# Must two non-isomorphic finite groups have a non-isomorphic Sylow $p$-subgroup?

I'm working through a problem in Rotman which asks for an example of two groups $G\not\cong H$ such that $G$ and $H$ have the same number of elements of order d, for every d.

(Rotman's question has been discussed on MSE, here, for example.)

Intuitively, it seems like I should be able to narrow down my search to groups of order $p^k$ by arguing that if $G$ and $H$ have the above property and $|G| = |H| = p_1^{k_1}\cdots p_n^{k_n}$, then $G$ and $H$ have, in particular, a non-isomorphic Sylow $p_j$-group.

The problem is, I don't know if this is true. Any proof ideas or counter-examples?

• If you narrow your search down to groups of order $p^k$, you will never find a counterexample, as the groups are then their own $p$-Sylow subgroups and therefore isomorphic by assumption. – j.p. Oct 2 '17 at 7:07

## 1 Answer

Compare the two groups of order 6.

• It seems this is the case for any $G$ and $H$ of order $pq$ where $p$ and $q$ are prime because groups of prime order are cyclic. – Peter Kagey Oct 1 '17 at 18:19
• @PeterKagey: There is a difference between $pq=6$ and $pq=15$. In both cases all Sylow subgroups are cyclic, hence $G$ and $H$ have isomorphic Sylow subgroups for all primes, but for $pq=6$ there exist non-isomorphic groups of order $6$, whereas there exists up to isomorphisms only one group of order $15$. So Ted couldn't have written 15 instead of 6. – j.p. Oct 2 '17 at 6:55
• @PeterKagey: If you look at groups of order $2^2\cdot 3$ and $3^2\cdot 5$ you will get the following: for groups of order $12$ there are non-isomorphic groups that have all $p$-Sylow subgroups isomorphic, whereas groups of order $45$ are isomorphic if and only if their $p$-Sylow subgroups are isomorphic for all primes $p$. – j.p. Oct 2 '17 at 6:59
• @PeterKagey: If both groups $G$ and $H$ are abelian (of the same order), then they are isomorphic if and only if their $p$-Sylow subgroups are isomorphic for all primes $p$. So if all groups of the given order are abelian, then you get the implication "all $p$-Sylow subgroup isomorphic" $\implies$ "groups isomorphic", otherwise you can find counterexamples to this statement for the given order. – j.p. Oct 2 '17 at 7:04