Factor Groups: Conjugation & 2 Conjugates

[Ala']

Factor groups!

Please I just want to ask about factor groups.. how could a factor group G/A acts on A by conjugation, knowing that A is a normal & abelian subgroup of G..

and what do we mean when we say that an element in a group has jus 2 conjugates??

thanks in advance :)

 

[Deveno]

the only possible way that makes sense is to define:

(gA)(a) = gag^-1. note that gag^-1 is in A, since A is normal.

however, we have to check that:

if g'A = gA, then g'ag'^-1 = gag^-1

(that is, that our action is well-defined).

now if g'A = gA, g' = ga' for some a' in A. thus:

g'ag'^-1 = (ga')a(ga')^-1

= ga'aa'^-1g^-1

but since A is assumed abelian, aa'^-1 = a'^-1a, so:

ga'aa'^-1g^-1 = ga'a'^-1ag^-1

= geag^-1 = gag^-1, which was to be proven.

*******

to say an element x has just two conjugates, means that gxg^-1 has just two possible values (one of which is obviously x), no matter what g is.

for example, consider the group Q = {1,-1,i,-i,j,-j,k,-k} under quaternial multiplication.

(i² = j² = k² = -1, ij = k, jk = i, ki = j, and
ji = -k, kj = -i, ik = -j)

(1)(i)(1) = i
(-1)(i)(-1) = i
(i)(i)(-i) = i
(-i)(i)(i) = i
(j)(i)(-j) = (-k)(-j) = kj = -i
(-j)(i)(j) = (-j)k = -i
(k)(i)(-k) = j(-k) = -i
(-k)(i)(k) = (-k)(-j) = kj = -i

the only conjugates i has is i, and -i.

 

[Ala']

Thank you very much for your clear answer, it was really very useful for me :)