Let $H$ be a subgroup of a finite group $G$.Given an irreducible representation $\pi$ of $G$,we may decompose its restriction to $H$ into irreducible $H$- representations.Show that every irreducible representation of $H$ can be obtained in this way.
My initial idea was to use induced representations,but later I wanted prove this result without appealing to that concept.I am stuck with this problem for quite some time.Please help.Thanks.
[Math] Irreducible Representation by Restriction
representation-theory
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The irreducible $(p-1)$-dimensional $\mathbb{Q}G$-module is not induced from a rational representation of $C_p$. But twice that is. One way of seeing this is to work over $\mathbb{C}$. Over $\mathbb{C}$, the $(p-1)$-dimensional guy decomposes as a sum of $\frac{p-1}{2}$ irreducible 2-dimensional representations. Each is induced from a one-dimensional representation of $C_p=H$. However, if $\chi$ is a non-trivial one-dimensional character of $C_p$, then $\mathrm{Ind}_{G/H}\chi = \mathrm{Ind}_{G/H}\bar{\chi}$. That follows from Clifford theory: $H$ is normal in $G$ and $\chi$ and $\bar{\chi}$ lie in one orbit under the action of $G$ on the irreducible characters of $H$.
Thus, if you induce the sum of all non-trivial one-dimensional representations of $H$ to $G$, you hit each irreducible 2-dimensional complex representation of $G$ twice.
Another way of obtaining the $(p-1)$-dimensional guy through induction is by inducing the trivial representation of $C_2\leq G$. This induction will be $p$-dimensional and will have the trivial $G$-representation as a direct summand (this uses Maschke's theorem, which of course works just fine over $\mathbb{Q}$). The remaining $(p-1)$-dimensional chunk is irreducible over $\mathbb{Q}$, and is the representation you are looking for.
I find this a natural question. Particularly because I recently heard that I would be teaching group theory for advanced undergrads next year (or the year after), so I want to test my motivational skills here :-)
Indeed, one of the aspects of representation theory is to study the groups being represented. It is nice that we can take some abstract group, and replace its elements with objects that are easier to compute with: matrices (as in linear representations) or permutations (as in groups acting on sets). To that end it is helpful (if not imperative) that the representation is faithful lest we draw mistaken conclusions about identities of elements of the said group. Even if the action is not faithful we can slice off the kernel of the action as a part of a divide-and-conquer approach to understanding the group better, if/when so inclined.
On the other hand we also want to understand what kind of objects can be acted upon by a given group (either via permutations or linear transformations). There the focus is different. If we look at objects being permuted by a group, the first divide-and-conquer step is to study the objects one orbit at a time, as the orbits don't interact (but e.g. their sizes are constrained by the size of the acting group). If we look at vector spaces acted upon by a group of linear transformations, then the divide-and-conquer approach tells us to study subspaces stable under the group, and check whether we can build the whole space from atomic subspaces (often we can).
So representation theory is in some sense a two-way street: extract information about the group by studying its possible actions, and OTOH extract information about a collection of objects given that they are acted upon by a given group in some meaningful way.
Possibly the best known result in the former direction is the Burnside's $p^aq^b$-theorem telling us that a finite group whose order has only two prime divisors is necessarily solvable. This is likely covered in all texts. Going the other way let me lead off with a simple example. Let $V$ be the space of complex functions on the real line with period $2\pi$. If $f(t)$ is such a function, then so is the function $g\cdot f$ defined by 180 degree phase shift: $(g\cdot f)(t)=f(t+\pi)$. As the functions in $V$ have period $2\pi$, we see that $g\cdot(g\cdot f)=f$ for all $f$ meaning that the cyclic group $C_2=\langle g\rangle$ of order two generated by $g$ acts on $V$. What can we say about this action? Representation theory tells us that there are the following kind of special functions in relation to this action. The space $$ V^+=\{f\in V\mid g\cdot f=f\} $$ of functions that have $\pi$ as a period, and the space $$ V^-=\{f\in V\mid g\cdot f=-f\}, $$ i.e. the function $f$ with the property $f(t+\pi)=-f(t)$ for all $t$. Representation theory also tells us that $V=V^+\oplus V^-$. It is an easy exercise that given any function $f\in V$, then the function $$ f^+(t)=\frac{f(t)+f(t+\pi)}2\in V^+, $$ and $$ f^-(t)=\frac{f(t)-f(t+\pi)}2\in V^-. $$ Furthermore $f=f^++f^-$. Once you reach the idempotents of the group ring in your studies, this should give you a deja vu -experience :-) At that point an easy exercise for you is to figure what happens, if $g$ acts by a phase shift by $2\pi/n$ for some natural number $n>2$.
A more high-browed example of this latter direction comes from particle physics. There it is imperative that the symmetries of the nature are not broken. Quantum physics uses vectors in some space to represent elementary particles. What then happens is that particles can only react in such a way that the tensor product of those vectors has a non-trivial component acted upon trivially by the group. IOW: "trivial representation = invariants" is a very important representation - fully agree with Mariano.
Caveat: I did go on a hyperbole with the last example, and the painted picture may not be quite accurate. Anyway, they were handing out Nobel prizes in physics for that stuff in the 60s and 70s :-)
Best Answer
I'm not really sure why you'd insist you don't want to use induced representations here; Frobenius reciprocity means it's the natural approach, and this is an immediate corollary of Frobenius reciprocity...
But OK. Suppose there is an irreducible representation $\sigma$ of $H$ which isn't contained in the restriction of an irreducible representation of $G$. Then $\sigma$ doesn't occur in the restriction to $H$ of the regular representation $\Bbb{C}[G]$ of $G$, which contains a copy of $\Bbb{C}[H]$, and hence of $\sigma$, so you're done.