement is false but I didn't find counterexample yet, I think it's false because I thought about dividing the group to p-sylow subgroups or write her has a direct product of them and then do a uoion or direct product of the isomorphisms but there is no diviton og G1 and G2 to her p-sylow so I couldn't prove it so I tried to find counterexample but I didn't find one.

If you found one please help me :)






finite-groups sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
    $endgroup$
    – Brian Tung
    Jul 27 '18 at 5:06






  • 5




    $begingroup$
    Think about groups of order $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jul 27 '18 at 6:01






  • 1




    $begingroup$
    Terribly non informative title.
    $endgroup$
    – Did
    Jul 27 '18 at 6:18










  • $begingroup$
    I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
    $endgroup$
    – user579852
    Jul 31 '18 at 11:56










  • $begingroup$
    You don't need to assume that $|G_1| = |G_2|$.
    $endgroup$
    – the_fox
    Nov 29 '18 at 10:47
















0












$begingroup$


Decide whether the following staement is true of false. If true, prove it. If false, provide a counterexample



Let G1, G2 be a finite groups such as for all prime p, p-sylow subgroups of G1 isomorpic ($cong$) to p-sylow subgroups of G2 and |G1|=|G2| then G1$cong$G2.



I think this statement is false but I didn't find counterexample yet, I think it's false because I thought about dividing the group to p-sylow subgroups or write her has a direct product of them and then do a uoion or direct product of the isomorphisms but there is no diviton og G1 and G2 to her p-sylow so I couldn't prove it so I tried to find counterexample but I didn't find one.

If you found one please help me :)






finite-groups sylow-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
    $endgroup$
    – Brian Tung
    Jul 27 '18 at 5:06






  • 5




    $begingroup$
    Think about groups of order $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jul 27 '18 at 6:01






  • 1




    $begingroup$
    Terribly non informative title.
    $endgroup$
    – Did
    Jul 27 '18 at 6:18










  • $begingroup$
    I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
    $endgroup$
    – user579852
    Jul 31 '18 at 11:56










  • $begingroup$
    You don't need to assume that $|G_1| = |G_2|$.
    $endgroup$
    – the_fox
    Nov 29 '18 at 10:47














0












0








0





$begingroup$


Decide whether the following staement is true of false. If true, prove it. If false, provide a counterexample



Let G1, G2 be a finite groups such as for all prime p, p-sylow subgroups of G1 isomorpic ($cong$) to p-sylow subgroups of G2 and |G1|=|G2| then G1$cong$G2.



I think this statement is false but I didn't find counterexample yet, I think it's false because I thought about dividing the group to p-sylow subgroups or write her has a direct product of them and then do a uoion or direct product of the isomorphisms but there is no diviton og G1 and G2 to her p-sylow so I couldn't prove it so I tried to find counterexample but I didn't find one.

If you found one please help me :)






finite-groups sylow-theory







share|cite|improve this question











$endgroup$




Decide whether the following staement is true of false. If true, prove it. If false, provide a counterexample



Let G1, G2 be a finite groups such as for all prime p, p-sylow subgroups of G1 isomorpic ($cong$) to p-sylow subgroups of G2 and |G1|=|G2| then G1$cong$G2.



I think this statement is false but I didn't find counterexample yet, I think it's false because I thought about dividing the group to p-sylow subgroups or write her has a direct product of them and then do a uoion or direct product of the isomorphisms but there is no diviton og G1 and G2 to her p-sylow so I couldn't prove it so I tried to find counterexample but I didn't find one.

If you found one please help me :)






finite-groups sylow-theory




finite-groups sylow-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 10:05









Nicky Hekster

28.4k53456




28.4k53456










asked Jul 27 '18 at 4:56









user579852user579852

335




335












  • $begingroup$
    It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
    $endgroup$
    – Brian Tung
    Jul 27 '18 at 5:06






  • 5




    $begingroup$
    Think about groups of order $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jul 27 '18 at 6:01






  • 1




    $begingroup$
    Terribly non informative title.
    $endgroup$
    – Did
    Jul 27 '18 at 6:18










  • $begingroup$
    I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
    $endgroup$
    – user579852
    Jul 31 '18 at 11:56










  • $begingroup$
    You don't need to assume that $|G_1| = |G_2|$.
    $endgroup$
    – the_fox
    Nov 29 '18 at 10:47


















  • $begingroup$
    It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
    $endgroup$
    – Brian Tung
    Jul 27 '18 at 5:06






  • 5




    $begingroup$
    Think about groups of order $6$.
    $endgroup$
    – Lord Shark the Unknown
    Jul 27 '18 at 6:01






  • 1




    $begingroup$
    Terribly non informative title.
    $endgroup$
    – Did
    Jul 27 '18 at 6:18










  • $begingroup$
    I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
    $endgroup$
    – user579852
    Jul 31 '18 at 11:56










  • $begingroup$
    You don't need to assume that $|G_1| = |G_2|$.
    $endgroup$
    – the_fox
    Nov 29 '18 at 10:47
















