Subsection 1.3.1 Representations
Now, we will be exploring some unique representations and their uses specifically within the context of abelian groups. We will introduce and derive them from examples and additionally prove that every cyclic group, \(C_{m} = \{g\,|\,g^{m}=1\}\text{,}\) has unique one-dimensional representations of the its group action in the form of,
\begin{equation*}
\rho: C_{m} \mapsto \mathbb{C}.
\end{equation*}
We will prove that these representations can be given by the m-th roots of unity which may be intuitive by \(g^{m} = 1\text{.}\) This will then be generalized to abelian groups. Within this exploration we will examine this through the lens of other areas of algebra such as invariant theory. Now, let us start with an example of representations which we will keep track of throughout the paper with reference to what we are exploring.
\(GL_{n}\)\(S_{3}\text{.}\)\(GL_{3}\)
\begin{equation*}
\begin{align*}S_{n}&= \{(),(1 \,2),(1\,3),(2\,3),(1\,2\,3),(1\,3\,2)\} \\&\mapsto \{I, \begin{pmatrix}0 & 1& 0\\ 1 & 0& 0\\ 0 & 0& 1\end{pmatrix},\begin{pmatrix}0 & 0& 1\\ 0 & 1& 0\\ 1 & 0& 0\end{pmatrix},\begin{pmatrix}1 & 0& 0\\ 0 & 0& 1\\ 0 & 1& 0\end{pmatrix},\begin{pmatrix}0 & 1& 0\\ 0 & 0& 1\\ 1 & 0& 0\end{pmatrix},\begin{pmatrix}0 & 0& 1\\ 1 & 0& 0\\ 0 & 1& 0\end{pmatrix} \}\\&= \left\langle \begin{pmatrix}0 & 1& 0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}, \begin{pmatrix}0 & 1& 0 \\ 1&0&0 \\ 0&0&1\end{pmatrix} \right\rangle.\end{align*}
\end{equation*}
\(\langle \rho (1\,2\,3)\rangle = \left\langle \begin{pmatrix}0 & 1& 0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}\right\rangle.\)Now let us examine some preliminaries that will be necessary for further exploration of the topic.
faithfully\(Stab(x)\)\(x_{0} \in X\)
\begin{equation*}
Stab(x_{0}) = \{gx_{0}=x_{0}|g \in G\}
\end{equation*}
\(n\text{,}\)\(GL_{n}\text{.}\)
Proof.
Say we have a cycle \(c\) of length \(n\text{,}\) then as a cycle will permute any element to one and only one element uniquely. Thus for \(c\) we can give a representation
\begin{equation*}
\rho (c) \mapsto \sigma I.
\end{equation*}
We can permute the identity matrices columns to give a representation of any cycle. This representation will be with \(GL_{n}\) by definition. In addition, observe that for \(\rho\) we have a homomorphism so the identity maps to the identity.
Now we use something called circulant matrices to represent our cyclic groups and these matrices have some special properties.
\(n\times n\)
\begin{equation*}
C =\begin{pmatrix}a_{1}&a_{2}&\cdots&a_{n}\\ a_{n}&a_{1}&&\vdots \\ \vdots&&\ddots\\ a_{2}&\cdots&&a_{1}\end{pmatrix}
\end{equation*}
\(tr(C) = a_{1}*n\text{.}\)For our permutation matrices there will be a one in every row that is shifted once every row. We will be using this result implicitly for much of this paper.
\begin{equation*}
\begin{pmatrix}0 & 1 & 0\\ 0& 0& 1 \\ 1& 0& 0\end{pmatrix}
\end{equation*}
\((1 \ 2 \ 3)\)We now introduce new machinery into our representations! Complex numbers, specifically primitive roots, provide very good representations of especially abelian groups. We can start to derive results by examining this next corollary.
Proof.
We will derive our result through examining cyclic groups. Say we have the cyclic group \(C_{m} = \{g\,|\,g^{m}=1\}\) then we will prove that we have a representation of the form
\begin{equation*}
\rho: C_{m} \mapsto \mathbb{C}.
