# Show that the matrix representation of the dihedral group D4 by M is irreducible.

• July 28th 2011, 02:42 PM
blueyellow
Show that the matrix representation of the dihedral group D4 by M is irreducible.
Show that the matrix representation of the dihedral group D4 by M is irreducible.

You are given that all of the elements of a matrix group M can be generated
from the following two elements,

A=
|0 -1|
|1 0|

B=
|1 0|
|0 -1|

in the sense that all other elements can be written A^n B^m for integer m, n >or= 0.
Find the remaining elements in M.
• July 28th 2011, 03:39 PM
NonCommAlg
Re: Show that the matrix representation of the dihedral group D4 by M is irreducible.
Quote:

Originally Posted by blueyellow
Show that the matrix representation of the dihedral group D4 by M is irreducible.

You are given that all of the elements of a matrix group M can be generated
from the following two elements,

A=
|0 -1|
|1 0|

B=
|1 0|
|0 -1|

in the sense that all other elements can be written A^n B^m for integer m, n >or= 0.
Find the remaining elements in M.

the representation is irreducible because $A$ and $B$ have no common eigenvectors. see the theorem in my blog.
you should do the second part of the problem yourself. note that $0 \leq m \leq 1$ and $0 \leq n \leq 3.$
• July 31st 2011, 07:54 PM
Drexel28
Re: Show that the matrix representation of the dihedral group D4 by M is irreducible.
Quote:

Originally Posted by blueyellow
Show that the matrix representation of the dihedral group D4 by M is irreducible.

You are given that all of the elements of a matrix group M can be generated
from the following two elements,

A=
|0 -1|
|1 0|

B=
|1 0|
|0 -1|

in the sense that all other elements can be written A^n B^m for integer m, n >or= 0.
Find the remaining elements in M.

Alternatively, you could do the second part first (which is trivial) and then (assuming you are talking about $\mathbb{C}$-representations) appeal to the common theorem that a representation is irreducible if and only if the inner product of the character with itself is one, i.e. if $\displaystyle \frac{1}{|G|}\sum_{g\in G}|\chi_M(g)|^2=1$.