$begingroup$
It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
$endgroup$
– Brian Tung
Jul 27 '18 at 5:06




$begingroup$
It might help (and it would probably go some way to avoiding close votes) if you could articulate why you think the statement is false.
$endgroup$
– Brian Tung
Jul 27 '18 at 5:06




5




5




$begingroup$
Think about groups of order $6$.
$endgroup$
– Lord Shark the Unknown
Jul 27 '18 at 6:01




$begingroup$
Think about groups of order $6$.
$endgroup$
– Lord Shark the Unknown
Jul 27 '18 at 6:01




1




1




$begingroup$
Terribly non informative title.
$endgroup$
– Did
Jul 27 '18 at 6:18




$begingroup$
Terribly non informative title.
$endgroup$
– Did
Jul 27 '18 at 6:18












$begingroup$
I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
$endgroup$
– user579852
Jul 31 '18 at 11:56




$begingroup$
I thought about what you said but how I can prove that every p-sylow subgroup of $G_1$ isomorpic to p-sylow subgroup of $G_2$??
$endgroup$
– user579852
Jul 31 '18 at 11:56












$begingroup$
You don't need to assume that $|G_1| = |G_2|$.
$endgroup$
– the_fox
Nov 29 '18 at 10:47




$begingroup$
You don't need to assume that $|G_1| = |G_2|$.
$endgroup$
– the_fox
Nov 29 '18 at 10:47










1 Answer
1






active

oldest

votes


















0












$begingroup$

You are right, the statement is false.



Let $p_1,ldots, p_n$ be distinct primes and $G_1, G_2$ be any two groups of order $p_1cdots p_n$ each. It follows that a Sylow $p$-subgroup (for $p$ one of $p_1,ldots,p_n$) of each one of them is of order $p$, meaning it is $mathbb{Z}_p$ up to isomorphism. So any Sylow $p$-subgroup of $G_1$ is isomorphic to any Sylow $p$-subgroup of $G_2$ and vice versa.



At this point all you have to do is find two such groups that are not isomorphic. The easiest choice is $G_1=mathbb{Z}_6$ and $G_2=S_3$ both of order $6=2cdot 3$.






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    $begingroup$

    You are right, the statement is false.



    Let $p_1,ldots, p_n$ be distinct primes and $G_1, G_2$ be any two groups of order $p_1cdots p_n$ each. It follows that a Sylow $p$-subgroup (for $p$ one of $p_1,ldots,p_n$) of each one of them is of order $p$, meaning it is $mathbb{Z}_p$ up to isomorphism. So any Sylow $p$-subgroup of $G_1$ is isomorphic to any Sylow $p$-subgroup of $G_2$ and vice versa.



    At this point all you have to do is find two such groups that are not isomorphic. The easiest choice is $G_1=mathbb{Z}_6$ and $G_2=S_3$ both of order $6=2cdot 3$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      You are right, the statement is false.



      Let $p_1,ldots, p_n$ be distinct primes and $G_1, G_2$ be any two groups of order $p_1cdots p_n$ each. It follows that a Sylow $p$-subgroup (for $p$ one of $p_1,ldots,p_n$) of each one of them is of order $p$, meaning it is $mathbb{Z}_p$ up to isomorphism. So any Sylow $p$-subgroup of $G_1$ is isomorphic to any Sylow $p$-subgroup of $G_2$ and vice versa.



      At this point all you have to do is find two such groups that are not isomorphic. The easiest choice is $G_1=mathbb{Z}_6$ and $G_2=S_3$ both of order $6=2cdot 3$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        You are right, the statement is false.



        Let $p_1,ldots, p_n$ be distinct primes and $G_1, G_2$ be any two groups of order $p_1cdots p_n$ each. It follows that a Sylow $p$-subgroup (for $p$ one of $p_1,ldots,p_n$) of each one of them is of order $p$, meaning it is $mathbb{Z}_p$ up to isomorphism. So any Sylow $p$-subgroup of $G_1$ is isomorphic to any Sylow $p$-subgroup of $G_2$ and vice versa.



        At this point all you have to do is find two such groups that are not isomorphic. The easiest choice is $G_1=mathbb{Z}_6$ and $G_2=S_3$ both of order $6=2cdot 3$.






        share|cite|improve this answer











        $endgroup$



        You are right, the statement is false.



        Let $p_1,ldots, p_n$ be distinct primes and $G_1, G_2$ be any two groups of order $p_1cdots p_n$ each. It follows that a Sylow $p$-subgroup (for $p$ one of $p_1,ldots,p_n$) of each one of them is of order $p$, meaning it is $mathbb{Z}_p$ up to isomorphism. So any Sylow $p$-subgroup of $G_1$ is isomorphic to any Sylow $p$-subgroup of $G_2$ and vice versa.



        At this point all you have to do is find two such groups that are not isomorphic. The easiest choice is $G_1=mathbb{Z}_6$ and $G_2=S_3$ both of order $6=2cdot 3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 9:25

























        answered Nov 29 '18 at 14:49









        freakishfreakish

        11.8k1629




        11.8k1629






























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