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Given $F$ a field and $H$ a finite group and $H_n=\sum_{h \in H}h \in F[H]$, I'm trying to show that the left ideal $(H_n)\subset F[H]$ is a simple submodule.

Now I'm not really sure what the ideal looks like, I think it will be just the elements of $H$ but I don't see how simplicity follows.

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It indeed follows from looking at how the ideal looks. Consider what happens when you multiply $H_n$ by an element of $H$ from the left. This clearly (after a substitution) leaves us with $H_n$ again (it is invariant). Now consider an element $$\sum_{h \in H}{f_h h } \in F[H]$$ Then using the above remark, we find that the left multiplication of this element with $H_n$ gives us $$ (\sum_{h \in H}{f_h} H_n)$$ This proves that the left ideal generated by $H_n$ is $$ (H_n) = FH_n$$ Suppose we are given a non-trivial subideal $I$ of this left ideal, take a non-zero element $r \in I$. This can always be written (for some $f \in F \setminus \left\lbrace 0 \right\rbrace $) as $$ r= f H_n $$ Now we use that $F$ is a field to see that $$ H_n = f^{-1} fH_n \in I $$ Thus we find that $$ (H_n) \subseteq I \subseteq (H_n) $$ Which means that $(H_n)$ is a simple $F[H]$-module.

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    $\begingroup$ Thank you, I'm just a little unsure how we can easily see that $H_n$ will be in the subideal $\endgroup$ – ForTheGlory Jan 30 '16 at 20:19
  • $\begingroup$ I'll update it, so it will be more complete! Hold on... $\endgroup$ – A. Van Antwerpen Jan 30 '16 at 20:27
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    $\begingroup$ Thank you, this has helped a great deal $\endgroup$ – ForTheGlory Jan 30 '16 at 20:38

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