# A naive question about composition factor of a representation

Let $G$ be a Lie group, and $(\pi,V)$ is a continuous representation of $G$ which has finite composition series. A question I have which might be somehow naive is that: for any irreducible representation $(\sigma,W)$ of $G$, is it true that $(\sigma,W)$ occurs as one composition factor if and only if the set $Hom_G(V,W)$ is nonzero?

I have no idea how difficult or how easy this question might be, and any reference or answer is appreciated.

Edit: Thanks a lot for all of your answers, comments and examples. Now if $G$ is real reductive, $(\pi,V)$ is smooth admissible. Is there a way to determine all of the composition factors of $V$?

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I'm not too familiar with the etiquette on MO myself, but since your original question was already answered, you might want to make a separate question about real reductive groups. –  Kimball Mar 21 '11 at 14:48

No. This holds only if $V$ is semisimple. Consider the case when $V$ has two composition factors. This means that one is an invariant subspace and the other is the quotient of $V$ by this invariant subspace. If there is also an invariant subspace isomorphic to the quotient then $V$ is the direct sum of these two representations and so $V$ is decomposable.
Bruce's answer is perfectly satisfactory, but you might want to see an explicit example. Let $B$ be the group of all upper triangular matrices in $GL_2({\mathbb R})$ and let $\pi:B\to GL_2({\mathbb R})$ be the inclusion map, which you might consider as a representation on $\mathbb R^2$. Let $\chi_1,\chi_2:B\to GL_1({\mathbb R})$ be the representations given by $$\chi_j\left(^{a_1}\ ^x_{a_2} \right)=a_j.$$ Then $\chi_1$ is a subrepresentation and $\chi_2$ is a quotient of $\pi$. So both are subquotients, but $Hom_B(\pi,\chi_1)$ is zero.