Central Units of the Integral Group Ring ℤA_{5}

Yuanlin Li and M.M. Parmenter

Abstract

There are very few cases known of nonabelian groups G where the group of central units of ℤG, denoted Z(U(ℤG)), is nontrivial and where the structure of Z(U(ℤG)), including a complete set of generators, has been determined. In this note, we show that the central units of augmentation 1 in the integral group ring ℤA_{5} form an infinite cyclic group ⟨u⟩, and we explicitly find the generator u.

1 Introduction

Let ℤG denote the integral group ring of a group G, U(ℤG) the group of units of such a group ring and V (ℤG) the subgroup of units of augmentation 1. We will use the term trivial units to describe the subgroup ±G of U(ℤG).

For any group it is possible to define the important families of bicyclic and Bass cyclic units in V (ℤG) (see [6] for the definitions). It turns out that for many finite groups, these families, taken together, generate a subgroup of finite index in V (ℤG) ([4], [6]). The problem of finding the full structure of V (ℤG), including a complete set of generators, seems to be more difficult and has been settled for only a small number of special cases (see [6] for an excellent survey).

Even less is known about Z(U(ℤG)), the group of central units of ℤG. When G is finite, Ritter and Sehgal ([3], [6]) proved the following theorem giving necessary and sufficient conditions for Z(U(ℤG)) to be trivial.

Theorem 1 Let G be a finite group. All central units of ℤG are trivial if and only if for every x ϵ G and every natural number j relatively prime to |G|, x^{ȷ} is conjugate to x or x^{−1}.

The second author was supported in part by NSERC grant A8775.

1991 Mathematics Subject Classification. Primary 16U60, 20C05.

Also, when G is finite Ritter and Sehgal [5] constructed a finite set of generators for a subgroup of finite index in Z(U(ℤG)), while Jespers, Parmenter and Sehgal [1] found a different set of generators which works for finitely generated nilpotent groups (and some others as well). In the latter case, the generators were constructed from Bass cyclic units in ℤG and the construction depended on the existence of a very well behaved finite normal series in G. In general, however, there is no simple procedure known for constructing examples of central units in ℤG (even when Theorem 1 guarantees their existence). Also there are very few cases of nonabelian groups where Z(U(ℤG)) is nontrivial and where a complete set of generators has been obtained for Z(U(ℤG)).

In this paper we make some progress on these questions. Since A_{5}, the alternating group on 5 letters, is a simple group, the procedure in [1] cannot be used to construct central units in ℤA_{5}. However, if α = (12345) then α and α^{−1} are conjugate to each other but not to α^{2} = α^{7}, so Theorem 1 says that Z(U(ℤA_{5})) is nontrivial. We will show that Z(U(ℤA_{5})) = ±⟨u⟩ where ⟨u⟩ is an infinite cyclic group. More significantly, we will explicitly find the generator u, thus obtaining a complete description of Z(U(ℤA_{5})).

The work described here will form part of the first author’s doctoral dissertation.

2 Main Results

Recall that whenever R is a commutative ring with 1, the centre of RG is a free R–module with basis consisting of the finite conjugacy class sums in RG. A_{5} has 5 distinct conjugacy classes, and we will denote the corresponding class sums by C_{0}, C_{1}, C_{2}, C_{3}, C_{4} where C_{0} = 1, C_{1} is the sum of elements conjugate to (12345), C_{2} is the sum of elements conjugate to (13524), C_{3} is the sum of all 3–cycles and C_{4} is the sum of all elements which are the product of 2 disjoint transpositions. We will need to use the following identities:

C_{1}^{2} = 12 + 5C_{1} + C_{2} + 3C_{3}
C_{1}C_{2} = C_{1} + C_{2} + 3C_{3} + 4C_{4}
C_{1}C_{3} = 5C_{1} + 5C_{2} + 3C_{3} + 4C_{4}
C_{1}C_{4} = 5C_{2} + 3C_{3} + 4C_{4}
C_{2}^{2} = 12 + C_{1} + 5C_{2} + 3C_{3}
C_{2}C_{3} = 5C_{1} + 5C_{2} + 3C_{3} + 4C_{4}
C_{2}C_{4} = 5C_{1} + 3C_{3} + 4C_{4}
C_{3}^{2} = 20 + 5C_{1} + 5C_{2} + 7C_{3} + 8C_{4}
C_{3}C_{4} = 5C_{1} + 5C_{2} + 6C_{3} + 4C_{4}
C_{4}^{2} = 15 + 5C_{1} + 5C_{2} + 3C_{3} + 2C_{4}

