# Show this is the only $2$-dimensional $\mathbb{C}$-representation of $Q_8$ up to isomorphism.

I'm very new to representation theory and trying to wrap my head around it. Specifically, I'm finding it hard to prove negative statements i.e. of the form "there does not exist X" because I'm finding the computations not too difficult but struggling with the theory.

By considering $$Q_8$$ as a subgroup of $$\mathbb{H}$$ which can be viewed as a $$2$$-dimensional vector space over $$\mathbb{C}$$ with basis $$\{1, j \}$$ I have arrived at the representation

$$A=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix},B=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

of $$Q_8$$. ($$A$$ is the matrix representing $$i$$, $$B$$ represents $$j$$)

And then to show this is irreducible follows from the fact that there's no eigenvectors in common between these two matrices.

But how do I prove the negative of there's no more $$2$$-dimensional irreducible $$\mathbb{C}$$-representations up to isomorphism? If you can present it in such a way, a method that works in more general circumstances would be very helpful.

• One way to proceed: every irreducible representation $V$ appears in the regular representation $\dim(V)$ times. There is the trivial representation, 3 sign representations, and 2 copies of the 2-dimensional representation you gave above. $1 + 3 + 2\cdot 2 = 8 = \#Q_8$, so those fill up the regular representation, showing that there's no room for other irreducible representations. Commented Oct 17, 2023 at 16:03
• @ViktorVaughn Thanks. I've not seen the result that every irreducible representation $V$ appears in the regular representation $\mathrm{dim}(V)$ times yet but after that it makes sense. I'll try think about how that could be proven. Commented Oct 17, 2023 at 19:24
• Another way is to use the fact that tensor powers of any faithful representation will contain every irreducible representation, decompose $\rho^{\otimes 2}$ into irreducibles (for $\rho$ the $2$-dimensional representation) and argue by induction. More generally, if you're trying to show that a set $S$ of representations contains all the irreducibles, it suffices to show that for some faithful $\rho$ and for each $\sigma \in S$, $\rho \otimes \sigma$ can be decomposed into elements of $S$. Commented Oct 18, 2023 at 7:30

Here's a hands-on approach. We will use one representation theory fact: recall for any complex finite-dimensional representation of a finite group $$G$$, every operator $$\rho(g)$$ is diagonalizable. (This follows from using Jordan canonical form from linear algebra.) Moreover, if $$g$$ has order $$n$$ then $$\rho(g)$$'s eigenvalues must evidently be $$n$$th roots of unity.

First let's consider $$\rho(-1)$$. Its eigenvalues must be $$\pm1$$. This gives three cases:

• $$\rho(-1)=\big[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\big]$$. This would mean all $$\rho(g)$$s commute (why?) hence must be scalar transformations (why?) so all 1D subspaces are subreps. Contradiction.
• $$\rho(-1)=\big[\begin{smallmatrix}1&\phantom{-}0\\0&-1\end{smallmatrix}\big]$$, up to change-of-basis. Since $$-1$$ commutes with all $$g$$s, $$\rho(-1)$$ commutes with all $$\rho(g)$$s, so this tells us the coordinate axes are subreps (why?). Contradiction.
• $$\rho(-1)=\big[\begin{smallmatrix}-1&\phantom{-}0\\\phantom{-}0&-1\end{smallmatrix}\big]$$. (Conclusion.)

Next let's consider $$\rho(\mathbf{i})$$. Its eigenvalues must be $$\pm i$$. We again have cases:

• $$\rho(\mathbf{i})=\pm\big[\begin{smallmatrix}i&0\\0&i\end{smallmatrix}\big]$$. This is central, so commutes with $$\rho(\mathbf{j})$$, a contradiction. (How?)
• $$\rho(\mathbf{i})=\big[\begin{smallmatrix}i&\phantom{-}0\\0&-i\end{smallmatrix}\big]$$, up to change-of-basis. (Basis change cannot affect $$\rho(-1)=\big[\begin{smallmatrix}-1&\phantom{-}0\\\phantom{-}0&-1\end{smallmatrix}\big]$$.)

