Universal properties of certain crossed products So I was wondering if there are any nice universal properties that the crossed product $C^*$ algebra, $C(\mathbb{T})\times_\alpha \mathbb{Z}_2$ satisfies, where $\alpha$ is the action of conjugation. The reason I am asking is that I am primarily interested in finding an isomorphism between this algebra and $C^\ast(\mathbb{Z}_2\ast\mathbb{Z}_2)$.
 A: Let me list some facts about the algebra $A=C^\ast(\mathbb{Z}_2\ast\mathbb{Z}_2)$. Denote by $s,t\in\mathbb{Z}_2\ast\mathbb{Z}_2$ the canonical generators.
$\bullet$ $A$ is the universal algebra generated by two reflections: if $S,T\in B(H)$ are two reflections, i.e. $S=S^*,T=T^*,S^2=I,T^2=I,$ then there exists a unique homomorphism from $A$ onto the $C^*$-algebra $A_1\subseteq B(H)$ generated by $S,T$ such that $s,t$ are mapped onto $S,T$ respectively. 
$\bullet$ $A$ is the universal algebra generated by two projections: if $P,Q\in B(H)$ are two orthogonal projections, then there exists a unique homomorphism from $A$ onto the $C^*$-algebra $A_1\subseteq B(H)$ generated by $P,Q$, such that $s,t$ are mapped onto $2P-I,2Q-I$ respectively. 
$\bullet$ The group $\mathbb{Z}_2\ast\mathbb{Z}_2=\langle s,t\ |\ s^2=t^2=1\rangle$ contains the subgroups $\langle st\rangle\simeq\mathbb Z$ and $\langle s\rangle\simeq\mathbb Z_2$ which define an isomorphism $\mathbb Z_2*\mathbb Z_2\simeq \mathbb Z\rtimes\mathbb Z_2$ mentioned in Martin Argerami's answer. In particular
$$
C(\mathbb T)\times_\alpha\mathbb Z_2\simeq C^*(\mathbb Z\rtimes\mathbb Z_2)\simeq C^*(\mathbb Z_2*\mathbb Z_2).
$$
$\bullet$ All irreducible representations of $A$ are $1$- or $2$-dimensional. Moreover, $A$ is isomorphic to the $C^*$-algebra of continuous functions $f:[0,1]\to M_2(\mathbb C)$ such that $f(0),f(1)$ are diagonal matrices.
A: Group algebras are not my thing, but here is an attempt. You have, according to Wikipedia, that $\mathbb Z_2*\mathbb Z_2\simeq \mathbb Z\rtimes\mathbb Z_2$, so 
$$
C(\mathbb T)\times_\alpha\mathbb Z_2\simeq C^*(\mathbb Z\rtimes\mathbb Z_2)\simeq C^*(\mathbb Z_2*\mathbb Z_2).
$$
The first isomorphism holds due to the fact that  $\mathbb Z\rtimes\mathbb Z_2$ is amenable, so any faithful representation generates the full C $^*$-algebra.
