ERIC EJ1054466: A Method to Determine of All Non-Isomorphic Groups of Order 16

2012·0.46 MB·English

by  ERIC

A method to determine of all non-isomorphic groups of order 16 65
Volume 5, Number 4, 2012
A METHOD TO DETERMINE OF ALL
NON-ISOMORPHIC GROUPS OF ORDER 16
Dumitru Vălcan
Abstract. Many students or teachers ask themselves: Being given a natural number n,
how many non-isomorphic groups of order n exists? The answer, generally, is not yet given.
But, for certain values of the number n have answered this question. The present work
gives a method to determine of all non-isomorphic groups of order 16 and gives descriptions
of all these groups. The results obtained here can be used by students in the senior year,
studentsincoursesorseminars,orotherMathematicsteachersareindifferentstagesoftheir
professional development.
Key words: non-isometric groups of order n
The problem of determining all non-isomorphic groups of a given order is not new. She appeared
in the latter part of the nineteenth century and part was solved by Cayley and Frobenius - abelian
case. The problem is still open. In 1905, Dickson presents some general solutions, and in 1992,
Jungnickel solve the problem of uniqueness of the groups of order n.
In [6] Purdea has determined all non-isomorphic groups of order n≤10, and I have determined in
[9] all these groups of order n∈{p,q,pq,p2,p3}, where p and q are two distinct prime numbers.
In this work we will present a method to determine of all non-isomorphic groups of order 16. In
this context, throughout this paper by group we mean a group (denoted by G) of order 16 in
multiplicative notation and we will denote: by 1 the identity (the "neutral") element of G, by
ord(g) the order of the element g∈G and by |A| the cardinal of the set A. If A is a subgroup of G,
then |A| is (also) the order of A.
Followingthesamereasoningasin[9],respectively[10],wewillprovethemainresultofthispaper:
Theorem: There are 14 non-isomorphic groups of order 16.
Proof: So, letGbeagroupoforder16andZ(G)hiscenter. AccordingtoLagrange’stheoremand
to Burnside’s theorem we have the following cases:
Case I: |Z(G)|=16. In this case G is commutative and by [3,8.4] and [5,2.2 (p. 86) and 6.1 (p.
97)] it follows that we have the following possibilities:
1) G=G ∼=Z ; so, there is an element x∈G such that ord(x)=16 and
1 16 1
G ={1,x,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12,x13,x14,x15}.
1
For this group see Table 1;
2) G=G ∼=Z ×Z ; so, there are x,y∈G such that ord(x)=8, ord(y)=2, xy=yx and
2 2 8 2
G ={1,x,x2,x3,x4,x5,x6,x7,y,xy,x2y,x3y,x4y,x5y,x6y,x7y}.
2
For this group see Table 2;
3) G=G ∼=Z ×Z ; so, there are x,y∈G such that ord(x)=4, ord(y)=4, xy=yx and
3 4 4 3
G ={1,x,x2,x3,y,y2,y3,xy,x2y,x3y,xy2,xy3,x2y2,x2y3,x3y2,x3y3}.
3
For this group see Table 3;
Volume 5, Number 4, 2012
66 Dumitru Vălcan
4) G=G ∼=Z ×Z ×Z ; so, there are x,y,z∈G such that ord(x)=4, ord(y)=ord(z)=2,
4 2 2 4 4
xy=yx, xz=zx, yz=yz and
G ={1,x,x2,x3,y,xy,x2y,x3y,z,xz,x2z,x3z,yz,xyz,x2yz,x3yz,x3y3}.
4
For this group see Table 4;
5) G=G ∼=Z ×Z ×Z ×Z ; so, there are x,y,z,t ∈G such that ord(x)=ord(y)=ord(z)=
5 2 2 2 2 5
ord(t)=2, xy=yx, xz=zx, xt =tx, yz=zy, yt =ty, zt =tz and
G ={1,x,y,z,t,xy,xz,xt,yz,yt,zt,xyz,xyt,xzt,yzt,xyzt}.
5
For this group see Table 5.
Case II: |Z(G)|=8. Then |G/Z(G)|=2 and since the group G/Z(G) is cyclic, by [5,2.2 (p. 143)]
it follows that G is commutative-contradiction to the hypothesis. So, there is no group G of order
16 with |Z(G)|=8.
Case III: |Z(G)|=4. In this case |G/Z(G)|=4 and according to [8,5.5] the group G/Z(G) is
