Schur's Di-Lemma: finite and Lie groups different?

For a finite group it's nothing special if two one-dimensional irreps pop up in a product, e.g. for $C_{3v}$ symmetry, $E\bigotimes{E}=A_1\bigoplus{A_2}\bigoplus{E}$ or in dimensions, $2*2=1+1+2$. Schur's Lemma merely forces that only one $A_1$ can be in the product.
Now I never saw a Lie group irrep with dimension $1$ which is not the trivial irrep. Have I just not looked around far enough? :-) (Maybe e.g. beyond semisimple ones?)
(Background: Playing around with tensors for 3-nodes and crossings, I found a six-dimensional irrep which gives an invariant for tangled graphs, the Clebsch-Gordan series being $6*6=1+1+6+8+8+12$. The second $1$ is kind of an "antisymmetric one", like $A_2$. In every respect (I could check), this invariant behaves like the invariants coming from Lie groups.)

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Semisimple Lie groups coincide with their own commutator subgroups, so they can't have any non-trivial 1-dimensional representation. But, for example, $\mathrm S^1$ has infinitely many. –  Angelo Sep 27 '12 at 15:07
This question is not research-level; it would be better asked on math.stackexchange.com. –  Qiaochu Yuan Sep 27 '12 at 16:29
Here are some examples of non-trivial 1-dim rep's of Lie groups which pop up naturally: $A\mapsto \det(A)$ on $GL_n(\mathbb{R})$; or $A\mapsto (\det A)^k$ on $U(n)$... You may also look at the $ax+b''$-group, and take the linear part of your affine transformation. As soon as your group is not perfect, you have tons of examples. –  Alain Valette Sep 27 '12 at 17:08
Thanks, I follow the leads. @Qiaochu: I hesitated to ask at all, but on Stackexchange noone answered a similar question. (If only the LiE software would be more versatile, I would pester MO less often...) –  Hauke Reddmann Sep 28 '12 at 9:21