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For the quaternion group $Q_8$ we have the presentation $$Q_8 = \big<a,b : a^4 = 1, a^2=b^2, b^{-1}ab=a^{-1} \big>$$.

Now knowing that for matrix $$ A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} $$ we have that $A^4 = I_2$ and also knowing that

$$ B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $$ we have that $B ^2 = A^2$. And we know that both $A,B \in GL(2,C)$ where $C$ is the field of complex numbers , we can define a representation $\phi:Q_8 \to GL(2,C)$ as follows $$\phi:a^ib^j \to A^iB^j \space \space(0 \leq i \leq 3,0 \leq j \leq 1)$$.

Now I want to show that this representation is irreducible by just observing the two matrices $A$ and $B$. Is it possible ?

Generally,My prof claimed that for any diagonal $n \times m$ matrix in the form $$ U = \begin{bmatrix} t_1 & 0 & 0 &..... & 0 \\ 0 & t_2 & 0 &..... & 0 \\ 0 & 0 & . &..... & 0 \\ 0 & 0 & 0 &. & 0 \\ 0 & 0 & 0 & 0 & t_n \\ \end{bmatrix} $$ where all $t_i's$ are all distinct and a Permutation $(n \times n)$ Matrix $P$.

Then if we have a group $G$ with generators $a,b$ and a representation $\delta$ such that $\delta:a^ib^j \to A^iP^j$ then $\delta$ is an irreducible represenation.

I just want to get why $\phi$ is irreducible first and then possibly attempt to generalise it.

Here is my attempt.

Since we are not interested in trivial vector subspaces $W$ so we disregard $\{0\}$ and $C^2$, so if a subspace was to be found then it should be of dimension $1$ and now the only possibility for a basis of $W$ is $u=(0,1)$ and $v=(1,0)$ and so we take $W = span(u)$ or $W = span(v)$

Now $ B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ which $\not\in W$ if we take $W = span(u)$ However we can check that $Av = \begin{bmatrix} 0 \\ -i \end{bmatrix}$ which $\in W$ and so as we can see it is variant under one matrix but not variant under the other and so the representation is irreducible because the same thing happen if we take $W = span(v)$ , it is true what I did ?

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If it was reducible, then there would be a $1$-dimensional subspace spanned by a common eigenvector of all group elements. The only eigenvectors of the matrix $A$ are multiples of $(1,0)$ and $(0,1)$ and these are not eigenvectors of $B$, so it must be irreducible.

This argument only generalizes to $\langle U,P \rangle$ if the $t_i$ are all distinct and the permutation matrix $P$ corresponds to an $n$-cycle. If the permutation had more than one cycle, then the basis vectors corresponding to that cycle would span a $G$-invariant subspace.

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