commutative. Againby[3,8.4]and[5,2.2(p. 86)and6.1(p. 97)]itfollowsthateitherG/Z(G)∼=Z
4
orG/Z(G)∼=Z ×Z . SincethegroupZ iscyclic,itfollowsthatonlythesecondpossibilityholds.
2 2 4
Therefore
G/Z(G)={Z(G),xZ(G),yZ(G),xyZ(G)}
, where x2 ∈Z(G), y2 ∈Z(G) and xyZ(G)=yxZ(G). If xy=yx then G is commutative, which is
impossible. So, xy(cid:54)=yx and there is an element b∈Z(G) such that yx=xyb. Then
y2x=y(yx)=y(xyb)=(yx)yb=xybyb=xy2b2.
Since y2∈Z(G) it follows that b2=1. On the other hand, Z(G) being a (sub)group of order 4 (of
G) we distinguish the following subcases:
Subcase 1: Assume Z(G)∼=Z , so Z(G)={1,a,a2,a3}, with a∈G. From the above proved facts
4
it follows that yx=xya2. Then xy=yxa2 and we distinguish the following possibilities:
a)x2=y2=1. ThenthereisagroupG =(cid:104)x,y,a(cid:105),withx,yandasatisfyingtheaboveconditions;
6
see Table 6.
b) x2=1 and y2=c∈{a,a2,a3}.
i) If y2 =a then there is a group G =(cid:104)x,y,a(cid:105)=(cid:104)x,y(cid:105), with x, y and a satisfying the above
7
conditions; see Table 7.
ii)Ify2=a2 thenthereisagroupG =(cid:104)x,y,a(cid:105),withx,yandasatisfyingtheaboveconditions;
8
see Table 8.
iii)Ify2=a3 thenthereisagroupG =(cid:104)x,y,a(cid:105),withx,yandasatisfyingtheaboveconditions;
8
see Table 9.
c) x2=c and y2=d, where c, d∈{a,a2,a3}.
i) If x2 =y2 =a then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above
10
conditions; see Table 10.
ii) If x2 =a and y2 =a2 then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the
11
above conditions; see Table 11.
iii) If x2 =a and y2 =a3 then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the
12
above conditions; see Table 12.
iv) If x2 =y2 =a2 then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above
13
conditions; see Table 13.
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A method to determine of all non-isomorphic groups of order 16 67
v) If x2 =a2 and y2 =a3 then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the
14
above conditions; see Table 14.
vi) If x2 =y2 =a3 then there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above
15
conditions; see Table 15.
Subcase 2: Assume Z(G)∼=Z ×Z , so Z(G)={1,a,b,ab}, with a, b and ab=ba, a2 =b2 =1.
2 2
Since xyZ(G)=yxZ(G), xy(cid:54)=yx and x2, y2∈Z(G)it follows that (say) xy=yxa, G=(cid:104)x,y,a,b(cid:105)and
we distinguish the following possibilities:
a) If x2=y2=1 then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
16
conditions; see Table 16.
b) x2=1 and y2=c∈{a,b,ab}.
i) If y2 =a then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
17
conditions; see Table 17.
ii) If y2 =b then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
18
conditions; see Table 18.
iii) If y2 =ab then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
19
conditions; see Table 19.
c) x2=d and y2=e, where d, e∈{a,b,ab}.
i) If x2=y2=a then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
20
conditions; see Table 20.
ii) If x2=a and y2=b then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the
21
above conditions; see Table 21.
iii) If x2=a and y2=ab then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying
22
the above conditions; see Table 22.
iv) If x2=y2=b then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the above
23
conditions; see Table 23.
v) If x2 =b and y2 =ab then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying
24
the above conditions; see Table 24.
vi) If x2=ab and y2=a then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying
25
the above conditions; see Table 25.
vii) If x2=ab and y2=b then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying
26
