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Let $V$ be a representation of a finite group $G$ with $V$ being finite dimensional. Fix a $g \in G$. Is it necessarily true that $\dim V \geq \operatorname{ord} g$?

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What about the trivial representation? – Alex Youcis Feb 14 '12 at 5:57
Right, but what if we assume that $V$ is not the trivial representation? – user24954 Feb 14 '12 at 6:01
$V$ could be the direct sum of 2 trivial representations, and $g$ could have order 3. Even if you insist that $V$ is irreducible, there are many groups that have irreducible representations of dimension 2, for example, all the dihedral groups, which can have elements of arbitrarily large order. – Ted Feb 14 '12 at 6:06
Still not true. $S_n$ has a non-trivial representation on spaces of dimension $2$ (induced by the quotient map $S_n\to C_2$, and letting the nontrivial element of $C_2$ act by exchanging coordinates). Even if the representation is faithful: $S_n$ acts on an $n$-dimensional space by acting on the coordinates, but for $n\gt 5$, contains elements of order larger than $n$ (e.g., $S_5$ has elements of order $6$). – Arturo Magidin Feb 14 '12 at 6:08
If $G$ is abelian then all its irreducible representations have dimension 1, but (unless it's the one-element group) it has elements of order greater than 1. – Gerry Myerson Feb 14 '12 at 6:30
up vote 1 down vote accepted

No, not at all. Let $V$ be a one-dimensional vector space, and let $G$ be the two-element cyclic group, where the generator for $G$ acts as multiplication by $-1$. Then $V$ is one-dimensional, but $G$ has an element of order two.

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