Let $K$ be a compact semisimple Lie group. The adjoint group $K'$ associated to $K$ is defined as the image of $K$ under the adjoint representation Ad:$K\to GL(\mathfrak{k})$.
I think that $K'\cong K/Z(K)$. I am trying to show that the center of $K'$ is trivial but I don't have any idea how to prove this. Any hints would be appreciated.
On
More generally, the following result can be found in the literature, e.g. here, Propostion $7.2$, with a proof on page $14$:
Theorem: Suppose that $G$ is a connected compact Lie group and that $C$ is the center of $G$. Then the center of $G/C$ is trivial.
Proof: Let $T$ be a maximal torus of $G$. Then $T$ contains $C$ and $T/C$ is a torus. The diffeomorphism $$ (G/C)/(T/C) ≃ G/T $$ shows that $\chi((G/C)/(T/C))$ is nonzero and thus that $T/C$ is a maximal torus for $G/C$. Suppose that $x$ is a non-identity element of $G/C$ and that $\tilde{x}\in G$ is an element which projects to $x$. Since $\tilde{x}\not\in C$, there is a maximal torus $T′$ for $G$ which does not contain $\tilde{x}$. Then $T′/C$ is a maximal torus for $G/C$ which does not contain $x$, and so $x$ does not belong to the center of $G/C$. Hence this center is trivial.
I think I have found a simpler proof as follows: Let $[a]\in Z(K')$, then it follows that for any $b\in K$ we have $$ a^{-1}b^{-1}ab\in Z(K). $$ Now we consider the continuous function $\phi_a:K\to K$ defined by $$ \phi_a(b)=a^{-1}b^{-1}ab. $$ Then $\phi_a$ maps $K$ into $Z(K)$. However, since $K$ is semisimple, the center of $\mathfrak{k}$ (the Lie algebra of K) is trivial. Therefore, $Z(K)$ is discrete and it follows that $\phi_a$ is a constant function and in particular equals identity. Therefore, $a\in Z(K)$.