Composition of representation with monomorphism is irreducible $\implies$ representation is irreducible

Question

Let $f:H \rightarrow G$ be group monomorphism and $(V, \rho)$ a $G$-representation. Show that, if the $H$-representation $(V, \rho \circ f)$ is irreducible, then $(V, \rho)$ is irreducible. I started by considering a $G$-invariant subspace $W$ and I would like to show that it is $H$-invariant, but I don't see how to use the injectivity to prove it.

Answer 1

You have exactly the right idea!

We want to show the contrapositive. That is, if the $G$-representation $(V,\rho)$ decomposes as $(V_0, \rho_0) \oplus (V_1, \rho_1)$, then the $H$-representation $(V, \rho f)$ decomposes as $(V_0, \rho_0 f) \oplus (V_1, \rho_1 f)$.

Of course, if $f : H \to G$ is mono, that means we can view $H$ as a subgroup of $G$! Moreover, in this case $\rho f$ is just the restriction of $\rho$ to $H$. Then since $V_0$ is $G$-invariant and $H \leq G$, $V_0$ must be $H$ invariant as well.

More formally, let $v \in V_0$ and $h \in H$. Then $(\rho f h)(v) = (\rho g)(v)$ for some $g$, but we know $(\rho g)(v) \in V_0$ since $V_0$ is $G$-invariant. Since this always works, we see that $V_0$ is $H$ invariant too. Notice this proof doesn't really use the fact that $f$ is mono anywhere. We usually restrict to the case $H \leq G$, but the theory goes through fine without that assumption.

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I hope this helps ^_^