# Classyfing the simple submodules of $\mathbb{C[Q_8]}$

I'm taking a course in abstract algebra, at the moment we are working on the Wedderburn theorem and I'm still unable to understand how to use it properly to solve exercises. I tried to solve a problem and explain my reasoning to see where are the holes in my understanding and to see if there's some other ways of thinking about it or suggestions for these type of problems.

The problem I tried to solve is the following:

Find the simple submodules of $$\mathbb{C[Q_8]}$$ (up to isomorphism).

First by Maschke's theorem I know that the group algebra is semisimple because char($$\mathbb{C}$$) doesn't divide char($$\mathbb{Q_8}$$). This means I can now use Wedderburn's theorem as it's a finite dimensional algebra. By this theorem I have the following decomposition: $$\mathbb{C[Q_8]} \simeq \prod_{i \in I} M_{n_i}(\mathbb{C})$$ As it's an isomorphism, they both must have the same dimensions over $$\mathbb{C}$$. The dimension of $$\mathbb{C[Q_8]}$$ over $$\mathbb{C}$$ is 8. This implies that the only possible dimensions of the matrices algebras are $$n_1=2,n_2=1,n_3=1,n_4=1,n_5=1$$ because otherwise if $$n_i = 1$$ for $$i \in \{ 1,...,8\}$$ this would imply $$\mathbb{C[Q_8]}$$ is commutative which would imply $$\mathbb{Q_8}$$ is abelian as a group. The other cases $$n_1 = 2, n_2 = 2$$ cannot happen as there is always a submodule of a group algebra of dimension 1 (the one generated by $$\sum_{h \in \mathbb{Q_8}} h$$, is this correct?).

To find these subspaces isomorphic to $$\mathbb{C}$$ I considered the $$x \in \mathbb{C[Q_8]}$$ such that $$i*x, j*x, k*x$$ is a complex multiple of $$x$$. From these I found the four subspaces, one of them being the one that's guaranteed to exist. My problem is I can't seem to find the subspace isomorphic to $$M_2(\mathbb{C})$$. Any tips and help would be greatly apreciated.

I tried to solve the problem in another way. I read on the internet but couldn't prove it that $$dim_{\mathbb{C}} Z(\mathbb{C[Q_8]}) = \# \text {conjugacy classes of \mathbb{Q_8}}$$. I haven't worked that much with conjugacy classes but I gave it a try in this problem as I have seen that also, $$dim_{\mathbb{C}} Z(\mathbb{C[Q_8]}) = \# \text {simples in the Wedderburn decomposition }$$. By this I understood the number of matrices algebras in the product that is isomorphic to the algebra $$\mathbb{C[Q_8]}$$. In this particular case, the conjugacy classes I found are 5: the conjugacy class of $$\{1\}$$ and $$\{-1\}$$ as they both are in the center of the group. For each $$i,j,k$$ I found they are on their own conjugacy class. So indeed, the number of conjugacy classes is the same as the number of matrices algebras in the product, i.e 5. One question I have is if from these conjugacy classes I can get more information about these matrices algebras or they just determine the number of them in the product?

I would very happy if someone could point me to notes or books that explain these theorem and if possible with exercises similar to the one I asked.

• The notation $\Bbb H$ is standard for the algebra of quaternions, and $Q_8$ is standard for the quaternion group of order $8,$ but I have never ever seen $\Bbb H$ used for the quaternion group of order $8$. Does the source of the exercise really use that strange notation? – anon May 30 '18 at 23:21
• Actually, you're right. The notation is not this "blackboard" H from latex I mistankly used but still it's an "cursive" H, if you will, which is the quaternion group of order 8. – Leo Lerena May 30 '18 at 23:33
• Maybe I should change it to $Q_8$ as to avoid further mistakes. – Leo Lerena May 30 '18 at 23:34

This is the standard notation that I will abide be:

• $\mathbb{H}$ denotes the quaternions, a $4$-dimensional $\mathbb{R}$-algebra
• $Q_8$ denotes the quaternion group of order $8$

Thus, $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ is a finite subgroup of $\mathbb{H}^{\times}$.

Over here is a nice proof that the multiplicity of an irrep $V$ in the decomposition of $\mathbb{C}[G]$ (as a representation) equals $\dim V$, which implies $\sum (\dim V)^2=|G|$, and also the formula

$$e_{\small V}=\frac{\dim V}{|G|}\sum_{g\in G}\chi_V(g^{-1})g$$

where $e_{\small V}$ is the element of $\mathbb{C}[G]$ which acts as the identity on $V$ and the zero map on all of the other irreps. Then the simple subalgebras of $\mathbb{C}[G]$ are generated by these (just as in $\bigoplus\mathrm{End}(V)$).