the above conditions; see Table 26.
viii) If x2 =y2 =ab then there is a group G =(cid:104)x,y,a,b(cid:105), with x, y, a and b satisfying the
27
above conditions; see Table 27.
Case IV: |Z(G)|=2. Then |G/Z(G)|=8 and according to [6,p.141-142] and [5,2.2 (p. 143)] either
G/Z(G)∼=Z ×Z or G/Z(G)∼=Z ×Z ×Z , or G/Z(G)∼=D , where D is the dihedral group of
4 2 2 2 2 4 4
order 4, or G/Z(G)∼=Q, where Q is the quaternions group.
Therefore Z(G)=(cid:104)a(cid:105)={1,a}, with a∈G and a2=1.
Subcase 1: G/Z(G)∼=Z ×Z . It follows that
4 2
G/Z(G)={Z(G),xZ(G),x2Z(G),x3Z(G),yZ(G),xyZ(G),x2yZ(G),x3yZ(G)}
, where x, y∈G, x4, y2 ∈Z(G) and xyZ(G)=yxZ(G). Thus G=<x,y,a> and xy=yxa. But in
these conditions yx2 =x2y and x2 ∈Z(G)-which is impossible; so, there is no exist group G with
the above properties.
Volume 5, Number 4, 2012
68 Dumitru Vălcan
Subcase 2: G/Z(G)∼=Z ×Z ×Z . It follows that
2 2 2
G/Z(G)={Z(G),xZ(G),yZ(G),zZ(G),xyZ(G),xzZ(G),yzZ(G),xyzZ(G)}
, where x, y, z∈G, x2, y2, z2 ∈Z(G) and xyZ(G)=yxZ(G), xzZ(G)=zxZ(G), yzZ(G)=zyZ(G).
Thus we have the following possibilities:
a) yx=xy, zx=xza and yz=zya. In these conditions
z(xy)=(xza)y=x(zya)=(xy)z
and xy∈Z(G)-which is impossible; so, it doesn’t exist group G with the above properties.
b) yx=xya, zx=xz and yz=zya. In these conditions
(xz)y=x(yza)=(yx)z=y(xz)
and xz∈Z(G)-which is impossible; so, it doesn’t exist group G with the above properties.
c) yx=xya, zx=xza and yz=zy. In these conditions
(yz)x=y(xza)=x(yz)
and yz∈Z(G)-which is impossible; so, it doesn’t exist group G with the above properties.
d) yx=xya, zx=xza and yz=zya. Then we obtain the following equalities:
x(xyz)=yz=(xyz)x
,
y(xyz)=xza=(xyz)y
and
z(xyz)=xy=(xyz)z
. Therefore xyz∈Z(G)-which is impossible. So, also in these conditions there is no group G of
order 16.
Subcase 3: G/Z(G)∼=D . It follows that
4
G/Z(G)={Z(G),xZ(G),x2Z(G),x3Z(G),yZ(G),xyZ(G),x2yZ(G),x3yZ(G)}
, where x, y∈G, x4, y2∈Z(G) and yxZ(G)=x3yZ(G). Thus in this subcase we have the following
possibilities:
a) yx=x3y.
i) If x4=1 then
yx2=(x3y)x=x3(x3y)=x2y
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
ii) If x4=a and y2=1 then
yx2=(x3y)x=x3(x3y)=x2ya
and there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above conditions; see Table 28.
28
iii) If x4=y2=a then again we obtain that
yx2=(x3y)x=x3(x3y)=x2ya
Acta Didactica Napocensia, ISSN 2065-1430
A method to determine of all non-isomorphic groups of order 16 69
and there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above conditions; see Table 29.
29
b) yx=x3ya.
i) If x4=1 then again we obtain that
yx2=(x3y)x=x3(x3y)=x2y
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
ii) If x4=a and y2=1 then
yx2=(x3ya)x=x3(x3ya)a=x2ya
and there is a group G =(cid:104)x,y,a(cid:105), with x, y and a satisfying the above conditions; see Table 30.
30
iii) If x4=y2=a then again we obtain that
yx2=(x3ya)x=x3(x3y)a=x2y
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
Subcase 4: G/Z(G)∼=Q. By [8, p.93] It follows that again
G/Z(G)={Z(G),xZ(G),x2Z(G),x3Z(G),yZ(G),xyZ(G),x2yZ(G),x3yZ(G)}
, where x, y∈G, x4, y4∈Z(G), x2Z(G)=y2Z(G) and yxZ(G)=x3yZ(G). Thus in this subcase we
have the following possibilities:
a) yx=x3y and x2=y2.
i) If x4=1 then
yx2=(x3y)x=x3(x3y)=x2y
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
ii) If x4=a then
yx2=(yx)x=x3(yx)=x2ya
and
y3=yy2=yx2=x2ya.
On the other hand,
y3=y2y=x2y
and a=1-which is impossible. So, in these conditions there is no group G of order 16.
b) yx=x3y and x2=y2a. If x4=a then
yx2=(x3y)x=x3(x3y)=x2ya
and
y3=yy2=yx2a=x2y
. On the other hand,
y3=y2y=x2ya
