# Covering Problems concerning Abelian Groups

```242
S. K. ZAREMBA
COVERING PROBLEMS CONCERNING ABELIAN GROUPS
S. K. ZABEMBAf.
The following theorem was proved independently by B. KuttnerJ,
J. G. Mauldon, [2]&sect;, and myself, [3]||.
THEOREM 1. Let 0 denote an Abelian Group with n base elements,
9v •••&gt; 9,&raquo; each &deg;f order p, where p is a prime, and let 8 be the set of the
Ifvisapower
v *= n{p—\)+\ elements l,gv ...,g{~1, ...,gn, ...,gl~lofO.
of p, there exists a sub-group H of G of order pn/v such that HS = G^.
The present note deals with some extensions of this theorem, and with
other problems directly connected with it. The main generalization
applies to the case when the order of the base elements of G is no longer a
prime, but a power of a prime. In order to obtain this result, we shall first
prove the following proposition:
2. Let V denote an Abelian group with kn base elements
,ju [i = i5 ...5 n; j •-- 1, ..., k), each of order p, where p is a prime, and let
]j be the set of the /x = n(pk—l)-\-l elements of the form g}[ g\l ... &lt;/\$, where
0 ^.yi^.p— 1 (j= 1, ..., k) arid i= 1, ..., n. If fj, is a power of pk, then
there exisU a sub-group X of T of order pknl^ such that F = X S .
PROPOSITION
Proof. Assuming /x = pkr, let A be an auxiliary Abelian group with
kr base elements, each of order p. As is well known (e.g. [4], pp. 289-291),
we can form a set (called a geometric set) of (p1*— l)j{pk— 1) = n sub-groups
Ax, . , Atl of order pfc of A, such that any element of A other than the unit
should belong to one, and only one, of these sub-groups. Let aiv ..., aik
be a set of base elements for the group A( (i= 1, ..., n). If IIg&pound;/ is any
element of V, let 11 ?&amp;?v be the corresponding element oiA, which can also be
represented as II a\$' .
This is clearly a honiomorphic mapping of T onto A.
Now, let X be the sub-group of V formed by the elements corresponding
to the unit in A.
In order to prove that T — XX, let us consider any element
g— 11 &lt;#' of 1\ If this element corresponds to lla^., in A, h = gY\.g~*&gt;/
tt
V
&quot;
V
^
J In a talk delivered at the Birmingham Meeting of the British Association in 1950.
&sect; The numbers in square brackets refer to the literature quoted at the end of this note.
|| [Added 12 October, 1951.] The same theorem was also found independently by
E. Mattioli, &quot; Sopra una particolare proprieta dei gruppi abeliani finiti &quot;, Ann. Scuola
Norm. Super. Pisa (3), 3 (1950), 59-65.
If A and B being two aub-sets of Q, Ab denotes, as usual, the set of all the products ab,
where a belongs to A, and b belongs to B.
COVERING PROBLEMS CONCERNING ABELIAN GROUPS.
243
corresponds to the unit, i.e. belongs to X. Thus g = h U gB*, where h belongs
to X and II g^ belongs to E.
This is the only way of decomposing an
element of F into an element of X and an element of 2. Indeed, in view
of the homomorphic mapping of F onto A, since the factor belonging to X
corresponds to the unit in A, the factor belonging to 2 must correspond
to II aB&raquo; , and therefore must be II g*&quot;. Hence also the factor belonging to
X, and consequently the whole factorization, is uniquely determined.
This also shows that the order of X is pkn\\i.
Hence we deduce the following theorem:
THEOREM 3. Let 0 denote an Abelian group with n base elements,
gv ..., gn, each of order pk, where p is a prime, and let S denote the set of the
elements 1, gv ..., gf~l, ..., gn, ..., gg^'of 0. If p is a
H = n(pk-l)+l
power of pk, then there exists a subset H of 0 composed of pknjii elements,
and satisfying 0 = HS.
Proof. Let /x be equal to p**.
Any integer c^ satisfying 0 ^ a, ^ pk— 1
k
can be represented in a unique way as 2 p*'1^,
where 0
ii
i
(j = 1, ..., k). Hence, if we map the element gp ... gi&quot;&raquo; of 0 on the element
II0\$* of the group F of Proposition 2, this is a one-to-one (though not
isomorphic) mapping of 0 onto F. Let H be the sub-set of 0 corresponding
to X, and therefore composed of pknjfi elements.