\end{equation*}
This is in some ways pretty intuitive by introducing the roots of unity. Observe that if the subgroup is cyclic of prime order then it has a generating element \(g\) such that \(\langle g\rangle = G\) and \(rho\) being a faithful homomorphism implying that its representation group within \(GL_{n}\) will also be cyclic generated by \(\rho (g)\) such that \(\langle \rho(g)\rangle = \rho(G)\text{.}\) From this we can examine the eigenvalues and vectors of the generating matrix and these will hold for all elements of the group due to it being cyclically generated. Now we need to prove that these eigenvalues will be roots of unity and the eigenvectors will extend this representation. Say for \(\rho(g) = A\) which will imply that for \(\rho(g^{n}) = A^{n}\) this means that if we have eigenvectors \(\vec v\) and eigenvalues \(\lambda\) of this matrix we then have,
\begin{equation*}
A^{n} \vec v = \lambda^{n} \vec v
\end{equation*}
as the eigenvectors remain directionally invariant. Thus we have \(\rho(g^{n}) = \lambda^{n}\) and we have a representation \(\rho: C_{m} \mapsto \mathbb{C}\) as eigenvalues are complex by nature of \(\rho\) being a homomorphism and the identity mapping to the identity
\begin{equation*}
\rho(g)^{m} = \rho(g^{m}) = 1,
\end{equation*}
so then \(G\) must be generated by the m-th root of unity as solutions to the characteristic polynomial.
Another proof of this lies in this next theorem.
\begin{equation*}
\det \left ( \begin{pmatrix}a_{1}&a_{2}&\cdots&a_{n}\\ a_{n}&a_{1}&&\vdots \\ \vdots&&\ddots\\ a_{2}&\cdots&&a_{1}\end{pmatrix}\right ) = \displaystyle\prod_{j=0}^{n-1}(a_{1} + \lambda^{j} a_{2} + \cdots+\lambda^{(n-1)j}a_{n})
\end{equation*}
\(\lambda \in \mathbb{C}\)\(nth\)The proof of this theorem is solely a calculation of the determinant and will thus be excluded. However, we will use this result to show our above corollary once more.
Proof.
Say we have a permutation matrix that generates a cyclic group, then we will have a circulant matrix where in each row there is precisely one entry which is a one. This gives that the determinant will then be a root of unity and by the fact that \(\rho\) is homomorphism the representation is injective. Thus we have a one dimensional representation that is unique.
\begin{equation*}
G = \left\langle \begin{pmatrix}0 & 1& 0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}\right\rangle\text{.}
\end{equation*}
\(\begin{bmatrix}a \\ b \\ c\end{bmatrix}\text{.}\)
\begin{equation*}
\det \left( \begin{pmatrix}0 & 1& 0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}- \lambda I \right) = \lambda^{3}-1 = (\lambda -1)(\lambda^{2} + \lambda+1)= 0
\end{equation*}
\(\lambda = 1, e^{\frac{2\pi}{3}i},e^{\frac{4\pi}{3}i}= \langle e^{\frac{2\pi}{3}i}\rangle\)\(G\)
\begin{equation*}
\vec v = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}e^{\frac{4\pi}{3}i}\\ e^{\frac{2\pi}{3}i}\\ 1\end{bmatrix},\begin{bmatrix}e^{\frac{2\pi}{3}i}\\ e^{\frac{4\pi}{3}i}\\ 1\end{bmatrix}.
\end{equation*}
Proof.
Observe that because we have a finite abelian group \(g,k \in G\) implies \(gk= kg\) which gives that every subgroup is normal as
\begin{equation*}
\begin{align*}gk&=kg\\ gkg^{-1}&= gg^{-1}k \\&= k.\end{align*}
\end{equation*}
Thus all of its Sylow P-subgroups \(P_{i}\) are normal. This gives that we can write the group as a product of its Sylow P-subgroups,
\begin{equation*}
G = \prod P_{i}
\end{equation*}
Additionally, the order \(|P_{i}| = p^{k}\) for some prime \(p\text{.}\) From here we take that every \(P_{i}\) has subgroups and all of these subgroups are normal as \(G\) is abelian. This gives that we can represent the group as a product of its subgroups. However, a subgroups order can only ever divide the order of the group by Lagrange’s theorem. Thus all of the subgroups will inductively have minimal proper subgroups of order \(p\text{.}\) Any group of order \(p\) is cyclic. This means that the whole group can be written as a product of its minimal subgroups which are cyclic.
