Let $Γ$ be a discrete group and suppose $C^∗(Γ)$ is simple i.e. it contains no non-trivial closed and two-sided ideals. Show that $Γ$ is the one-element group.
I tried to show this by contradiction method. Is there any other alternative method of this problem? Also i am not sure about my answer.
Suppose $Γ$ is not the one-element group, which means it has at least two distinct elements. Let's consider, $g$ and $h$, in $Γ$ such that $g ≠ h$. Since $Γ$ is discrete, there exists an open neighborhood $U$ of $g$ that does not contain $h$. We can also find an open neighborhood $V$ of $h$ that does not contain $g$.
Now, consider the functions $f_U$ and $f_V$ in $C^∗(Γ)$ defined as follows:
$f_U= \begin{cases} 1, & \text{if $x ∈ U$ } \\ 0, & \text{if $x ∉ U$ } \end{cases}$
$f_V= \begin{cases} 1, & \text{if $x ∈ V $ } \\ 0, & \text{if $x ∉ V $ } \end{cases}$
Here, $f_U$ and $f_V$ are continuous functions on $Γ$, hence they belong to $C^∗(Γ)$. Moreover, these functions are non-zero, as they take the value 1 on their respective neighborhoods.
Now, let's consider the closed two-sided ideal $I$ generated by $f_U$ in $C^∗(Γ)$. This ideal consists of all elements of $C^∗(Γ)$ that can be obtained by convolving $f_U$ with other functions in $C^∗(Γ)$.
Since $C^∗(Γ)$ is simple, we know that $I$ must be either ${0}$ or $C^∗(Γ)$. However, $f_V$ is a non-zero element in $C^∗(Γ)$ that does not belong to $I$ because convolving $f_V$ with $f_U$ results in a zero function (since $U$ and $V$ are disjoint).
Therefore, a non-zero element $f_V$ in $C^∗(Γ)$ that does not belong to any non-trivial closed two-sided ideal. This contradicts the assumption that $C^∗(Γ)$ is simple.
Hence, our initial assumption must be false. Therefore, Γ is the one-element group.
Here is how I would prove this. The (universal) group $C^*$-algebra $C^*(\Gamma)$ has the universal property that unitary representations of $\Gamma$ uniquely lift to $*$-representations of $C^*(\Gamma)$. If you apply this to the trivial representation $\tau: \Gamma \to B(\mathbb{C})=\mathbb{C}: g \mapsto 1$, you find a unique $*$-homomorphism $$\tau: C^*(\Gamma)\to \mathbb{C}: g \mapsto 1.$$ Note that $\ker(\tau)$ is a non-trivial ideal of $C^*(\Gamma)$ if $\Gamma \ne \{1\}$, so $C^*(\Gamma)$ is not simple if $\Gamma\ne \{1\}$.
A much more interesting question to ask is when the reduced group $C^*$-algebra $C_r^*(\Gamma)$ is simple (discrete groups $\Gamma$ for which this is true are called $C^*$-simple). If $\Gamma$ is a non-trivial amenable group, then $\Gamma$ is clearly not $C^*$-simple. An example of a $C^*$-simple group is $\mathbb{F}_2$, the free group on two generators. This is a result by Powers.