Suppose u ϵ Z(V (ℤA_{5})). Let u = Σa_{i}C_{i} and u^{−1} = Σb_{i}C_{i} where a_{i}, b_{i} ϵ ℤ, 0 ≤ i ≤ 4. Since uu^{−1} = 1, the identities just stated can be used to give 5 equations, one for each C_{i} (some of the details will be omitted here as they are routine). The augmentation map tells us that a_{0} + 12a_{1} + 12a_{2} + 20a_{3} + 15a_{4} = 1 and similarly for the b_{i}. Substituting for a_{0} and b_{0}, we see that the equation arising from C_{0} can be ignored as it is a linear combination of the rest. The other 4 equations are:

(1 − 19a_{1} − 11a_{2} − 15a_{3} − 15a_{4})b_{1} + (−11a_{1} + a_{2} + 5a_{3} + 5a_{4})b_{2} +(−15a_{1} + 5a_{2} + 5a_{3} + 5a_{4})b_{3} + (−15a_{1} + 5a_{2} + 5a_{3} + 5a_{4})b_{4} = −a_{1}

(a_{1} − 11a_{2} + 5a_{3} + 5a_{4})b_{1} + (1 − 11a_{1} − 19a_{2} − 15a_{3} − 15a_{4})b_{2} +(5a_{1} − 15a_{2} + 5a_{3} + 5a_{4})b_{3} + (5a_{1} − 15a_{2} + 5a_{3} + 5a_{4})b_{4} = −a_{2}

(3a_{1} + 3a_{2} − 9a_{3} + 3a_{4})b_{1} + (3a_{1} + 3a_{2} − 9a_{3} + 3a_{4})b_{2}+ (1 − 9a_{1} − 9a_{2} − 33a_{3} − 9a_{4})b_{3} + (3a_{1} + 3a_{2} − 9a_{3} + 3a_{4})b_{4} = −a_{3}

(4a_{2} + 4a_{3} − 8a_{4})b_{1} + (4a_{1} + 4a_{3} − 8a_{4})b_{2} + (4a_{1} + 4a_{2} + 8a_{3} − 16a_{4})b_{3} +(1 − 8a_{1} − 8a_{2} − 16a_{3} − 28a_{4})b_{4} = −a_{4}

Adding these together, we obtain

(1 − 15(a_{1} + a_{2} + a_{3} + a_{4}))(b_{1} + b_{2} + b_{3} + b_{4}) = −(a_{1} + a_{2} + a_{3} + a_{4})

Since we are dealing with integers, this means that a_{1} + a_{2} + a_{3} + a_{4} = 0 and b_{1} + b_{2} + b_{3} + b_{4} = 0. Substituting for a_{1} and b_{1}, and ignoring the first equation which is then a linear combination of the others, we are reduced to

(1 + 4a_{2} − 8a_{3} − 8a_{4})b_{2} + (−8a_{2} − 4a_{3} − 4a_{4})b_{3}
+(−8a_{2} − 4a_{3} − 4a_{4})b_{4} + a_{2} = 0
(1 − 12a_{3})b_{3} + a_{3} = 0
(−8a_{2} − 4a_{3} − 4a_{4})b_{2} + (−4a_{2} − 12a_{4})b_{3}
+(1 − 4a_{2} − 12a_{3} − 12a_{4})b_{4} + a_{4} = 0

It follows from the second equation that 1 − 12a_{3} divides a_{3}, forcing a_{3} = 0 and b_{3} = 0. We are now reduced to

(1 + 4a_{2} − 8a_{4})b_{2} + (−8a_{2} − 4a_{4})b_{4} = −a_{2}
(−8a_{2} − 4a_{4})b_{2} + (1 − 4a_{2} − 12a_{4})b_{4} = −a_{4}

The determinant of the 2 × 2 matrix arising here is

D = 1 − 20(4a_{2}^{2} + 4a_{2}a_{4} − 4a_{4}^{2} + a_{4}) = −20(2a_{2} + a_{4})^{2} + (10a_{4} − 1)^{2}.