All irreps show up as simple submodules of $\mathbb{C}[G]$ up to isomorphism. (Not up to isomorphism there can be infinitely many. For instance, even though $\mathbb{R}^2\cong\mathbb{R}\oplus\mathbb{R}$ as an $\mathbb{R}$-module has only two summands in a direct sum decomposition, it has infinitely many isomorphic simple submodules.)

The irreps of a cyclic group are easy: they are just homomorphisms into $\mathbb{C}^{\times}$. We can also use these to build all irreps of a finite abelian group, using their classification. While $Q_8$ is not abelian, it has three normal subgroups of index two, namely $\langle i\rangle$, $\langle j\rangle$ and $\langle k\rangle$, and so the quotient is $C_2$, which has the sign representation. (Any rep of $G/H$ may be composed with the projection $G\to G/H$ to obtain a rep of $G$.) This yields three inequivalent $1$-dimensional representations.

Another representation is the "standard" one. We know $Q_8$ is a subgroup of the quaternion algebra, and also that $\mathbb{H}$ is a two-dimensional vector space over $\mathbb{C}$. (By "right" I mean we apply complex scalars on the right side of a quaternion in $\mathbb{H}$, so that it doesn't interfere with multiplying by elements of $Q_8$ on the left.) Can you verify this is an irrep?

Finally there's the trivial rep. Since $1^2+1^2+1^2+1^2+2^2=|Q_8|$, these are all of the irreps of $Q_8$.

By computing the character of the $2$-dim irrep, you can get the corresponding idempotent, then multiply it by elements to span a $4$-dimensional subalgebra of $\mathbb{C}[Q_8]$ isomorphic to $M_2(\mathbb{C})$.

By the way, $\dim Z(\mathbb{C}[G])$ is the number of conjugacy classes, because $x\in Z(\mathbb{C}[G])$ is equivalent to $gx=xg$ or in other words $gxg^{-1}=x$ for each $g\in G$, which implies the coefficients in $x=\sum c_gg$ are constant on orbits of conjugation-by-$g$, i.e. conjugacy classes, so the elements $\sum_{k\in K}k$ (where $K$ is a conjugacy classes) form a basis.

• Thanks for the answer!! I haven't studied this in the context of representations so I'm still trying to understand the solutions (in my course we defined them but we haven't used them much). I will get back to you and your answer whenever I understand this better, it seems very interesting! – Leo Lerena May 31 '18 at 0:45
• I have been reading from this in order to better understand your answer: maths.gla.ac.uk/~abartel/docs/reptheory.pdf – Leo Lerena May 31 '18 at 2:56
• @LeoLerena $e_{\small V}$ is the element of $\Bbb C[G]$ that acts as the identity map on $V$ and the zero map on all the other irreps. (I.e. it corresponds to the element of $\bigoplus {\rm End}(W)$ which is $\rm Id$ in the $V$ summand and $0$ in all the other summands). By "do these calculations" do you mean compute the character of a representation? For the $2\times 2$ case, look at $\Bbb H$, form a basis for it as a right vector space over $\Bbb C$ (I'd choose $\{1,j\}$), compute the matrix corresponding to the elements, then find their traces. – anon May 31 '18 at 3:21
• If you do the same thing for left-multiplication-by-$i$, you should get the matrix $\left[\begin{smallmatrix} i & 0 \\ 0 & -i\end{smallmatrix}\right]$ which also has trace $0$. In fact all but $\pm1\in Q_8$ will have trace $0$, so the idempotent is $(1)+(-1)$ (where $-1$ is interpreted as an element of $Q_8$, not the additive inverse of $1$). Perhaps we should use $\epsilon$ for the order-two element of $Q_8$, so we can call the idempotent $1+\epsilon$. Multiply it by each element of $Q_8$ and delete copies and you'll have a basis for a copy of $M_2(\Bbb C)$ in $\Bbb C[Q_8]$. – anon May 31 '18 at 3:31
• Right, irrep=irreducible representation, and having no (proper, nontrivial) invariant subspaces is equivalent to every element of the module generating the whole thing. (Hint: you can get $1\in\Bbb H$ from any $x\in\Bbb H$ by multiplying by a quaternion, since every $x$ is invertible in $\Bbb H$, and then you can find an element of $\Bbb R[Q_8]$ which also turns $x$ into $1$...) – anon May 31 '18 at 3:57