and a=1-which is impossible. So, in these conditions there is no group G of order 16.
c) yx=x3ya and x2=y2.
i) If x4=1 then
yx2=(x3ya)x=x3(x3ya)a=x2y
Volume 5, Number 4, 2012
70 Dumitru Vălcan
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
ii) If x4=a then
yx2=(yx)x=x3(yx)a=x2ya
and
y3=yy2=yx2=x2ya.
On the other hand,
y3=y2y=x2y
and a=1-which is impossible. So, in these conditions there is no group G of order 16.
d) yx=x3ya and x2=y2a.
i) If x4=1 then
yx2=(x3ya)x=x3(x3ya)a=x2y
and x2∈Z(G)-which is impossible. So, in these conditions there is no group G of order 16.
ii) If x4=a then
yx2=(yx)x=x3(yx)a=x2ya
and
y3=yy2=yx2a=x2y.
On the other hand,
y3=y2y=x2ya
and a=1-which is impossible. So, in these conditions there is no group G of order 16.
It follows that in the conditions from this subcase doesn’t exist group G of order 16.
Thereforewehavedetermined30groupsoforder16. Afterwardswearegoingtoshowthefollowing
isomorphisms:
G ∼=G ∼=G ,
6 8 13
G ∼=G ∼=G ∼=G ∼=G ∼=G ,
9 10 11 12 14 15
G ∼=G
16 17
and
G ∼=G ∼=G ∼=G ,
18 19 24 26
G ∼=G ∼=G ∼=G ∼=G ∼=G .
20 21 22 23 25 27
By the tables of group it follows the Table 0, which presents all the elements of order 2, 4 and 8 in
each group determined above.
Indeed, using the Table 0, it is straightforward to verify that:
α) the map f :G →G defined by f(x)=x, f(y)=xy and f(a)=a is an isomorphism of groups
6 8
and G ∼=G ;
6 8
β) the map g:G →G defined by g(x)=xa, g(y)=y and g(a)=a is an isomorphism of groups
8 13
and G ∼=G ;
8 13
γ) so, the map g◦ f :G →G defined by (g◦ f)(x)=xa, g(◦f)(y)=xya and (g◦ f)(a)=a is an
6 13
isomorphism of groups and G ∼=G .
6 13
Similarly, check the other isomorphisms above.
Therefore only 14 groups from these above presented are not isomorphic, namely: G , G , G , G ,
1 2 3 4
G , G , G , G , G , G , G , G , G , G - the first five are abelian and last nine non-abelian.
5 6 7 9 16 18 20 28 29 33
Now the theorem is completely proved.
At the end of this paper we present the Table 0 - above mentioned and the (multiplication) tables
of these 30 groups which have been determined.
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References
[1] Călugăreanu, G., Introduction to Abelian Groups Theory, (in Romanian), Editura Expert,
Cluj-Napoca, 1994.
[2] Dickson, J., E., Definitions of a group and a field by independent postulates, Trans. Amer.
Math. Soc., 6(1905), p. 198-204.
[3] Fuchs, L., Infinite Abelian Groups Theory, Vol. I, Academic Press, New York and London,
Pure and Applied Mathematics, 36, 1970.
[4] Jungnickel, D., On the Uniqueness of the Cyclic Group of Order n, Amer. Math. Monthly,
6(1992), p. 545-547.
[5] Popescu, D., Vraciu, C., Elements of Finite Group Theory, (In Romanian), Editura Şiinţifică
şi Enciclopedică, Bucureşti, 1986.
[6] Purdea, I., On the groups of order n≤10, Selected paper from "Didactica Matematicii",
Vol. 1984-1992, "Babeş-Bolyai" University, Faculty of Mathematics and Computer Science,
Research Seminars, Cluj-Napoca, 1992, p. 133-144.
[7] Purdea, I., Pic, G., Algebra, (In Romanian), "Babeş-Bolyai" University, Faculty of Mathe-
matics and Mechanics, Lit., Cluj-Napoca, 1973.
[8] Rotmann, J., J., The Theory of Groups: An introduction, Allyn and Bacon, Inc., Boston,
1968.
[9] Vălcan, D., On some groups of finite order, (In Romanian), Lucrările Seminarului de "Didac-
tica Matematicii" Vol. 13(1998), p. 177-182.
[10] Vălcan, D., A method to determine of all non-isomorphic groups of order 12, Creative Math.,
Nr. 14 (2005), p. 57-66.
Authors
Dumitru Vălcan, Babes-Bolyai University, Cluj-Napoca (Romania), E-mail: [email protected]
Volume 5, Number 4, 2012
72 Dumitru Vălcan

