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For a general group $G$ and subgroup $H$, decomposing the permutation representation $Ind$ $1^G_H$ into irreducibles of $G$ seems difficult (from what I've read it seems to follow in general from some combination of Mackey theory and Frobenius Reciprocity?)

That said, representation theory often becomes much simpler in the abelian case, so I was wondering whether there exists a general decomposition formula for $G$ an abelian group?

I am particularly interested in the subgroups which are simply the cyclic components of G, that is for \begin{align*} G=\mathbb{Z}_{k_1} \oplus \ldots \oplus \mathbb{Z}_{k_n} \end{align*} Let $x \in \{0,1\}^n$ with 1 in coordinates $m_1,\ldots,m_k$ denote the subgroup $\mathbb{Z}_{m_1} \oplus \ldots \oplus \mathbb{Z}_{m_k}$.

How does $Ind$ $1^G_x$ decompose into irreducibles?

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    $\begingroup$ I am on a phone right now, so I will just comment for now that this simply gives the regular representation of the quotient by that subgroup. So the irreps are those whose character is identically 1 on the subgroup. $\endgroup$ – Tobias Kildetoft Feb 4 '18 at 10:15
  • $\begingroup$ Ah I see this is the case for all normal subgroups. Then the coefficients are just given by the dimension of the irreducibles which is always 1 in this case. Thanks! $\endgroup$ – Max Hopkins Feb 4 '18 at 18:12
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    $\begingroup$ No, it is not quite this for any normal subgroup if the group is not abelian. Even there it is more subtle, and known as Clifford theory. Note that the reason the action of the subgroup on the induced representation is trivial is that the subgroup is central. Without the subgroup being central, it need not act trivially on the induced representation. $\endgroup$ – Tobias Kildetoft Feb 4 '18 at 19:08
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As I already mentioned in a comment, the representation $1_H^G$ is just the regular representation of $G/H$ (seen as a representation of $G$), no matter which subgroup we choose. (though this assumes that we are working over a field for which we both have Schur's lemma and Maschke's theorem, as otherwise the argument will break down).

To prove this, we need a few things:
First, we need to know that all the irreducibles of abelian groups are $1$-dimensional (so this is where we need Schur's lemma to hold).
Next, we need Frobenius reciprocity: An irreducible module $L$ is a constituent of $1_H^G$ iff the restriction of $L$ to $H$ has $1$ as a constituent.

Now, putting these together, we see that the constituents of $1_H^G$ precisely are those which restrict to $1$ on $H$ (since they are already $1$-dimensional), so the constituents are precisely those with $H$ in their kernel, or in other words, they are the irreducible representations of $G/H$. Since they each occur with multiplicity $1$ (again by Frobenius reciprocity), this shows that the induced module is precisely the regular representation of $G/H$.

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