We have now to show that 0 = HS or, in other terms, that if we represent the elements of 0 as products of powers of the base elements, then to
any element of 0 there corresponds an element of H differing from it by
the exponent of at most one of the base elements. Let g be any element of
0, and g* the corresponding element of F. According to Proposition 2,
there exists an element h* of X and an element g]l ...&lt;7&lt;| of S satisfying
g*~—h*g\\ ... &lt;/?&pound;. This means that if we represent g* and K* as products
of powers of the base elements of F, all the exponents coincide, except,
perhaps, for those of some of the elements giv ..., gik. Therefore, if h is
the element of H corresponding to A*, and if we represent it as a product of
powers of g1} ..., gn, all the exponents, with the possible exception of that
of git coincide with those of the analogous representation of g. Hence
the proof is complete.
The question can be asked whether, or not, the set H can be so chosen
as to form a sub-group of 0. In order to obtain the answer, which is
negative (unless, obviously, k = 1, in which case Theorem 1 is applicable),
we shall first prove the following lemma.
LEMMA
4.
Let 0 be an Abelian group with n base elements, gv ..., gn,
B2
244
S. K . ZAREMBA
each of order q, and let S denote the set of the n{q—1) + 1 elements
If there exists a sub-set H of 0 containing the unit element and such
that any element of G can be represented in a unique way as the product
of an element of H and oj an element of S, then H contains, apart from the unit
element, no product of powers of fewer than three base elements of G. On the
other hand, any product of the type gx&quot;gyp is a factor of a product of the type
&lt;//&lt;//&lt;// belonging to H.
Proof. If H contained gf, this element could be represented as l&lt;/,&deg;
or yf I. where in both cases the first factor belongs to / / . and the second to #.
If// contained a product like &lt;//&lt;//, we should similarly find for gxu the
two decompositions yx — 1 yx* and gxa ----- (&lt;//&lt;//) y^p, in both of which the
first factor belongs to H and the second to S.
()n the other hand, clearly yxyf cannot be obtained as a product of an
element of // and of an element of 8, in which the element of H is either
the unit or a product of more than three base elements; therefore it must be
obtainable as (gtagy*gsy)(fcy&gt; where yxay/gsy belongs to H.
The answer to the question asked above is contained in the following
corollary of the preceding lemma.
( OROLLARY 5. On the assumptions of Lemma 4, if q is not a prime,
ti cannot be a group.
Proof. Assume q = q1 &lt;/2, where neither of the integers qx and q2 is 1.
Then, according to Lemma 4, H must contain an element g = gig%ig*.
If H is a group, this implies that the element gq* = g\ig\q* must also belong
to H, which contradicts the lemma.
When the existence of a set, or group, H satisfying G = HS (so that
the corresponding decomposition of the elements of G should be unique)
can be proved, we may seek all the possible sets, or all the possible groups, H.
As far as groups are concerned, the solution is given below:
PROPOSITION 6. On the assumptions of Proposition 2, the method of
obtaining the group X outlined in the proof of this proposition gives all the
solution*, i.e. all the sub-groups of V of wder pkll/fi satisfying F = XL.
Proof. The groups A and F/X are obviously isomorphic, the image
of u-ai} being Xj/^. Hence, if X is given, we can take F/X as the auxiliary
group A which yields this particular solution.
Since Theorem 1 is a special case of Proposition 2, corresponding to
k •• I, this answers a question proposed in [3],
COVERING PROBLEMS CONCERNING ABBLIAN GROUPS.
245
The problem offindingthe most general set H is much more complicated.
If gr^ ..., gn are the base elements of the given group 0, and the conditions
of Theorem 1 are fulfilled, it is easy to see that, if fi(a), ..., fn(a) are
arbitrary functions permuting among themselves the remainders modulo
p, the transformation mapping grji... g&pound;* into g^J ... g^an) transforms any
of our groups H into a set H of as many elements as H, such that 0 = HS.
