How many groups are there with 2023 elements?

There’s obviously at least one: *Z*_{2023}, the integers mod 2023.

Now 2023 = 7 × 289 = 7 × 17 × 17 and so we could also look at

*Z*_{7} + *Z*_{17} + *Z*_{17}

where + denotes direct sum. An element of this group has the form (*a*, *b*, *c*) and the sum

(*a*, *b*, *c*) + (*a*′, *b*′, *c*′)

is defined by

((*a* + *a*)′ mod 7, (*b* + *b*′) mod 17, (*c* + *c*)′ mod 17).

Is this a different group than *Z*_{2023}? Are there any other groups of order 2023?

Let’s first restrict our attention to Abelian groups. The classification theorem for finite Abelian groups tells us that there are two Abelian groups of order 2023:

*Z*_{7} + *Z*_{289}

and

*Z*_{7} + *Z*_{17} + *Z*_{17}

But what about *Z*_{2023}? There’s a theorem [1] that says

*Z*_{mn} ≅ *Z*_{m} + *Z*_{n}

if and only if *m* and *n* are relatively prime. Since 7 and 289 are relatively prime, t

*Z*_{2023} ≅ *Z*_{7} + *Z*_{289}.

The theorem also says that *Z*_{17} + *Z*_{17} is not isomorphic to *Z*_{289} and it follows that their direct sums with *Z*_{7} are not isomorphic.

So we’ve demonstrated two non-isomorphic Abelian groups of order 2023, and a classification theorem says these are the only Abelian groups. There are no non-Abelian groups of order 2023, though that’s harder to show, and so we’ve found all the Abelian groups with 2023 elements.

## More group theory posts

[1] Sketch of proof. Let *d* be the greatest common divisor of *m* and *n*. If *d* > 1 then every element of *Z*_{m} + *Z*_{n} has order *mn*/*d* < *mn* and so *Z*_{m} + *Z*_{n} if cannot be isomorphic to *Z*_{mn}. On the other hand, if *d* = 1, then *Z*_{m} + *Z*_{n} has an element of order *mn* and so is cyclic.