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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 $\mathbb{Z}_2 \times \mathbb{Z}_2$?

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  • $\begingroup$ I think $\theta=\theta_1+\cdots+\theta_n$ where $\theta_i=a_i\cdot \frac{2\pi}{p_i}$, right? $\endgroup$ – freakish Apr 15 at 10:15
  • $\begingroup$ I think that doesn't always give you a faithful representation. $\endgroup$ – MBolin Apr 15 at 10:16
  • $\begingroup$ @freakish I have added an example explaining why I think that doesn't always give you a faithful representation. $\endgroup$ – MBolin Apr 15 at 10:21
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    $\begingroup$ 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/… $\endgroup$ – freakish Apr 15 at 10:22
  • $\begingroup$ @freakish OK I think I lacked that bit of information $\endgroup$ – MBolin Apr 15 at 10:23
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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.

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