As special cases, corresponding to /,(a) = a-j-8,- (mod p) (i = 1, ..., n), where
hx, ..., 8n are arbitrary constants, wefindall the co-sets of H as solutions. If
p = 2 or p = 3, permutations of exponents do not yield anything else but
co-sets of H. Can all the possible sets be obtained in the aforesaid
manner? This question seems to be difficult to answer in general. In
some very simple cases it can be seen that the answer is affirmative.
This has already been noticed by 0. Taussky and J. Todd, [1], for the case
p = 3, n = 4, not to mention here the rather trivial case of p = 2,
n = 3. It is perhaps worth showing that the same applies to the case
p = 2, n = 7. This is implied in the following proposition.
PROPOSITION 7.
If 0 is the Abelian group generated by the 7 base elements gx, g2, &lt;73, 9t, g5, &lt;76, g?, each of order 2, and if S denotes the set of the 8
elements 1, gx, g2, &lt;73, &lt;/4, &lt;75, g%, g7, then any sub-set H of O satisfying HS = O,
containing the unit, and composed altogether of 16 elements, is necessarily a
group.
Proof. The conditions of Lemma 4 are now satisfied. On the other
hand, if gxgvgz belongs to H, gtgvgt (t^z) cannot belong to H, since
otherwise we should find two distinct decompositions of gxgv into an
element of H and an element of 8. Therefore it is clear that by a suitable
choice of notations we can make sure that
* i = y I 9% 06.
K = 0i 03 05&gt;
^3 = 0i 04 07
belong to H. But H must contain a product &lt;7203 0t&gt; a n d from what has
just been said it follows that &pound; ?&amp; 1, 2, 3, 5, 6, i.e. i = 4 or i = 7. Since the
permutation of &lt;74 and g7 does not affect hv h2, hz, we can assume that the
notations were so chosen that H contains also
^4 = 02 03 04-
In view of Lemma 4, the inclusion of hv h2, hz, h^ in H determines entirely
the remaining products of three base elements of O included in H. These
can only be:
^2^3^4; ^6 = 0308 07 = A fyA J ^7
The remaining elements of H are found by means of the following;
246
COVERING PROBLEMS CONCERNING A B E L I A N GROUPS.
LEMMA
8.
Ifgrg,gf and grgugv belong to H, then gsgtgugv belongs to H.
Indeed, none of the products gtgugv, gtgugv, g,gtgu, 0,0,0, can belong
to H, since each of them has two factors in common either with grgtgt or
with grgugv. They can therefore be obtained as a product of an element
of H and of an element of 8 only if the element of H is a produot of four
base elements, so that H must contain gs gu gv g{, gt gu gv gp g, g, gu gk, gt gtgv gt.
If Qi9t9u9v did not belong to H, each of the indices i,j,k,l would be
different from s, t, u, v. But they also should be pairwise distinct, since
otherwise two elements of H composed of four base elements of 0 would
have three factors in common, which is clearly impossible, as the produot
of these three factors would appear in HS twice. In other terms, the 8
indices s,t,u,v,i,j,k,l
should be all distinct, which is impossible when
n = 7. Hence the proof of the lemma is complete.
Accordingly, H must contain the following products of four base
elements:
^ 3
= 0103 04 06'
= 02 04 06 07'
^ 2 = 02030606'
^2^4 = 0102 04 05'
^3^4 = 0102 03 07'
h2 hB = hx h2 h3 hi =
Thus, including the unit, we have found 15 elements of H. The
remaining element must produce, after multiplication with the various
elements of S, all the products of six base elements, as well as g1 gz g3 g&plusmn; g5 gr6 g7,
and therefore can only be
Hence H is the group generated by hv h2, h3, hx.
References.
1. O. Taussky and J. Todd, &quot; Covering theorems for groups &quot;, Ann. Soc. Polon. Math.,
21 (1949), 303-308.
2 . J. Q. Mauldon, &quot; Covering theorems for groups &quot;, Quart. J. of Math., (Second Oxford
Series), 1 (1950), 284-287.
3 . S. K. Zareraba, &quot; A covering theorem for Abelian groups &quot;, Journal London Math. Soc.,
26 (1950), 71-72.
4 . R. D. Carmichael, Introduction to the theory of groups of finite order (Ginn &amp; Co., Boston,
1937).
Polish University College,
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