Finally, consider $$\rho(\mathbf{j})=\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]$$. Apply $$\rho$$ to $$\mathbf{ij}=-\mathbf{ji}$$ and solve for $$a,b,c,d$$ for $$\rho(\mathbf{j})=\pm\big[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}\big]$$. Use $$\rho(\mathbf{i})$$ as a change-of-basis matrix if necessary to ensure $$\rho(\mathbf{j})=\big[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}\big]$$. Since $$\mathbf{i},\mathbf{j}$$ generate $$Q_8$$, this determines $$\rho$$, and we conclude $$\rho$$ must be the given matrix representation in some choice of coordinates.

Expanding on my comment: one can use fundamental results about the regular representation $$\newcommand{\Vreg}{V_{\operatorname{reg}}} \Vreg$$ of a group $$G$$ to show that there are no other $$2$$-dimensional irreducible representations of $$Q_8$$. (For reference, see this post or $$\S18.2$$ of Dummit and Foote.)

• The number of irreducible representations of $$G$$ (up to isomorphism) is the equal to the number of conjugacy classes of $$G$$.
• Every irreducible representation $$V$$ of $$G$$ occurs in $$\Vreg$$ with multiplicity $$\dim(V)$$. So letting $$V_1, \ldots, V_r$$ be the irreducible representations of $$G$$ with dimensions $$\dim(V_i) = n_i$$, then $$\Vreg \cong V_1^{\oplus n_1} \oplus \cdots \oplus V_r^{\oplus n_r} \, .$$ In particular, taking dimensions yields \begin{align} \#G &= \dim(\Vreg) = \dim\left(V_1^{\oplus n_1} \oplus \cdots \oplus V_r^{\oplus n_r}\right) = n_1 \dim(V_1) + \cdots + n_r \dim(V_r)\\ &= n_1^2 + \cdots + n_r^2 \, . \tag{*} \label{eqn:irreps} \end{align}

In the case of $$Q_8$$, $$\Vreg$$ has dimension $$\#Q_8 = 8$$. The $$2$$-dimensional representation you gave above accounts for $$2 \cdot 2 = 4$$ dimensions, and the trivial representation for $$1$$ more, leaving a $$3$$-dimensional space yet to be determined. Another $$2$$-dimensional irrep would give another $$4$$-dimensional space, which is too big. Thus all other irreps must be $$1$$-dimensional. Or we could instead conclude by noting that $$Q_8$$ has $$5$$ conjugacy classes, hence $$5$$ irreps. Thus equation (\ref{eqn:irreps}) becomes $$8 = 1^2 + n_2^2 + n_3^2 + n_4^2 + 2^2 \implies 3 = n_2^2 + n_3^2 + n_4^2 \,$$ and hence we must have $$n_2 = n_3 = n_4 = 1$$.

We can even see more directly that $$Q_8$$ has four $$1$$-dimensional irreps. Since $$\operatorname{GL}_1(\mathbb{C}) = \mathbb{C}^\times$$ is abelian, every $$1$$-dimensional representation of $$G$$ factors through the abelianization $$\newcommand{\Gab}{G^{\operatorname{ab}}} \Gab = G/[G,G]$$. One can show that $$[Q_8, Q_8] = \langle -1 \rangle$$ (for instance, $$i j i^{-1} j^{-1} = ijij = i(-ij)j = -i^2 j^2 = -1$$ and one can compute the other commutators similarly), so $$(Q_8)^{\operatorname{ab}} \cong C_2 \times C_2$$. As we can choose either the trivial or sign representation on each factor of $$C_2 \times C_2$$, we conclude that $$Q_8$$ has four $$1$$-dimensional irreducible representations (including the trivial representation).