This theorem is going to be critical for proving the case of general abelian groups as now we have the ability to see abelian groups through two lenses:
As their product of cyclic groups.
As the character of a matrix and eigenvalues and vectors associated with them.
Both of these are simply different things on a surface level. Computationally, it is easier to use matrices and eigenvalues. Specifically, the characteristic polynomial is very familiar,
\begin{equation*}
p(\lambda) = \det(A - \lambda I)
\end{equation*}
Generally, this gives us some polynomial of indeterminant \(\lambda\) that will have solutions in the complex field. However, specifically for permutation matrices we have a correlation where we will generally have eigenvalues of the form \(1, e^{\frac{2\pi}{n}i},...\) for n being the degree of the matrix. Similarly, the eigenvectors will have entries of the same form.
Now, there are a few special things to mention that lead us to the action eigenvalues cause the eigenvectors to be invariant under then forms their stabilizer group. Eigenvectors corresponding to different eigenvalues of a symmetric matrix are orthogonal. Moreover, the eigenvectors span a ”one-dimensional” space which is helpful for simplifying representations. This means that we will have irreducible one dimensional complex representations for any abelian group!
\(G\)\((V,\rho)\)\(G\)
\(V\) is a finite basis
\(G\) is finitely generated
\(V\)Proof.
First, lets state all of the preliminaries for this proof
Thus as we have proven it for the cyclic case and then we can use induction to show that it is true for the products of cyclic groups which by the fundamental theorem of finite abelian groups are all abelian groups. Say we have an abelian group generated by cyclic groups of prime order, \(C_{p} = \{g\,|\,g^{p}=1\}\text{,}\) then we have,
\begin{equation*}
G = \langle g_{1}\rangle\times \langle g_{2}\rangle\times...\times\langle g_{n}\rangle
\end{equation*}
and define \(\rho: C_{p} \mapsto \mathbb{C}\) then for \(\rho\langle g_{i}\rangle = \mathbb{C}_{i}\) for \(1\leq i \leq n\text{.}\) This gives
\begin{equation*}
\rho(G) = \langle \mathbb{C}_{1}\rangle\times \langle \mathbb{C}_{2}\rangle\times...\times\langle \mathbb{C}_{n}\rangle
\end{equation*}
as each of these are roots of unity we know have,
\begin{equation*}
\rho(G) = \langle \mu_{1} \rangle\times \langle \mu_{2}\rangle\times...\times\langle \mu_{n}\rangle
\end{equation*}
for \(\mu_{i}\) being distinct roots of unity. Now we know that \(\mu_{i}\times \mu_{j} = \mu_{k}\text{,}\) thus abelian group representations are made of only roots of unity. Observe for distinct roots of unity of prime order their product will be distinct or \(e^{\frac{2\pi}{k_{1}}i}*e^{\frac{2\pi}{k_{2}}i}= e^{\frac{2\pi(k_{1}+k_{2})}{k_{1} k_{2}}i}\) for \(k_{1},k_{2}\) being primes, thus as the bottom product is unique, \(\rho\) is injective.
\begin{equation*}
\mathbb{Z}_{2} \times \mathbb{Z}_{2} \cong G = \{e, (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4) (2 \, 3)\}
\end{equation*}
\(\rho_{1}: G \mapsto\text{,}\)
\begin{equation*}
\rho(G)= \left\{ I,\begin{pmatrix}0 & 1& 0 &0\\ 1&0&0&0 \\ 0&0&0& 1\\ 0 &0&1&0\end{pmatrix},\begin{pmatrix}0 & 0& 1 &0\\ 0&0&0&1 \\ 1&0&0& 0\\ 0 &1&0&0\end{pmatrix}, \begin{pmatrix}0 & 0& 0 &1\\ 0&0&1&0 \\ 0&1&0& 0\\ 1 &0&0&0\end{pmatrix} \right\}
\end{equation*}
\(t^{2} -1\)\(\rho_{2}: \rho_{1}(G) \mapsto \mathbb{C}\)\(\pm 1\)\(\begin{bmatrix}1 \\ 1 \\ 1\\ 1\end{bmatrix},\begin{bmatrix}1 \\ -1 \\ 1\\ -1\end{bmatrix},\begin{bmatrix}1 \\ 1 \\ -1\\ -1\end{bmatrix},\begin{bmatrix}-1 \\ 1 \\ 1\\ -1\end{bmatrix}.\)
\begin{equation*}
\mathbb{Z}_{3} \times \mathbb{Z}_{2} \cong G = \{(e, (1\, 3\, 5)(2\, 4\, 6),(1\ 5\ 3)(2\ 6\ 4),(1\ 4)(2\ 5)(3\ 6),(1\ 6\ 5\ 4\ 3\ 2),(1\ 2\ 3\ 4\ 5\ 6)\}.