We note that D ≠ 0 and also

b_{2} = \frac{(2a_{2} + a_{4})^{2} − (5a_{4}^{2} + a_{2})}{D}
b_{4} = \frac{10a_{4}^{2} − a_{4} − 2(2a_{2} + a_{4})^{2}}{D}

Since 2b_{2} + b_{4} = \frac{−2a_{2} − a_{4}}{D} is an integer, we have that D|(2a_{2} + a_{4}). The equation for b_{4} then says that D|a_{4}(10a_{4} − 1). We conclude from the equation for D that gcd(D, a_{4}) = 1, so D|(10a_{4} − 1). Setting 2a_{2} + a_{4} = Du and 10a_{4}−1 = Dv, we obtain D = D^{2}(v^{2}−20u^{2}). Since D ≠ 0 and D ≠ −1, it follows that D = 1.

We then have that (10a_{4} − 1)^{2} − 20(2a_{2} + a_{4})^{2} = 1, and this can be rewritten as (2a_{2} + a_{4})^{2} = (5a_{4} − 1)a_{4}. It follows that a_{4} is an even number and that both a_{4} and 5a_{4} − 1 are ± (perfect squares). Let a_{4} = ±4Y^{2} and 5a_{4} − 1 = ±X^{2}. If a_{4} > 0, we get X^{2} − 20Y^{2} = −1. Since the left hand side is 0 or 1 (mod 4), this equation has no solution.

If a_{4} ≤ 0, we have the Pell’s equation X^{2} − 20Y^{2} = 1. Working back through the identities which have been developed, we have proved Proposition 2 Z(U(ℤA_{5})) = {±u|u = (1 + 12Y^{2})C_{0} + (±XY + 2Y^{2})C_{1} + (∓XY +2Y^{2})C_{2}−4Y^{2}C_{4} where X, Y range through all solutions of the Pell’s equation X^{2} − 20Y^{2} = 1}.

Note that if X, Y is any solution of the above Pell’s equation, then the solution −X, Y gives the same units, so we may assume X and Y are nonnegative. Also, if X, Y is a particular solution of the equation, then the 2 units obtained from this solution are inverse to each other – i.e., if u = (1 + 12Y^{2})C_{0} + (XY + 2Y^{2})C_{1} + (−XY + 2Y^{2})C_{2} − 4Y^{2}C_{4}, then u^{−1} = (1 + 12Y^{2})C_{0} + (−XY + 2Y^{2})C_{1} + (XY + 2Y^{2})C_{2} − 4Y^{2}C_{4}.

For example, the solution X = 9, Y = 2 gives the inverse pair v = 49 + 26C_{1} − 10C_{2} − 16C_{4}, v^{−1} = 49 − 10C_{1} + 26C_{2} − 16C_{4}. In fact, our main theorem shows that this particular v is more than just an isolated example.

Theorem 3 Z(U(ℤA_{5})) = ±⟨v⟩ where v is as defined above.

A careful discussion of solutions to Pell’s equation can be found in [2], but for our purposes the crucial result is

Lemma 4 ([2], Theorem 7.26) Consider the Pell’s equation x^{2} −dy^{2} = 1 where d is a positive integer which is not a perfect square. Let X_{1}, Y_{1} be the least positive solution to the equation. Then all positive solutions are given by X_{n}, Y_{n} for n = 1, 2, 3 ⋯, where X_{n} and Y_{n} are the integers defined by X_{n} + Y_{n}\sqrt{d} = (X_{1} + Y_{1}\sqrt{d})^{n}.

Proof of Theorem 3

By Proposition 2 and the subsequent remark, we are considering nonnegative solutions to the equation X^{2} − 20Y^{2} = 1. When Y = 0 we get u = 1, while there is no solution when X = 0, so we may assume X, Y > 0.

