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I apologize in advance if my question sounds naive to those who are experts in representation theory.

Let $H$ be a subgroup of some finite group $G$. Let $\pi: G \rightarrow GL_n(C)$ be some (finite-dimensional) complex irreducible representation of $G$. Now, can we say the following for the restriction of $\pi$ to the subgroup $H$

  1. $\pi|_H$ is a irreducible representation of H.
  2. or $\pi|_H$ can be expressed directly as direct sum of irreducible representations of $H$. For example, $\pi|_H = \phi_1 \bigoplus \phi_2$, where, $\phi_1$ and $\phi_2$ be two irreducible representations of $H$.

According to my understanding, we can say so if $H$ is a normal subgroup. But what about any subgroup $H$?

Edit: For point 2, I am not asking about the equivalence to the direct sum of irreducible representation. I am asking if the restriction $\pi|_H$ can be expressed directly.

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2 Answers 2

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Consider a 2-dimensional irreducible representation of $S_3$ with $$(1,2,3) \mapsto \left(\begin{array}{rr}-1&1\\-1&0\end{array}\right),\ \ \ \ (1,2) \mapsto \left(\begin{array}{rr}0&1\\1&0\end{array}\right).$$ The restriction to $H := \langle (1,2,3) \rangle$ is equivalent to the sum of two $1$-dimensional complex representation. But it is irreducible over ${\mathbb R}$ and it is not equal to such a sum.

Note that $H$ is a normal subgroup of $G=S_3$, so your belief that this property holds for normal subgroups is wrong.

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  • $\begingroup$ Thanks for the answer. Then how should I interpret Clifford theory [en.wikipedia.org/wiki/Clifford_theory] $\endgroup$ Jan 6 at 23:02
  • $\begingroup$ Clifford's theorem says that the restriction of an irreducible representation to a normal subgroup is equivalent to a direct sum of irreducible representations. That result applies to representations of finite groups over an arbitrary field, not just the complex numbers. $\endgroup$
    – Derek Holt
    Jan 6 at 23:06
  • $\begingroup$ (12)(123)(21) = (1,3,2). How can H = <(1,2,3)> be a normal sub group? $\endgroup$ Jan 6 at 23:24
  • $\begingroup$ The question says $\pi$ is a complex irreducible representation of $G$. I don't understand why you mention that $\pi|_H$ is irreducible over $\mathbb R$. It is, but we're not working over $\mathbb R$. It's a direct sum of the two invariant subspaces of $\pmatrix{-1&1\\-1&0}$, which are the (complex) multiples of $\pmatrix{1\\-\exp(2\pi\mathrm i/3)}$ and $\pmatrix{1\\-\exp(4\pi\mathrm i/3)}$, respectively. $\endgroup$
    – joriki
    Jan 6 at 23:30
  • $\begingroup$ @MdAshiqurRahman: Your example illustrates why $H$ is a normal subgroup. You conjugated an element of $H$ with $(12)$ and again got an element of $H$. For permutations, conjugation is relabelling. A relabelling of a three-cycle is again a three-cycle, and the non-trivial elements of $H$ are exactly the three-cycles. $\endgroup$
    – joriki
    Jan 6 at 23:33
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This question has generated an unusual number of attempts at clarification, with several high-reputation users disagreeing both on the answer and on whether the question is clear.

It seems to me that the cause of all this was that you thought of representations as assigning to each group element a matrix, and consequently as the direct sum of representations as being formed by forming direct sums of the matrices, whereas many of the people seeking clarification (including me) think of representations as assigning to each group element an automorphism of a vector space, and consequently of the direct sum of representations as being formed by forming the direct sum of the vector spaces. See e.g. Wikipedia here and here or Groupprops for more on this.

If a representation is defined as a group action on a vector space and a direct sum of representations is accordingly defined as the direct sum of the underlying vector spaces equipped with the group action induced by the projections to the summands, the answer to your question is that since $\pi|_H$ is a representation of $H$, it is irreducible or a direct sum of irreducible representations.

If a representation is defined as a homomorphism from a group to a group of invertible matrices and a direct sum of representations is accordingly defined by forming the direct sums of the matrices, Derek Holt’s example shows that the answer is negative and the matrices of $\pi|_H$, while equivalent to irreducible matrix representations, need not be the direct sum of matrix representations.

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  • $\begingroup$ Thanks, @joriki, for the elaborate answer. I indeed meant the second one, i.e., when the representation is a homomorphism from group to group of invertible matrices. One further clarification question: Apart from the group $G$ being abelian, is there any other situation that will satisfy conditions 1 and 2? $\endgroup$ Jan 7 at 12:27

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