\end{equation*}
\(\rho_{1}: G \mapsto\)
\begin{equation*}
\rho(G) = \left\langle \begin{pmatrix}0 & 0& 1& 0 &0&0\\ 0 & 0& 0 &1 &0&0\\ 0 & 0& 0&0 &1&0\\ 0 &0&0&0 & 0& 1\\ 1 &0&0&0 & 0& 0\\ 0 &1&0&0 & 0& 0\end{pmatrix}, \begin{pmatrix}0 & 0& 0& 1 &0&0\\ 0& 0& 0 &0 &1&0\\ 0 & 0& 0&0 &0&1\\ 1 &0&0&0 & 0& 0\\ 0 &1&0&0 & 0& 0\\ 0 &0&1&0 & 0& 0\end{pmatrix}\right\rangle
\end{equation*}
\(\rho_{2}: \rho_{1}(G) \mapsto \mathbb{C}\)
\begin{equation*}
\rho_{2}\begin{pmatrix}0 & 0& 1& 0 &0&0\\ 0 & 0& 0 &1 &0&0\\ 0 & 0& 0&0 &1&0\\ 0 &0&0&0 & 0& 1\\ 1 &0&0&0 & 0& 0\\ 0 &1&0&0 & 0& 0\end{pmatrix} \mapsto\lambda = \pm 1 \,\,\,\,\,\,\,\,\vec v = \begin{bmatrix}1 \\ 0 \\ 1\\ 0 \\ 1\\ 0\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 0\\ 1 \\ 0\\ 1\end{bmatrix},\begin{bmatrix}0 \\ e^{\frac{2\pi}{3}i}\\ 0\\ e^{\frac{4\pi}{3}i}\\ 0\\ 1\end{bmatrix},\begin{bmatrix}0 \\ e^{\frac{4\pi}{3}i}\\ 0\\ e^{\frac{2\pi}{3}i}\\ 0\\ 1\end{bmatrix},\begin{bmatrix}e^{\frac{2\pi}{3}i}\\ 0 \\ e^{\frac{4\pi}{3}i}\\ 0 \\ 1\\ 0\end{bmatrix},\begin{bmatrix}e^{\frac{4\pi}{3}i}\\ 0 \\ e^{\frac{2\pi}{3}i}\\ 0 \\ 1\\ 0\end{bmatrix}
\end{equation*}
\begin{equation*}
\rho_{2}\begin{pmatrix}0 & 0& 0& 1 &0&0\\ 0& 0& 0 &0 &1&0\\ 0 & 0& 0&0 &0&1\\ 1 &0&0&0 & 0& 0\\ 0 &1&0&0 & 0& 0\\ 0 &0&1&0 & 0& 0\end{pmatrix} \mapsto\lambda = \langle e^{\frac{2\pi}{3}i}\rangle\,\,\,\,\,\,\,\, \vec v = \begin{bmatrix}1 \\ 0 \\ 0\\ 1 \\ 0\\ 0\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ 0\\ -1 \\ 0\\ 0\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 0\\ 0 \\ 1\\ 0\end{bmatrix},\begin{bmatrix}0 \\ -1 \\ 0\\ 0 \\ -1\\ 0\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\\ 0 \\ 0\\ 1\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\\ 0 \\ 0\\ -1\end{bmatrix}
\end{equation*}
Within this paper we have used a matrix representation of the group action to then use the determinant to calculate eigenvalues and eigenvectors of another representation over a vectors space. We can pretty effectively calculate the eigenvalues of the system by saying the characteristic polynomial of the permutation has roots that are equal to eigenvalues and thus a representation. We run into the issue of how to view the vector space that it acts on. We can make some conclusions by examining the amount of elements being permuted and the disjoint cycles they are in. Then we can transfer this information into the representation of a matrix. From there we can use the eigenvalues to compute the vector space of the representation.
