Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?


[Edited for clarity. -- PLC]



How big can the irreps of a finite group be (over an arbitrary field)?
**EDIT:** Part 4 added. **EDIT2:** Second proof of Part 4 added.


**1.** The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$.


**2.** Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step: 


If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective, 
since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).


**3.** Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups).


**4.** There is a way to improve **2.** so that it comes a bit closer to **3.**:


**Theorem 1.** If $V\_1$, $V\_2$, ..., $V\_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V\_1+\dim V\_2+...+\dim V\_m\leq\dim A$.


(Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get **2.** as a consequence.)


*First proof of Theorem 1.* At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V\_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V\_i^{\ast}$ and $a\in A$) is irreducible (since $V\_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V\_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V\_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V\_i^{\ast}$ and therefore equal to $V\_i^{\ast}$ due to the irreducibility of $V\_i^{\ast}$). Hence, by duality, $V\_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V\_i$ actually *is* a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.


Now, let us prove that the vector subspaces $V\_1$, $V\_2$, ..., $V\_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V\_1+V\_2+...+V\_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V\_1+V\_2+...+V\_{m-1}$ is already a direct sum. It remains to prove that $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)=0$. In fact, assume the contrary. Then, $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)=V\_m$ (since $V\_m\cap \left(V\_1+V\_2+...+V\_{m-1}\right)$ is a nonzero subrepresentation of $V\_m$, and $V\_m$ is irreducible). Thus, $V\_m\subseteq V\_1+V\_2+...+V\_{m-1}$. Consequently, $V\_m$ is isomorphic to a subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ (because the sum $V\_1+V\_2+...+V\_{m-1}$ is a direct sum, according to our induction assumption).


Now, according to Theorem 2.2 and Remark 2.3 of [Etingof's "Introduction to representation theory"](http://math.mit.edu/~etingof/replect.pdf), any subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ must be a direct sum of the form $r\_1V\_1\oplus r\_2V\_2\oplus ...\oplus r\_{m-1}V\_{m-1}$ for some nonnegative integers $r\_1$, $r\_2$, ..., $r\_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$ must be one of the representations $V\_1$, $V\_2$, ..., $V\_{m-1}$. Since we know that $V\_m$ is isomorphic to a subrepresentation of the direct sum $V\_1\oplus V\_2\oplus ...\oplus V\_{m-1}$, we conclude that $V\_m$ is isomorphic to one of the representations $V\_1$, $V\_2$, ..., $V\_{m-1}$. This contradicts the non-isomorphy of the representations $V\_1$, $V\_2$, ..., $V\_m$. Thus, we have proven that the sum $V\_1+V\_2+...+V\_m$ is actually a direct sum. Consequently, $\dim V\_1+\dim V\_2+...+\dim V\_m=\dim\left(V\_1+V\_2+...+V\_m\right)\leq \dim A^{\ast}=\dim A$, and Theorem 1 is proven.


*Second proof of Theorem 1.* I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in [Crawley-Boevey's "Lectures on representation theory and invariant theory"](http://www.amsta.leeds.ac.uk/~pmtwc/repinv.pdf):


Let $0=A\_0\subseteq A\_1\subseteq A\_2\subseteq ...\subseteq A\_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A\_i/A\_{i-1}$ of $A$ is irreducible.


Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A\_I/A\_{I-1}$ (as representations of $A$).


In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A\_iT\neq 0$ (such elements $i$ exist, because $A\_kT=AT=T\neq 0$). Then, $A\_IT\neq 0$, but $A\_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A\_It\neq 0$ (such a vector $t$ exists, because $A\_IT\neq 0$), and consider the map $f:A\_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A\_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A\_{I-1}$ to $0$ (because $f\left(A\_{I-1}\right)=A\_{I-1}t\subseteq A\_{I-1}T=0$), it gives rise to a map $g:A\_I/A\_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A\_I/A\_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A\_I/A\_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A\_I/A\_{I-1}\right)=f\left(A\_I\right)=A\_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A\_I/A\_{I-1}$.


So we have just proven that


**(1)** For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A\_I/A\_{I-1}$ (as representations of $A$).


Denote this $I$ by $I\_T$ in order to make it clear that it depends on $T$. So we have $T\cong A\_{I\_T}/A\_{I\_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V\_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V\_i\cong A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I\_{V\_1}$, $I\_{V\_2}$, ..., $I\_{V\_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I\_{V\_i}=I\_{V\_j}$ would yield $V\_i\cong A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}=A\_{I\_{V\_j}}/A\_{I\_{V\_j}-1}\cong V\_j$, but the representations $V\_1$, $V\_2$, ..., $V\_m$ are pairwise nonisomorphic), and thus


$\sum\limits\_{i=1}^{m}\dim\left(A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\right)=\sum\limits\_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I\_{V\_i}}}\dim\left(A\_j/A\_{j-1}\right)$
$\leq \sum\limits\_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A\_j/A\_{j-1}\right)$ (since $\dim\left(A\_j/A\_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=\sum\limits\_{j=1}^{k}\dim\left(A\_j/A\_{j-1}\right)=\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)$.


Since $\dim\left(A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\right)=\dim V\_i$ for each $i$ (due to $A\_{I\_{V\_i}}/A\_{I\_{V\_i}-1}\cong V\_i$) and $\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits\_{j=1}^{k}\left(\dim A\_j-\dim A\_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A\_k-\dim A\_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits\_{i=1}^{m}\dim V\_i\leq\dim A$. This proves Theorem 1.


