# Irreducible representation of finite Abelian group

I have seen that every finite Abelian group $$G$$ is isomorphic to a product of cyclic groups of prime power order, that is

$$G = \mathbb{Z}_{p_1} \times \mathbb{Z}_{p_2} \times ... \times \mathbb{Z}_{p_n} ,$$

where $$\mathbb{Z}_{p_i}$$ is the group $$\{ 0, 1, ..., p_i-1 \}$$ with addition modulo $$p_i$$. So if I understand it correctly, you can "represent" (it is not a representation, but a way to think about) the elements of $$G$$ as arrays

$$(a_1, a_2, ..., a_n)$$

with $$a_i \in \mathbb{Z}_{p_i}$$, and the group operation corresponds to vector addition. This "representation" establishes a group isomorphism.

My question: how do you match this "representation" of $$G$$ with a true irreducible representation of $$G$$, which, since $$G$$ is Abelian, must be a one-dimensional representation ? In other words, if you represent each element $$g\in G$$ as a phase $$e^{i \theta}$$, how do you write $$\theta$$ in terms of the array $$(a_1, a_2, ..., a_n)$$?

NOTE that I am only interested in representations $$\rho: G \rightarrow GL(V)$$ with $$V$$ a vector space over $$\mathbb{C}$$ or $$\mathbb{R}$$. In particular, the above one-dimensional representation is over a complex vector field.

EXAMPLE: Consider the group $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ (in the above decomposition of $$G$$ repetition of primes is allowed). This group has four elements of order 2 under addition:

$$\mathbb{Z}_2 \times \mathbb{Z}_2 = \{ (0,0), (0,1), (1,0), (1,1) \}.$$

On the other hand we know $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ is Abelian, so we should be able to represent $$g_i = e^{i \theta_i}$$. But each element has order two, so the only option is that each $$g$$ be represented by either 1 or $$e^{i \pi}$$, and this is not a faithful representation. What is the faithful irreducible representation of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$?

• I think $\theta=\theta_1+\cdots+\theta_n$ where $\theta_i=a_i\cdot \frac{2\pi}{p_i}$, right? – freakish Apr 15 at 10:15
• I think that doesn't always give you a faithful representation. – MBolin Apr 15 at 10:16
• @freakish I have added an example explaining why I think that doesn't always give you a faithful representation. – MBolin Apr 15 at 10:21
• Not sure where "faithful" came from? Irreducible representations need not be faithful. In fact a finite abelian group has a faithful irreducible representation if and only if it is cyclic: mathoverflow.net/questions/57129/… – freakish Apr 15 at 10:22
• @freakish OK I think I lacked that bit of information – MBolin Apr 15 at 10:23

Irreducible representations need not be faithful. In fact a finite abelian group has a faithful irreducible representation if and only if it is cyclic. And so $$\mathbb{Z}_2\times\mathbb{Z}_2$$ does not have a faithful irreducible representation.

Now let $$\rho:G\to GL(V)$$ be a representation with $$\dim V=1$$ over $$\mathbb{C}$$. So $$GL(V)\simeq\mathbb{C}\backslash\{0\}$$ with the standard multiplication. Now as you've said every element of $$G$$ can be written as $$(a_1,\ldots,a_n)\in\mathbb{Z}_{p_1}\times\cdots\times\mathbb{Z}_{p_n}$$. Put $$e_i=(0,\ldots,0,1,0,\ldots,0)$$ where $$1$$ is on the $$i$$-th position. With this $$\rho$$ is uniquely determined by values on $$\{e_i\}_{i=1}^n$$. And any such value will generate appropriate representation if $$\rho(e_i)^{p_i}=1$$. And so all we need to know is $$\rho(e_i)$$.

Since $$\rho(e_i)$$ has to be an element of $$\mathbb{C}\backslash\{0\}$$ of order dividing $$p_i$$ (which let me remind is a prime power, not a prime number) then the only choice is $$\rho(e_i)=exp(i \frac{2\pi}{p_i}k_i)$$ for some integer $$k_i$$. And therefore

$$\rho(a_1,\ldots, a_n)=exp\big(i(\frac{2\pi}{p_1}k_1a_1+\cdots+\frac{2\pi}{p_n}k_na_n) \big)$$

Meaning your $$\theta$$ can be written as

$$\theta=\frac{2\pi}{p_1}k_1a_1+\cdots+\frac{2\pi}{p_n}k_na_n$$

for some fixed integers $$k_1,\ldots,k_n\in\mathbb{Z}$$. And any such integers will generate a valid representation.