Conjugacy Classes in Groups

I am a math hobbyist working by myself. I am currently reading Beachy and Blair: "Abstract Algebra" on conjugacy in groups. I wish to ensure I have understood the notion of conjugacy classes. To do this I am considering the following group

$\displaystyle D_8$ = {1, r, $\displaystyle r^2$, $\displaystyle r^3$, s, sr, s$\displaystyle r^2$, s$\displaystyle r^3$}

and am looking at what would be required to determine the conjugacy class of r.

Beachy and Blair's definition (adapted for r and $\displaystyle D_8$) of the conjugacy class of r is

{g$\displaystyle \in$G| there exists an element a such that g = $\displaystyle {ar}a^{-1}$}

Now as far as I can see to determine the conjugacy class of r from first principles is quite lengthy - we are looking for g such that there exists an a such that g = $\displaystyle {ar}a^{-1}$ for some a

So start looking at possibiity g = 1

So we search through the elements of $\displaystyle D_8$ for an a such that 1 = $\displaystyle {ar}a^{-1}$

Try a = 1

1 = 1.r.$\displaystyle 1^{-1}$ ... No

Try a = r

a = r.r..$\displaystyle r^{-1}$ ... No

and so on to test all elements of $\displaystyle D_8$

Then go through the same process for g = r and so on through all the elements of $\displaystyle D_8$

Have I interpreted the definition correctly ... can someone affirm this or correct me please.

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Another issue is that the text by Eie and Chang: A Course in Abstract Algebra on page 87 write:

"The equivalence class of a under conjugation is called the conjugacy class containing a in G and is given by

[a] = {$\displaystyle {ga}g^{-1}$$\displaystyle \in$G | g$\displaystyle \in$G}"

This does not seem to be the same definition as Beachy and Blair since the set of all g such that g = $\displaystyle {ax}a^{-1}$ does not seem to me to be the same as the set of all $\displaystyle {ax}a^{-1}$

I note also that in Nicholson's book: "Introduction to Abstract Algebra", on page 349 he defines the conjugacy class of a $\displaystyle \in$ G as follows:

class a = {x$\displaystyle \in$| x is conjugate to a} = {$\displaystyle {ga}g{-1}$ | g$\displaystyle \in$G}

The second part of this definition is similar to Eie and Changs and seems different to the first part of the definition???

Can someone please clarify this?

Peter

Re: Conjugacy Classes in Groups

if a is in the conjugacy class of b then b is in the conjugacy class

$\displaystyle a = g b g^{-1}\Rightarrow g^{-1} a g = b $

let $\displaystyle h = g^{-1}$

$\displaystyle h a h^{-1} = b $

what dose that mean to find the conjugacy class of an element a we just have to find $\displaystyle gag^{-1} \; \forall g\in G $

for example to find the elements in conjugacy class of r

$\displaystyle 1 r 1 = r $ r in CC of r trivial

$\displaystyle sr^2 (r)sr^2 = sr^2 (r) r^2 s = srs = r^3 $ r^3 in CC of r

Re: Conjugacy Classes in Groups

the thing to remember is that conjugacy is an equivalence relation, so if there exists a,g such that gag^-1 = r, then a = g^-1rg.

so to find the conjugacy class of r, we can simply take grg^-1 for all g in G. now, there are some short-cuts we can take along the way:

if g is in Z(D8), then grg^-1 = rgg^-1 = r. in fact, we can even go further, if g is just in C(r), then we have grg^-1 = r as well.

now <r> is contained in C(r), so we needn't bother conjugating with any power of r, we know those all give the conjugate "r".

furthermore, we know that there are just as many distinct conjugates of r, as the index of the centralizer of r, which is [G:C(r)]

(this is what forms the basis of the "class equation").

now |C(r)| ≥ 4, which means that [G:C(r)] = 1 or 2. since s in not in C(r), C(r) ≠ D8, so C(r) = <r>.

this means we have just 2 distinct conjugates of r.

as Amer showed, r^3 is another conjugate of r, so the conjugacy class of r is: {r,r^3}.