Question

construct a group morphism of the first group table yo the second group table.
-write the images of the generators of your group .
- write the kernel
- find the image

Save Change table position ABCDE F GHI J KL MN OP QRSTU AKBCDE F GHI J KL MN O P QRS T U BB C D E F G KLMN H I STUOPQR CCDEFGABL MNHI J KPQRS TUO DDEFGABCNHI J KL MT UOP QR S EEF GABCDI J KL MNHQRS TUOP FFGABCDEKL MNHI J UOPQRST GABCDE F MNHI J KLRS TUOP Q HHI J KL MN O P QRS T U B C D E F G II J KL MNHQRSTUOPEFGABCD J J KL MNHI STUOPQRBCDE F G KKL MNHI JUOP QRS TE GABCDE L L MNHI J KP QRS TUOCDEF GAB MMNHI J KLRSTUOP QGABCDEF NNHI J KLMT UOP QR S DEF GABC ooP QRS TUABCDEF GHI J KL MN PPQRSTUOCDEF GABL MNHI JK QQRS TUOP EF GABCDI J KL MNH RRSTUOP QGABCDEF MNHI JKL SSTUOP QRBCDEF GAJ KL MNHI TTUOP QRS DEF GABCNHI J KL M Calculate Groep 2 ABCDE F GHI J KEMNOP CDBAGHFEKLJI OPN M DCABHGEF LKI J P OMN E F GHDCABNMP OKLJI EHGC DBAMNOP LKI J GHF E ABCDP OMNJ ILK HGEF BADCOP NMI J KL J KL MNOP ABCDEF GH ILKNMP C LJ I OPNMC DBA GH FE LKI J POMNDCABHGEF MNOP L KIJ FEHGCDBA NMPOKLİJ I EFGHDCA B OPNMI JKLİHGEFBADC P OMNJ ILKGHFEABCD MP OBADCFEHG

Answer 1

We need to first get the generators of the groups. Since its easier to get this for the second case(the table is symetric so it is an abelian group) Note the element indexed by A is the identity in both the cases. (In both the cases we shall use small scripts for the element indexed by its capital counterparts, in particular a is the identity). Denote the first table by G, and the second table by H Clearly e2-d,d' = b, b2-a → e8-a. Therefore we get the following (normal) subgroup of (rder 8: e3-h.e4-b, e-f, e-c, e-g. Now since there is no element of order 16, we know that there is a second generator of order 2 (order of the group is 16). We claim that t isi asi can't be generated by e and it is an element of order 2. Therefore the second group For the first group we know that there is no element of order 21, as the group is non-abelian. Therefore ba, ha are two generators of the group with the cominutor given by hbh 1 e = M. Hence the group is given Now we proceed to find φ : G → H, such that phi is a group honorphism To do this we only need to find the image of the generators of G. Now o(b)" = φ(67) = φ(a) = a. Therefore the order of (b) is either 7 or I. Since 6(b) H, the order of φ(b) divides 16. Hence φ(b-a. Silnilarly φ(h)-a as well because 3 X16. Hence the only homomorphism between the two groups is the trivial one. Therefore Ker(d) = G, Image(d)-(a) Note that two find a homomorphism from G to H, we didn't need to find H explicitly, only that the order of H is 16