Induced module and surjective morphism

Question

I am trying to solve the following question:

Let $G$ be a finite group and $H$ a subgroup of $G$. Let $A$ be a $G$-module. Show that $\pi: I^{H}_{G}(A)\to A$ defined by $\pi(f)=\sum_{g\in G/H}g\cdot f(g^{-1})$ is surjective morphism.

Here, $I^{H}_{G}(A)$ is the induced module from $H$ to $G$. It is $Hom_{H}(\mathbb{Z}[G],A)$.

I tried to understand the image of $\pi$, but I could only conclude that it is invariant by $H$. I don't have any other idea what to do.

Answer 1

Let's start with the following observation. If $H$ is a subgroup of $G$, then we have an isomorphism $f \colon \mathbb{Z}[G] \to \bigoplus_{G/H} \mathbb{Z}[H]$ of left $H$-modules. Here the direct sum is taken over the right cosets. For convenience, let's enumerate these cosets as $Hg_1,\ldots,Hg_r$, and let's write an element in the target as an $r$-tuple $(h_1,\ldots,h_r)$. The map $f$ is then the expected one: Start with an element $x \in G$, viewed as an element of $\mathbb{Z}[G]$. It lives in a unique $Hg_i$ for some $i$, and can thus be written as $hg_i$. Then $f(x) = (0,\ldots,h,\ldots,0)$, with the $h$ on the $i$-th index.

Now that we know this, we observe that $$\begin{split} \operatorname{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A) &= \operatorname{Hom}_{\mathbb{Z}[H]}(\bigoplus_{G/H} \mathbb{Z}[H],A)\\ &= \bigoplus_{G/H} \operatorname{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[H],A) \\ &= \bigoplus_{G/H} A\end{split}$$ Surely we should then expect a surjection onto $A$. The map $\pi$ you're interested in should, in this langauge, be precisely the 'expected' surjection $\bigoplus_{G/H} A \to A$, which is the identity on every component.

Edit. So I guess that if you meditate on these steps you should be able to 'see' an explicit pre-image of an element $a \in A$. Namely, split up $G$ into cosets $Hg_1,\ldots,Hg_r$, and assume WLOG that $g_1 = e$. Now consider the $H$-equivariant map $\alpha \colon \mathbb{Z}[G]$ sending $e$ to $a$ and hence $h$ to $ha$, while sending all the other cosets entirely to $0$. By a direct computation, $\pi(\alpha)$ should be $a$.

Answer 2

I think it is straightforward to check that $\pi$ is a morphism so I will just show that it is surjective. Let $a \in A$ and define $f:G \to A$ to be this function: for $h \in H$, define $f(h) = h \cdot a$ and for $g \notin H$, define $f(g)=0$. Then $f \in I_G^H(A)$. We see that $\pi(f)= 1 \cdot f(1)=a$. Therefore $\pi$ is surjective.