TABLE 0: with the elements of order 2, 4 and 8 in each of the 30 groups determined in the paper.

The group The elements of order 2 The elements of order 4 The elements of order 8
G x8 x4, x12 x2, x6, x10, x14
1
G x4, y, x4y x2, x6, x2y, x6y x, x3, x5, x7 xy, x3y, x5y, x7y
2
G x2, y2, x2y2 x, x3, y, y3, xy, x2y, x3y, xy2, xy3, x2y3, x3y2, x3y3 -
3
G x2, y, x2y, z, x2z, yz, x2yz x, x3, xy, x3y, xz, x3z, xyz, x3yz -
4
G all elements - -
5
G x, y, a2, xa2, ya2, xya, xya3 a, a3, xa, xa3, ya, ya3, xy, xya3 -
6
G x, a2, xa2 a, a3 y, xy, ya, ya2, ya3, xya, xya2, xya3
7
G x, xy, a2, xa2, ya, ya3, xya2 y, a, a3, xa, xa3, ya2, xya, xya3 -
8
G x, a2, xa2 a, a3, xa, xa3 y, xy, ya, ya2, ya3, xya, xya2, xya3
9
G xy, a2, xya2 a, a3, xya, xya3 x, y, xa, xa2, xa3, ya, ya2, ya3
10
G a2, ya, ya3 y, a, a3, ya2 x, xy, xa2, xa3, xya, xya2, xya3
11
G a2, xya, xya3 xy, a, a3, xya2 x, xy, xa, xa2, xa3, ya, ya2, ya3
12
G a2, xa, xa3, ya, ya3, xya, xya3 x, y, xy, a, a3, xa2, ya2, xya2 -
13
G a2, xa, xa3 x, a, a3, xa2 y, xy, ya, ya2, ya3, xya, xya2, xya3
14
G xy, a2, xya2 a, a3, xya, xya3 x, y, xa, xa2, xa3, ya, ya2, ya3
15
G x, y, a, b, ab, xa, xb, xab, ya, yb, yab xy, xya, xyb, xyab -
16
G x, xy, a, b, ab, xa, xb, xab, xya, xyb, xyab y, ya, yb, yab -
17
G x, a, b, ab, xa, xb, xab y, xy, ya, yb, yab, xya, xyb, xyab -
18
G x, a, b, ab, xa, xb, xab y, xy, ya, yb, yab, xya, xyb, xyab -
19
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xya, xyb, xyab -
20
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xya, xyb, xyab -
21
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xya, xyb, xyab -
22
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xya, xyb, xyab -
23
G xy, a, b, ab, xya, xyb, xyab x, y, xa, xb, xab, ya, yb, yab -
24
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xyab, xya, xyb, xyab -
25
G xy, a, b, ab, xya, xyb, xyab x, y, xa, xb, xab, ya, yb, yab -
26
G a, b, ab x, y, xy, xa, xb, xab, ya, yb, yab, xya, xyb, xyab -
27
G y, x2y, a, ya, x2ya x2, xy, x3y, x2a, xya, x3ya x, x3, xa, x3a
28
G xy, x3y, a, xya, x3ya x2, y, x2y, x2a, ya, x2ya x, x3, xa, x3a
29
G y, xy, x2y, x3y, a, ya, xya, x2ya, x3ya x2, x2a x, x3, xa, x3a
30

