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Definition: Exact Representation of finite group $G$ in some field $V$ - is injective homomorphism $$\rho : G\to GL(V).$$ (I don't know English terminology, so you may correct me; or probably exists other definition of "exact representation"(do it exist?)).

Question: Do exists non-exact (in my terminology) representations?

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The word you're looking for is faithful. Non-faithful representations exist of course. Take a representation of a proper quotient of $G$ and note that every representation of a quotient $G/N$ yields a representation of $G$ by composing it with the quotient map. – t.b. Aug 22 '11 at 22:07
@Geoff: Yes, we played "Schnipp Schnapp" in kindergarden (in Switzerland). It is the same game but we shouted schnipp! when laying down a card and schnapp! if it was a duplicate. Tears ensued for sure... (does snap also have the meaning of snatching something in English?) – t.b. Aug 22 '11 at 22:20
@Theo: The only time I can think of it having that meaning would when it's followed by "up", e.g. "the shark snapped up the fish", or "the shoppers snapped up every toy". – Zev Chonoles Aug 22 '11 at 22:28
For a period of time, a few years ago, kids in high school (my daughter's age then...) in the Midwestern U.S. said "snap" as synonym for "Haha, got you!", in the sense of "beating" someone at something, coming in slightly ahead of them, etc. Or in the sense of a "snappy comeback". – paul garrett Aug 22 '11 at 22:32
@Alexey: You can prove this by taking the vector space with basis $\{e_h\}_{h \in G}$ and define $\rho(g)e_h = e_{gh}$ (this is called the (left) regular representation of $G$). So you're essentially exploiting the fact that $G$ acts on itself on the left. But this essentially comes down to your argument, as Cayley's theorem is proved exactly the same way. – t.b. Aug 22 '11 at 22:59
up vote 3 down vote accepted

Let me summarize the comments (leaving out the snap-discussion):

An injective representation is usually called faithful while geometers often prefer to use effective (but more in connection with group actions, as far as I know).

It is easy to build non-faithful representations:

Take a normal subgroup $N \neq \{0\}$ and look at any representation $\rho: G/N \to \operatorname{GL}{(V)}$. Then the composition $\rho \circ \pi$ of $\rho$ with the canonical projection $\pi: G \to G/N$ will have $N \subset \ker{(\rho \circ \pi)}$. Every representation of $G$ is of this form by the homomorphism theorem. Note also that this shows that representation of a simple group $G$ is either trivial or faithful.

You asked in the comments how one can show that every (finite) group admits a faithful representation. You suggested to use the Cayley theorem, and that's essentially the way to go. You can (sort of) avoid it by taking the vector space $V = k[G]$ with a basis $\{e_{h}\}_{h \in G}$ and let $\lambda: G \to \operatorname{GL}{(V)}$ be the (left) regular representation $\lambda(g)e_h = e_{gh}$. Of course, this is only half avoiding Cayley's theorem, as the proof of Cayley's theorem also relies on the left action of $G$ on itself.

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In fact, there's a very important class of representations which are typically not faithful (non-exact, in your terminology): characters.

A character of a group $G$ is a homomorphism $$ \chi:G\longrightarrow{\Bbb C}^\times. $$ It corresponds to the case where $\dim V=1$.

The set $\hat G$ of caracters of $G$ is itself a group: you can multiply characters by $(\chi_1\cdot\chi_2)(g)=\chi_1(g)\chi_2(g)$

Some groups have very few characters. For instance, the symmetric group ${\cal S}_n$ has only two characters, the constant map and the sign map $$ {\rm sgn}:{\cal S}_n\longrightarrow\{\pm1\} $$ giving the sign of a permutation. The alternating group $A_n$ with $n\geq5$ has no non-trivial characters, as does every group lacking non-trivial normal subgroups.

On the other hand, if $G$ is abelian there are usually lots of characters. For instance, it s not too hard exercise to prove that if $G$ is finite, then ${\hat G}\simeq G$ (although non canonically).

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This is a situation where the numebr theory culture and the group theory culture clash a little. There is a different definition of character, which is the trace afforded by any (complex) representation. Such a character is not usually a group homomorphism, but retains much information about the unerlying representation. – Geoff Robinson Aug 23 '11 at 0:09
@Geoff Adding to the culture clash: in the context of Dirichlet characters an induced character means an object that is much simpler (IMHO) than its cousin from representation theory. – Jyrki Lahtonen Aug 23 '11 at 9:00

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