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Let $K$ be a field and $G$ a finite group. We can form the group ring $K[G]$. This is a ring via convolution product.

If I choose a $K[G]$-module $V$ in what way $V$ is a $K$-vectorspace? The statement $K\subset K[G]$ makes no sense. But for $k\in K$ I can form $k\chi_{1_G}\in K[G]$. Is this meant with $K\subset K[G]$?

If I now choose a submodul $U$ of the $K[G]$-module $V$ then I have as $K$-vectorspaces $V=U\oplus W$ for some $W$. If I choose $P:V\rightarrow U$ as the projection onto $U$, then this is in general not a $K[G]$-modulhom.

Why not?

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$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$For the first part, yes, this is what it is meant.

Let $G = \Span{g}$ be a group of order $2$, and let $K$ the field with $2$ elements. Consider a$K[G]$-module $V$ of dimension $2$ over $K$ and base $e_1, e_2$ such that $g$ acts on $V$ as $$ \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}. $$ It is easy to see that $U = \Span{e_1}$ is a $K[G]$-submodule (as $g e_1 = e_1$), and that $U$ is the only $K{G}$ submodule of dimension $1$. This is because $g e_2 = e_1 + e_2$ and $g(e_1 + e_2) = e_2$.

Now if the projection $P : V \to U$ were a $K[G]$-homomorphism, its kernel would be a $K[G]$-submodule of dimension $1$ distinct from $U$.

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    Thank you! So for the first part: Is $\phi:k\mapsto k\chi_{1_G}$ an isomorphism? So it is meant that $V$ is an $\phi(K)$-v.sp.? – 2017-01-05
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    Yes, as you may want to verify. – 2017-01-05
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    Okay, and if I define $L:=1/|G|\sum_{g\in G}\rho(g)P\rho(g^{-1})$ where $\rho(g)$ is the action of $g$ on $V$, this should be a $K[G]$-modulhom. But I don't understand: We write $v\in V$ as $v=u+w$. Then $P(v)=u=L(v)$, so $P=L$. – 2017-01-05
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    Note that $G$ has order $2$ in my example, and the field I have chosen has characteristic $2$, so you cannot divide by $2$ (which equals $0$ in $K$) to construct $L$. – 2017-01-05
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    Okay, but what if we have a field $K$, finite group $G$ with $ch(k)$ does not divide $|G|$? – 2017-01-05
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    See Maschke's Theorem. – 2017-01-05
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    I know this is what I am trying to prove. But I dont get why $L$ is a $K[G]$-modulhom. Is it true or false: $v=u+w$, then $u=L(v)$? – 2017-01-05