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.

  1. 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?

  2. Can a pair of non-isomorphic finite simple groups be Sylow-isomorphic?

Some results:

  1. 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.

  2. 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.

  • $\begingroup$ Do you know a pair of non-isomorphic, locally isomorphic groups? $\endgroup$
    – LSpice
    Commented May 8, 2023 at 3:06
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    $\begingroup$ No, but I've edited my post to include some thoughts about finding such a pair. $\endgroup$ Commented May 8, 2023 at 4:12
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    $\begingroup$ I'm sure you know this, but the theorem that finite solvable groups have a unique conjugacy class of nilpotent self-normalizing subgroups is due to R. Carter. Also, G. Glauberman proved that if $G$ is a finite group whose Sylow subgroups are all self-normalizing is a $q$-group for some prime $q$. The latter is Corollary 12.6 in Glauberman's article in the Conference Proceedings " Finite Simple Groups", eds M.B. Powell and G. Higman, Academic Press, 1971. $\endgroup$ Commented May 8, 2023 at 13:23
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    $\begingroup$ I didn't know about that theorem of Glauberman's! Thanks, I'll check it out. $\endgroup$ Commented May 9, 2023 at 4:41
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    $\begingroup$ (cont.) Once we know this, it follows that whenever $G$ is a finite group with all its (non-trivial) Sylow subgroups self normalizing, then $G$ has a normal subgroup of prime index (clear if $|G|$ has at most two prime divisors, and if more than two prime divisors, then $G$ has a self-normalizing Sylow $q$-subgroup for some prime $q \geq 5$, and $G$ has a normal subgroup $H$ of index $q$. But now $H$ ( and also $G$) must be a $q$-group, otherwise we take a orime $r \neq q$ dividing $|H|$, and we have $G = HN_{G}(R)$ for $R \in {\rm Syl}_{r}(H)$. Then $G = HR = H$, contradiction. $\endgroup$ Commented May 9, 2023 at 8:48

2 Answers 2


To answer your question 2, there are very few pairs of finite simple groups with the same order. There's $PSL_3(4)$ and $PSL(4,2)$, and there's $P\Omega_{2n+1}(q)$ and $PSp_{2n}(q)$ for $q$ odd. None of these are Sylow isomorphic in your sense.

As a contribution to your question 1, finite groups can be not only Sylow isomorphic but they can even be indistinguishable from the point of view of fusion at any prime. This is equivalent to the $\mathbb{Z}$-completion of the classifying spaces being homotopy equivalent. Probably the smallest example is $(\mathbb{Z}/3\rtimes\mathbb{Z}/4) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/2)$ and $(\mathbb{Z}/3\rtimes\mathbb{Z}/2) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/4)$, with the action of $\mathbb{Z}/4$ having the subgroup of index two in its kernel in both cases.

  • $\begingroup$ Thanks for the answer! May I know what it means for a pair of finite groups to be "indistinguishable from the point of view of fusion" at a prime? $\endgroup$ Commented May 8, 2023 at 8:13
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    $\begingroup$ It means that the fusion systems are equivalent. See the book of Craven, "The theory of fusion systems" or the book of Aschbacher, Kessar and Oliver, "Fusion systems in algebra and topology" for the relevant notions. Roughly speaking, it means that there's an isomorphism of the Sylow subgroups that respects (elementwise) conjugacy of subgroups in the ambient group. $\endgroup$ Commented May 8, 2023 at 8:16

Let $G_1$ and $G_2$ be nonisomorphic Sylow-isomorphic groups. For example let $G_1 = C_6$ and $G_2 = S_3$. Then for any finite group $H$, the groups $G_1 \times H$ and $G_2 \times H$ are nonisomorphic Sylow-isomorphic groups (by Krull--Schmidt). So there is not much limit on the answer to question 1.

Also all groups of the same square-free order are Sylow-isomorphic.


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