Acta Didactica Napocensia, ISSN 2065-1430
A method to determine of all non+isometric groups of order 16 73
OPERATIONS GROUP TABLES

TABLE 1: G ≅Z =<x>, x16=1; Z(G )=G .
1 16 1 1

1 x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15
1 1 x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15
x x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1
x2 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x
x3 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2
x4 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3
x5 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4
x6 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4 x5
x7 x7 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4 x5 x6
x8 x8 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4 x5 x6 x7
x9 x9 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4 x5 x6 x7 x8
x10 x10 x11 x12 x13 x14 x15 1 x x2 x3 x4 x5 x6 x7 x8 x9
x11 x11 x12 x13 x14 x15 1 x X2 x3 x4 x5 x6 x7 x8 x9 x10
x12 x12 x13 x14 x15 1 X x2 X3 x4 x5 x6 x7 x8 x9 x10 x11
x13 x13 x14 x15 1 x x2 x3 X4 x5 x6 x7 x8 x9 x10 x11 x12
x14 x14 x15 1 x x2 x3 x4 X5 x6 x7 x8 x9 x10 x11 x12 x13
x15 x15 1 x x2 x3 x4 x5 X6 x7 x8 x9 x10 x11 x12 x13 x14

Volume 5 number 4
74 Dumitru Vălcan

TABLE 2: G ≅Z ×Z =<x,y>, x8=y2=1, xy=yx; Z(G )=G .
2 8 2 2 2

1 x x2 x3 x4 x5 x6 X7 y xy x2y x3y x4y x5y x6y x7y
1 1 x x2 x3 x4 x5 x6 X7 y xy x2y x3y x4y x5y x6y x7y
x x x2 x3 x4 x5 x6 x7 1 xy x2y x3y x4y x5y x6y x7y y
x2 x2 x3 x4 x5 x6 x7 1 x x2y x3y x4y x5y x6y x7y y xy
x3 x3 x4 x5 x6 x7 1 x X2 x3y x4y x5y x6y x7y y xy x2y
x4 x4 x5 x6 x7 1 x x2 X3 x4y x5y x6y x7y y xy x2y x3y
x5 x5 x6 x7 1 x x2 x3 X4 x5y x6y x7y y xy x2y x3y x4y
x6 x6 x7 1 x x2 x3 x4 X5 x6y x7y y xy x2y x3y x4y x5y
x7 x7 1 x x2 x3 x4 x5 X6 x7y y xy x2y x3y x4y x5y x6y
y y xy x2y x3y x4y x5y x6y X7y 1 x x2 x3 x4 x5 x6 x7
xy xy x2y x3y x4y x5y x6y x7y y x x2 x3 x4 x5 x6 x7 1
x2y x2y x3y x4y x5y x6y x7y y xy x2 x3 x4 x5 x6 x7 1 x
x3y x3y x4y x5y x6y x7y y xy X2y x3 x4 x5 x6 x7 1 x x2
x4y x4y x5y x6y x7y y xy x2y X3y x4 x5 x6 x7 1 x x2 x3
x5y x5y x6y x7y y xy x2y x3y X4y x5 x6 x7 1 x x2 x3 x4
x6y x6y x7y y xy x2y x3y x4y X5y x6 x7 1 x x2 x3 x4 x5
x7y x7y y xy x2y x3y x4y x5y X6y x7 1 x x2 x3 x4 x5 x6

Acta Didactica Napocensia, ISSN 2065-1430