Help Wanted With New Sequence

[David Harden]

I have worked out the first 9 terms of this sequence, and it appears to be new (they do not give a match, but it may be nevertheless too close to something else to be new). First I give the definition:

This is the set of positive integers which never arise as the order of the unit group of a (not necessarily commutative) ring with unity.

One general criterion which helps:

Theorem. Let n>2 and suppose that gcd(n, phi(n))=1 (i.e., n is in A003277). Then n is in this sequence, unless it is expressible as a product of factors each of which is 1 less than a power of 2.

Proof. Let R be a ring (with unity) with a unit group of order n. We derive a contradiction by showing R actually has a larger multiplicative group.

Claim. R has characteristic 2.
Proof of Claim. Since gcd(n, phi(n)) = 1 and n>2, n is odd. If 1 and -1 are distinct elements of R, then they would form a subgroup of order 2 in the unit group of R. This cannot happen because n is odd. Thus 1=-1 so the Claim is proven.

Since gcd(n, phi(n))=1, all groups of order n are cyclic. Thus the unit group of R must be cyclic. Then let w be an element of order n in the unit group.
R has a subring F isomorphic to Z/(2). Let p(x) be the minimal polynomial for w over F. Write p(x) = r_1(x)...r_k(x) in F[x], where each factor on the right is irreducible.
Since n is odd, using derivatives shows the polynomial x^n-1 has no repeated factors in F[x]. Then p(x), which divides x^n-1, likewise has no repeated factors. Thus the r_i are distinct.
Then F[w] is isomorphic, via a Chinese remainder theorem argument, to the Cartesian product of F[x]/r_i(x) over all i. Since each r_i is irreducible, this is a field of order 2^(deg r_i), with unit group of order 2^(deg r_i) - 1. Then the unit group of R contains a subgroup of order equal to the product of 2^r_i - 1 over all i. This contains the subgroup <w> of order n, which must be proper unless n = the product of 2^r_i - 1 over all i, as claimed.

This gives the first six terms in the sequence: 5, 11, 13, 17, 19, 23. 
25 is in the sequence because any group of order 25 contains elements of order 5 but not elements of order 3. On the other hand, an element of order 5 in the unit group yields an element of order 15 in the unit group (since x^5-1 = (x-1)(x^4 + x^3 + x^2 + x + 1) and the second factor is irreducible in (Z/(2))[x] ) by the reasoning used above.
29 and 33 are in the sequence by the above theorem.

34 is the first even number which appears to be in this sequence. But testing even numbers for membership is a little more involved than testing odd numbers, since a multiplicative group of even order could arise in a ring with characteristic > 2. A ring with characteristic n has a subring isomorphic to Z/(n) and therefore its unit group has a subgroup of order phi(n). This means that if n is the characteristic of a ring whose unit group has order 34, phi(n) divides 34. So phi(n)= 1,2,17 or 34, which means phi(n)=1 or 2 so n= 1, 2, 3, 4 or 6.
n > 1 because our ring is not the zero ring.
n is not 2 because a group of order 34 has elements of order 17, which give rise to elements of order 255 in the unit group of a ring of characteristic 2.
n is not 3 because elements of order 17 give rise to elements of order 43046720 in the unit group of a ring of characteristic 3.
n is not 6 because the projection x --> (3x,4x) factors a ring of characteristic 6 into the Cartesian product of a ring of characteristic 2 and a ring of characteristic 3, neither of which can have elements of order 17 in its unit group and still have a unit group of order 34.
So n must be 4. But let R be a ring with a unit group of order 34 and characteristic 4. Since -1 is in the center of the unit group and it has order 2, the unit group of R is cyclic. Passing from R to R/(2) yields a ring of characteristic 2. What happens to the unit group when we do this? 1 and -1 get identified, so the new unit group has order 17 or 1. 17 was already shown to be impossible. But 1 means that all elements of order 17 in the unit group of R differ by multiples of 2. Then they are all congruent to 1 mod 2 in R. This shouldn't be a hard possibility to eliminate, but I haven't done this yet.

So here are the even numbers < 100 which appear to be in the sequence: 34, 38, 50, 68, 74, 76, 86, 94.

Here are further odd numbers < 100 which can be proven, via the kind of reasoning already explained, to be in the sequence: 
35, 37, 39, 41, 43, 47, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 95, 97, 99.
Also, I am uncertain about the membership status of 75.

---- David