The broad theme that underlies this question is: to what extent can the study of finite groups be reduced to the study of $p$-groups?

I imagine that it is possible for a pair of nonisomorphic finite groups $G$ and $H$ to have isomorphic Sylow subgroups. That is to say, for every prime $p$, $G$ and $H$ have isomorphic Sylow $p$-subgroups. In particular, $G$ and $H$ must have the same number of elements.

Let us call $G$ and $H$ **Sylow-isomorphic** if they have isomorphic Sylow subgroups.

For a given positive integer $n$ (that we may obviously assume

*isn't*a prime power), how many Sylow-isomorphic finite groups of order $n$ can there be?Can a pair of non-isomorphic finite simple groups be Sylow-isomorphic?

**Some results:**

Every finite nilpotent group is isomorphic to the external direct product of its Sylow subgroups, so a pair of finite nilpotent groups are Sylow-isomorphic if and only if they are actually isomorphic.

Let $C_n$ denote the cyclic group of order $n$. If $p>2$ is an

**odd**prime, $\text{Aut}(C_p)\cong (\mathbb{Z}/p\mathbb{Z})^\times\cong C_{p-1}$. We know that $C_p\rtimes C_{p-1}$ and $C_p\times C_{p-1}$ are non-isomorphic (the former is non-abelian while the latter is abelian), but clearly they are Sylow isomorphic. This shows that it is possible for a pair of non-isomorphic solvable groups to be Sylow-isomorphic.

**Additional question:**

I've heard that it is often useful to study the **normalizers** of $p$-groups in finite groups (this method is sometimes called **local analysis**, and normalizers of $p$-groups are called $p$-local subgroups).

Since the Sylow $p$-subgroups of a finite group $G$ are conjugate, their normalizers must also be conjugate, so we can unambiguously talk about the isomorphism type of a Sylow $p$-normalizer of a finite group. Suppose $G$ and $H$ have isomorphic Sylow $p$-normalizers for every prime $p$. This also means, by the way, that $G$ and $H$ must be Sylow-isomorphic. In this situation, let's call $G$ and $H$ **locally isomorphic**. Then what can we say about how $G$ and $H$ are related?

**Edit:** on finding a pair of non-isomorphic, but locally isomorphic groups

If $G$ has a normal Sylow $p$-subgroup for any prime $p$, then any group that is locally isomorphic to $G$ must be isomorphic to $G$, so we must restrict our attention to groups without any normal Sylow subgroups. This means we must ignore supersolvable groups altogether, since if $G$ is supersolvable and $p$ is the largest prime divisor of $|G|$, then a Sylow $p$-subgroup of $G$ is normal.

Also, one might consider a group in which the notion of "locally-isomorphic" coincides with "Sylow-isomorphic": that is, groups in which every Sylow subgroup is self-normalizing. Such a group (assuming it isn't of prime power order) cannot be solvable, because of the following theorem:

**Theorem:** (R. Carter) Let $G$ be a finite solvable group. Then $G$ has a self-normalizing nilpotent subgroup, and all such subgroups are conjugate (and thus isomorphic)

So if $G$ is solvable, then there is at most **one** prime $p$ such that a Sylow $p$-subgroup of $G$ is self-normalizing.

1more comment