All positive solutions are given by X_{n}, Y_{n} as stated in Lemma 4. For each such n, define

u_{n} = 1 + 12Y_{n}^{2} + (X_{n}Y_{n} + 2Y_{n}^{2})C_{1} + (−X_{n}Y_{n} + 2Y_{n}^{2})C_{2} − 4Y_{n}^{2}C_{4}

It is easy to see that X = 9, Y = 2 is the least positive solution of X^{2} − 20Y^{2} = 1, so u_{1} = v.

Using our earlier remarks on inverses, we will be finished if we can show that u_{n} = u_{1}^{n} for all n ≥ 1. This we will do by induction, the case n = 1 being obvious. Assume the result is true when n = k for some k ≥ 1. Since u_{1}^{k+1} is a central unit, Proposition 2 tells us that we only need prove that the identity coefficient of u_{1}^{k+1} equals 1 + 12Y_{k+1}^{2} and that the coefficient of C_{1} in u_{1}^{k+1} equals X_{k+1}Y_{k+1} + 2Y_{k+1}^{2}.

The identity coefficient of u_{1}^{k+1} = u_{1}u_{k} is 49(1 + 12Y_{k}^{2}) + 12(26)(X_{k}Y_{k} + 2Y_{k}^{2})+12(−10)(−X_{k}Y_{k} +2Y_{k}^{2})+15(−16)(−4Y_{k}^{2}) = 49+432X_{k}Y_{k} +1932Y_{k}^{2}. On the other hand, Lemma 4 says that 1 + 12Y_{k+1}^{2} = 1 + 12(9Y_{k} + 2X_{k})^{2} = 1 + 12(81Y_{k}^{2} + 36X_{k}Y_{k} + 4(20Y_{k}^{2} + 1)) = 49 + 432X_{k}Y_{k} + 1932Y_{k}^{2}, as desired.

The coefficient of C_{1} in u_{1}^{k+1} equals 49(X_{k}Y_{k^{+}}2Y_{k}^{2}) + 26(1 + 12Y_{k}^{2}) + 5(26)(X_{k}Y_{k}+2Y_{k}^{2})+26(−X_{k}Y_{k}+2Y_{k}^{2})+(−10)(X_{k}Y_{k}+2Y_{k}^{2})+(−10)(−X_{k}Y_{k}+ 2Y_{k}^{2}) + 5(−10)(−4Y_{k}^{2}) + 5(−16)(−X_{k}Y_{k} + 2Y_{k}^{2}) + 5(−16)(−4Y_{k}^{2}) = 26 + 233X_{k}Y_{k} + 1042Y_{k}^{2}. Lemma 4 says that X_{k+1}Y_{k+1} + 2Y_{k+1}^{2} = (9X_{k} + 40Y_{k})(2X_{k} + 9Y_{k}) + 2(2X_{k} + 9Y_{k})^{2} = 26(20Y_{k}^{2} + 1) + 233X_{k}Y_{k} + 522Y_{k}^{2} = 26 + 233X_{k}Y_{k} + 1042Y_{k}^{2}, and this completes the proof.

REFERENCES

1. E. Jespers, M.M. Parmenter and S.K. Sehgal, Central units of integral group rings of nilpotent groups. Proc. Amer. Math. Soc. (to appear)

2. Ivan Niven and H.S. Zuckerman, An Introduction to the Theory of Numbers (4th edition), Wiley, 1980.

3. J. Ritter and S.K. Sehgal, Integral group rings with trivial central units, Proc. Amer. Math. Soc. 108 (1990), 327–329.

4. J. Ritter and S.K. Sehgal, Construction of units in integral group rings of finite nilpotent groups, Trans. Amer. Math. Soc. 324(2) (1991), 603–621.

5. J. Ritter and S.K. Sehgal, Units of group rings of solvable and Frobenius groups over large rings of cyclotomic integers, Journal of Algebra 158 (1993), 116–129.

6. S.K. Sehgal, Units in Integral Group Rings, Longman, 1993.

Department of Mathematics and Statistics Memorial University of Newfoundland St. John’s, Newfoundland Canada A1C 5S7