I don't know in general, but this is certainly true when $Q$ is finite. If $K$ has a faithful linear representation, it is very easy to see that the induced representation of $G$ is also faitful.
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I don't in general, but this is certainly true when $Q$ is finite. If $K$ has a faithful linear representation, it is very easy to see that the induced representation of $G$ is also faitful. |
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