# Irreducibility of restricted representation $\rho|_H$ implies irreducibility of original representation $\rho$

I'm reading Mark Meckes' notes on representation theory and he has this exercise:

3.3. Let $$H < G$$ and let $$\rho: G \to GL(V)$$ be a representation of $$G$$.

(a) Show that if $$\rho|_{H}$$ is an irreducible representation of $$H$$, then $$\rho$$ is irreducible.

I thought about this for a while and also searched the internet for references, but at least based on the material in his notes, I'm not sure how to about proving the statement. It's also kind of counter-intuitive since $$H$$-invariant vector spaces are not necessarily $$G$$-invariant.

Could someone kindly explain the approach for this problem?

• it's not counterintuitive if you look at it the right way: if there's no nonzero proper $H$-invariant subspace then there's certainly no nonzero proper $G$-invariant subspace (since $G$-invariance implies $H$-invariance), so $H$-irreducible implies $G$-irreducible. Maybe it is easier to use the contrapositive: if $\rho$ is reducible then there's a nonzero proper $G$-invariant subspace, which is certainly $H$ invariant, so $\rho|_H$ is reducible. Apr 25, 2022 at 10:14

You're gonna kick yourself when you see how easy this is. Every $$G$$-invariant subspace of $$V$$ is also $$H$$-invariant, right? It follows that if $$V$$ had a nontrivial $$G$$-invariant subspace, then it would also have a nontrivial $$H$$-invariant subspace. In other words, if $$V$$ wasn't irreducible as a representation of $$G$$, then it couldn't be irreducible as a representation of $$H$$.

Hint. A $$G$$-invariant subspace is also $$H$$-invariant, so a direct sum of $$G$$-invariants subspaces is also a direct sum of $$H$$-invariants subspaces. Can you conclude from here ?

• I understand this but it's not clear to me what to conclude from here. Of course, $G$-invariant subspaces are $H$-invariant and the same holds for direct sums of such subspaces. But the question is asking for $G$-invariance given $H$-invariance. Could you elaborate a bit more?
– S.D.
Apr 24, 2022 at 16:03

Let us say that at representation $$V$$ is reducible if it is not irreducible, i.e., if it admits a subrepresentation apart from $$0$$ and $$V$$.

As you have observed, every $$G$$-invariant subspace of $$V$$ is also $$H$$-invariant. We therefore have the following implication: $$\text{V is reducible over G} \implies \text{V is reducible over H}$$ By modus tollens, this implication is equivalent to the implication $$¬(\text{V is reducible over H}) \implies ¬(\text{V is reducible over G}) \,,$$ which can better be expressed as $$\text{V is irreducible over H} \implies \text{V is irreducible over G} \,.$$

We see that when we talk about reducibility, when go from $$G$$ to $$H$$. But when we talk about irreducibiliby, then we can go from $$H$$ to $$G$$. This is because negation of logical statements reverses the direction of implications (i.e., the modus tollens rule).