# Does every irreducible representation of a finite group G embed into its group algebra?

Let $G$ be a finite group, and let $F$ be a field.
Is there a simple proof that every irreducible representation of $G$ embeds into the group algebra $F[G]$?
I am specially interested in the case when $gcd(|F|,G)\neq 1$?

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what do you mean by "embeds into"? what module action are you putting on $F[G]$? – Yemon Choi Aug 13 '11 at 20:58
Moreover, are you assuming from the outside that $F$ is a finite field? – Yemon Choi Aug 13 '11 at 21:00

A group algebra of a finite group over a field is a Frobenius algebra. See http://en.wikipedia.org/wiki/Frobenius_algebra

Essentially, it means there is a nice non-degenerate bilinear form on the algebra (send (a,b) to the coefficient of 1 in ab). In a Frobenius algebra the dual of the right regular module is isomorphic to to the left regular module. Since the injective indecomposables are the duals of the right projective indecomposables, it follows the injective indecomposables are direct summands in the left regular module. Since each simple embeds in its injective envelope (i.e., the dual of its right projective cover), it follows each simple module embeds in the left regular module.

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This is a nice, clean way of putting it. – Todd Trimble Aug 13 '11 at 23:52
Though it might be cleaner to say that the regular rep is self-dual, and surjects onto any simple (which is, of course, the same argument). – Ben Webster Aug 15 '11 at 2:11
@Ben, after I wrote my answer, I thought about re-editing to word it your way since it is indeed very clean. But then I decided I wanted to emphasize that simples always embed into the regular module for a self-injective finite dimensional algebra, so I left it as it was. – Benjamin Steinberg Aug 15 '11 at 11:07
Your argument thus shows, that one only needs the algebra to be selfinjective. – Julian Kuelshammer Aug 19 '11 at 11:34

When $F$ is algebraically closed (or, more generally a splitting field), the answer is positive in the sense that every irreducible right $FG$-module is isomorphic to a minimal right ideal of the group algebra $FG$ (I stick to my preferred notation). Possibly the simplest argument I know, which really dates back to Richard Brauer is as follows: (I assume the structure of semi-simple algebras known, which is reasonable for MO). Let $V$ be an irreducible right $FG$ module, and let $\sigma:FG \to {\rm End}_{F}(V)$ be the associated representation. Let $\tau$ be the $F$-valued trace afforded by this representation. Form the element $i(\tau) = \sum_{g \in G} \tau(g^{-1})g$, which is a central element of the group algebra $FG.$ Let $I_{\tau}$ denote the (two-sided) ideal $i(\tau)FG$ of $FG.$ Notice that for each $h \in G$, we have $i(\tau).h = \sum_{g \in G} \tau(g^{-1}h)g$. Hence for any $x \in FG$, we have $i(\tau)x = \sum_{g \in G} \tau(g^{-1}x) g.$ In particular, $i(\tau)x = 0$ whenever $x \in J(FG)$, since nilpotent endomorphisms have trace 0. More generally, the annihilator of $V$ (which is a maximal two-sided ideal of $FG$) annihilates $i(\tau)FG$. Since $\{ g\sigma: g \in G \}$ spans ${\rm End}(V)$, we see that an element $x$ of $FG$ annihilates $V$ if and only if $\tau(g^{-1}x) = 0$ for each $g \in G$, so the annihilator of $i(\tau)FG$ is no larger than the annihilator of $V$. Hence $i(\tau)FG$ is isomorphic to the simple algebra ${\rm End}_{F}(V)$ just as as right $FG$-module, and the latter module is isomorphic to the direct sum of ${\rm dim}_F(V)$ copies of $V$ as right $FG$-module. Hence a minimal right ideal of $FG$ contained in $i(\tau)FG$ is isomorphic to $V$.

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The result can be stated more generally than I have, but this is the essential case. – Geoff Robinson Aug 13 '11 at 22:08