Each irreducible representation is a subrepresentation of induced one I'm learning what irreducible representation is and need some examples. One of them is as follows:
Let $G$ be any group and $H$ - it abelian subgroup. How to prove that each irreducible representation of $G$ is a subrepresentation of someone, induced from one-dimensional representation of  $H$ (this is a problem in the problem sheet)?
 A: Let's prove a more general fact: If $H\subseteq G$ is a subgroup, then every irreducible representation $\rho$ of $G$ is a subrepresentation of a representation induced by an irreducible representation of $H$ (in the special case $H$ is abelian, every irreducible representation is $1$-dimensional). The proof below definitely works for finite groups (but maybe there are some issues if $G$ is infinite?) 
Given $\rho: G\to \operatorname{GL}(V)$ a representation, we first note that $\rho$ is a subrepresentation of the regular representation $\rho_{\operatorname{reg}}$ of $G$. Now you can check that $\rho_{\operatorname{reg}}$ is induced by the regular representation of $H$, which I will denote it by $\theta$. So $\theta$ is the regular representation of $H$. Now decompose $\theta$ into irreducible representations $\theta_1, ..., \theta_{m}$ of $H$. Thus,
$$
\theta = \theta_{1} \oplus \theta_{2} \oplus \cdots \oplus \theta_{m}
$$
Therefore,
$$
\operatorname{Ind}_{H}^{G}(\theta)= \operatorname{Ind}_{H}^{G}\left(\theta_{1} \oplus \theta_{2} \oplus \cdots \oplus\theta_{m}\right)
$$
But $\operatorname{Ind}_{H}^{G}$ commutes with direct sums, so we get
$$
\rho_{\operatorname{reg}} = \operatorname{Ind}_{H}^{G}(\theta)= \operatorname{Ind}_{H}^{G}(\theta_{1}) \oplus \operatorname{Ind}_{H}^{G}(\theta_{2}) \oplus \cdots \oplus \operatorname{Ind}_{H}^{G}(\theta_{m})
$$
Now $\rho$ is contained in $\rho_{\operatorname{reg}}$, so that
$\rho$ is contained in $\operatorname{Ind}_{H}^{G}(\theta_{1}) \oplus \operatorname{Ind}_{H}^{G}(\theta_{2}) \oplus \cdots \oplus \operatorname{Ind}_{H}^{G}(\theta_{m})$. 
But $\rho$ is irreducible! Thus, $\rho$ must be contained entirely in $\operatorname{Ind}_{H}^{G}(\theta_{i})$ for some $i$. In other words, $\rho$ is a subrepresentation of some rep of $G$ induced by irreducible rep of $H$. 
