# Is there a notion of induced representation that works not only for subgroups?

Given a group $$G$$ an a subgroup $$H, a representation of $$H$$ on $$V$$ is a pair $$(\rho, V)$$ where $$\rho \colon H \to \mathrm{GL}(V)$$ where $$V$$ is a vector space over a field $$K$$. We can the construct an induced representation of $$G$$ by $$\mathrm{Ind}_H^G \, \rho = K[G] \otimes_{K[H]} V \, .$$

Is there a way to generalize this for an arbitrary group homomorphism $$f \colon H \to G$$ such that when applying the construction for subgroups and the inclusion map $$i \colon H \to G$$ one recovers the original induced representation? That is, if instead of $$H$$ being a subgroup of $$G$$ we consider only a group homomorphism between $$H$$ and G.

At first I thought that, at least for the algebraic construction for finite groups, one could just change $$H$$ for $$\mathrm{im}\, f$$ and get done with it, but seems a too naive approach to work in general.

• Arnaud is right. Google "change of rings" for the general setting. – Matthew Towers Nov 4 '18 at 20:39
• @ArnaudD. Thank you for the comment. So a group homomorphism $f \colon H \to G$ would induce a ring homomorphism $K[G] \to K[H]$ by precomposing composition with $f$ and then to make $K[G]$ a $K[H]$-module I use extension of scalars. Does this make sense? – user314159 Nov 5 '18 at 9:56
• No, $f$ induces a ring homomorphism $K[H]\to K[G]$, and then you can make $K[G]$ a $K[H]$-module by restriction of scalars. – Arnaud D. Nov 5 '18 at 10:51
• @ArnaudD. Haha, just the opposite of what I wrote. Thank you very much! Maybe you can write that as an answer so the question is no longer unanswered. – user314159 Nov 5 '18 at 13:10
• @user314159 I've written an answer based on my comments (I've corrected a mistake in one of them by the way). – Arnaud D. Nov 5 '18 at 14:00

In order for the tensor product $$K[G]\otimes_{K[H]} V$$ to define a $$K[G]$$-module, all you need is that $$K[G]$$ is a $$(K[G],K[H])$$-bimodule. For this, you can use any group homomorphism $$f:H\to G$$ to define an action of $$H$$ on $$G$$ on the right by $$g\cdot h=gf(h)$$, and this will be compatible with the translation action of $$G$$ on itself on the left. Then it suffices to extend this bilinearly to get a bimodule, and then the tensor product $$K[G]\otimes_{K[H]} V$$ (where $$g\otimes \rho(h)v=gf(h)\otimes v$$) is a left $$K[G]$$-module, pretty much as in the case where $$f$$ is the inclusion of a subgroup.
Moreover, we still have a natural correspondance between maps $$\mathrm{Ind}_H^G (V,\rho)\to (W,\sigma)$$ and maps $$(V,\rho) \to \mathrm{Res}_H^G(W,\sigma)=(W,\sigma \circ f)$$, for all representations $$(W,\sigma)$$ of $$G$$.
You mention that you tried to take the same construction as in the case where $$H$$ is a subgroup but with $$H$$ replaced by $$im(f)$$. That is indeed a bit too simple to work, but it's not that far from correct either : if you factorize $$f$$ as a $$H\to \frac{H}{\ker(f)}\to G$$, then since $$im(f)\cong \frac{H}{\ker(f)}$$ your suggestion actually does half of the job! So it is enough to find the representation induced by the quotient map $$H\to \frac{H}{\ker(f)}$$, i.e. the $$K\left[\frac{H}{\ker(f)}\right]$$-module $$K\left[\frac{H}{\ker(f)}\right]\otimes_{K[H]}V$$. This is actually isomorphic to the module of coinvariants $$V_{\ker(f)}$$, which is defined as the quotient of $$V$$ by the ideal generated by terms of the form $$v-\rho(k)v$$, where $$v\in V$$ and $$k\in \ker(f)$$.
Thus you can construct $$K[G]\otimes_{K[H]}V$$ as $$\bigoplus_{g\operatorname{im}(f)\in G/\operatorname{im}(f)} V_{\ker(f)},$$ with the action defined as in the case where $$H$$ is just a subgroup.