Abstract Algebra: Finding Conjugates

[The_Iceflash]

Homework Statement

Consider this group of six matrices:

Let G = {I, A, B, C, D, K}, Matrix Multiplication>

$$I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$ $$A =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$ $$B =\begin{bmatrix}0 & 1\\-1 & -1\end{bmatrix}$$

$$C =\begin{bmatrix}-1 & -1\\0 & 1\end{bmatrix}$$ $$D =\begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix}$$ $$K =\begin{bmatrix}1 & 0\\-1 & -1\end{bmatrix}$$

Operation Table for this group:

_|I A B C D K
I |I A B C D K
A|A I C B K D
B|B K D A I C
C|C D K I A B
D|D C I K B A
K|K B A D C I

Define $$f:G\rightarrow$$ $$\left\langle\(R^{*}, \bullet\right\rangle$$ by f(x) = det(x) for any Matrix x $$\in$$ G.Finally, the question. Haha.

Find all the conjugates of A:
Find all the conjugates of B:

Homework Equations

N/A

The Attempt at a Solution

Now I know this isn't that hard of a concept and I understand what a conjugate is but I don't know how to find them. Any help on how I would go about finding them is greatly appreciated.

 

[ABarrios]

For conjugates of A,

G={I, A, B, C, D, K}

I^-1*A*I=A
B^-1*A*B=D*C=K
C^-1*A*C=C*B=K
D^-1*A*D=B*K=C
K^-1*A*K=K*D=C

So K and C are conjugates of A.

Do same procedure to find conjugates of B.

 

[Martin Rattigan]

Or given the fact that you've worked out the multiplication table, you can see by inspection that it's a nonabelian group of of order 6. (You don't need to check the associative law because that's taken care of by the elements being real matrices under multiplication.)

That means it's isomorphic to S₃ and the conjugacy classes are
(i) the identity
(ii) the elements of order 2
(iii) the rest

The identity is I of course.
The elements of order 2 can be read off by finding I on the main diagonal, so A,C,K.
And the rest is B,D.

Of course you need to have proved that there is only one nonabelian group of order 6, and determined the conjugacy classes in S₃ first.

 

[The_Iceflash]

Ok I understand that. Thanks. One question though, can A or B be considered a conjugate of either x^-1Ax or x^-1Bx respectively?

 

[ABarrios]

Yes. For take the identity element, I^-1*A*I=A So A is a conjugate of A, and likewise B is a conjugate of itself.