Universal property: Let $u:\Gamma \rightarrow B(H)$ be any unitary representation of $\Gamma$. Then, there is a unique $*-$homomorphism $\pi:C^{*}(\Gamma) \rightarrow B(H)$ such that $\pi_{u}(s)=u_{s}$ for all $s\in \Gamma$. (Here, $\Gamma$ is a discrete group here and the $C^{*}(\Gamma)$ is the reduced $C^{*}$-algebra.)
There is a quotation below:
Let $\Lambda\subset\Gamma$ be a subgroup, then, by universality, there is a canonical $*$-homomorphism $\pi: C^{*}(\Lambda)\rightarrow C^{*}(\Gamma)$.
In my view, there is a unitary representation: $\Lambda \rightarrow\Gamma\rightarrow B(l^{2}(\Gamma))$ (the first map is inclusion map and the second map is the left regular representation). Then, by universality, we only have a unique $*$-homomorphism $\pi: C^{*}(\Lambda)\rightarrow B(l^{2}(\Gamma))$. But I do not know how to get a canonical $*$-homomorphism $\pi: C^{*}(\Lambda)\rightarrow C^{*}(\Gamma)$.
What you do is you think of $C^*(\Gamma)$ as already represented. Then the inclusion $\Lambda\to\Gamma$ is a unitary representation, and you can apply the universality you defined.