# $G_1, G_2$ finite groups, for all primes $p$, Sylow $p$-subgroups of $G_1$ and $G_2$ are isomorpic and $|G_1|=|G_2|$ then $G_1 \cong G_2$.

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.

• 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. – Brian Tung Jul 27 '18 at 5:06
• Think about groups of order $6$. – Lord Shark the Unknown Jul 27 '18 at 6:01
• Terribly non informative title. – Did Jul 27 '18 at 6:18
• 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 $G_2$?? – user579852 Jul 31 '18 at 11:56
• You don't need to assume that $|G_1| = |G_2|$. – the_fox Nov 29 '18 at 10:47

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 $$6=2\cdot 3$$.