Orders of ${\rm PSL}(n,q)$ for different $(n,q)$

Question

Every undergraduate or beginner-graduate knows that the family projective linear groups ${\rm PSL}(n,q)$ over finite field of order $q=p^k$ is a family of simple groups (with $(n,q)\neq (2,2)$ and $(n,q)\neq (2,3)$).

It is easy to compute orders of these groups, and these groups are "different" (non-isomorphic) from $A_n$, with finitely many exceptions.

The problems I will consider here is about the orders of these groups;

for $(n,q)\neq (n_1,q_1)$ does the groups ${\rm PSL}(n,q)$ and ${\rm PSL}(n_1,q_1)$ have different orders with finitely many exceptions? Proof?

Note that for different values of $(n,q)$, it is easy to prove that the corresponding groups are non-isomorphic (with one or two exceptions), whereas I am considering here only orders of them.

After comments by many: some modification is done in question.

I think this MO question: http://mathoverflow.net/questions/107620/non-isomorphic-finite-simple-groups will probably answer all your questions. — John Gowers, Sep 30 '15 at 12:38
@Donkey: Thanks for the reference. The link answers that in lower dimensions, some equality of orders holds. But, what about other orders? Are they all distinct orders? — Groups, Sep 30 '15 at 12:43
The link asserts that the examples given in lower dimensions are the only examples of nonisomorphic finite simple groups of the same order. — John Gowers, Sep 30 '15 at 12:44
Since the groups $PSL(n,q)$ considered here are finite simple groups, this Question is a subset of the broader one, For which numbers there is only one simple group of that order?. — hardmath, Sep 30 '15 at 14:11

score 5 · Accepted Answer · edited Jun 12 '20 at 10:38

5

$\text{PSL}(3,4)$ and $\text{PSL}(4,2)\cong A_8$, though non-isomorphic, both have order $20160$. This is the only example if we disregard the isomorphisms $\text{PSL}(2,4)\cong\text{PSL}(2,5)$ and $\text{PSL}(2,7)\cong\text{PSL}(3,2)$

Quoting from https://mathoverflow.net/a/107660/70015:

the only nonisomorphic finite simple groups with the same orders are

$A_8 \cong {\rm PSL}(4,2)$ and ${\rm PSL}(3,4)$ of order 20160.

The groups ${\rm P \Omega}_{2n+1}(q)$ and ${\rm PSp}_{2n}(q)$ for all odd prime powers $q$ and $n \ge 3$. These have order

$$(q^{n^2} \Pi_{i=1}^n (q^{2i}-1))/2$$

Since $\text{PSL}(n,q)$ is a finite simple group for all $n,p$

edited Jun 12 '20 at 10:38

Community

1

answered Sep 30 '15 at 12:33

John Gowers

25,678

I mentioned it ...with one or two exceptions. – Groups Sep 30 '15 at 12:35
@Groups OK...is the second example I just added any better? – John Gowers Sep 30 '15 at 12:36
I don't understand what "no" means here. The Question asks when the groups $PSL(n,q)$ have different orders. – hardmath Sep 30 '15 at 12:36
that's OK. I would also like to see whether these are exceptional cases. – Groups Sep 30 '15 at 12:36
@hardmath The question in yellow in the OP is a yes-or-no question, which I interpret to mean, "Does $(n, q) \neq (n', q')$ imply that $#PSL(n, q) \neq #PSL(n', q')$?" I have to say, I don't quite understand OP's statement after this question, however. – Travis Willse Sep 30 '15 at 12:39
@Donkey, hardmath, Travis: to avoid wrong confusions, shall I slightly modify question? Donkey, you may keep your answer as it is. – Groups Sep 30 '15 at 12:46
Yes please.$;;;;$ – John Gowers Sep 30 '15 at 12:49
Thanks all. Sorry for inconvenience. – Groups Sep 30 '15 at 12:53

Orders of ${\rm PSL}(n,q)$ for different $(n,q)$

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