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 :)

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

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 $p_1\cdots p_n$ 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 $6=2\cdot 3$.


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