$G_1, G_2$ finite groups, for all primes $p$, Sylow $p$-subgroups of $G_1$ and $G_2$ are isomorpic and...
$begingroup$
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
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
$endgroup$
add a comment |
$begingroup$
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
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
$endgroup$
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
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
edited Nov 29 '18 at 10:05
Nicky Hekster
28.4k53456
28.4k53456
asked Jul 27 '18 at 4:56
user579852user579852
335
asked Jul 27 '18 at 4:56
user579852user579852
335
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $6=2cdot 3$.
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
$endgroup$
add a comment |
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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 $6=2cdot 3$.
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
$endgroup$
add a comment |
$begingroup$
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 $6=2cdot 3$.
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
$endgroup$
add a comment |
$begingroup$
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 $6=2cdot 3$.
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
$endgroup$
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 $6=2cdot 3$.
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
edited Nov 30 '18 at 9:25
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
answered Nov 29 '18 at 14:49
freakishfreakish
11.8k1629
11.8k1629
add a comment |
add a comment |
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$begingroup$
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