Why $\mathbb QG\simeq \mathbb Q\oplus \mathbb Q \Delta G$ in representation decomposition? 
Let $G$ be a finite group, $\mathbb Q G$ is the regular representation of $G$ over $\mathbb Q$, then it has a decomposition $\mathbb Q\oplus \mathbb Q \Delta G$ where $\mathbb Q$ represents the trivial representation and $\Delta G$ is the augmented ideal of $G$.(i.e. kernel of the map $\mathbb Z G\rightarrow \mathbb Z$ by $\sum z_gg\rightarrow \mathbb \sum z_g$.) 

How to prove the above statement? I have some knowledge on representation theory over $\mathbb C$ but not much over $\mathbb Q$.
(If it's a well known fact, is there some good references?)
 A: Note: I will use "$G$-representation" and "$kG$-module" words interchangeably which outside of pure algebra might not be the same. But here it is fine.
More generally consider a finite group $G$ and a field $k$. Then we have a short exact sequence of $kG$-modules:
$$0\to \ker(A)\to kG\xrightarrow{A} k\to 0$$
where $A$ is the augmentation map $A\big(\sum \lambda g\big)=\sum\lambda$.
Now let $v:=\sum_{g\in G} g\in kG$ which makes sense since $G$ is finite. This special element has the property that $hv=v$ for any $h\in G$. This shows that for any $\lambda\in k$ the linear map
$$f_{\lambda}:k\to kG$$
$$f_{\lambda}(1)=\lambda v$$
is actually a $kG$-module homomorphism. Now if $|G|x=1$ has a solution in $k$ (i.e. when $char(k)$ does not divide $|G|$) then $f_x$ is a splitting map for the augmentation map $A$ since $A(v)=|G|1$. So by the splitting lemma $kG$ is the direct sum of representations $k\oplus \ker(A)$. Assuming $char(k)$ doesn't divide $|G|$. This argument can be further generalized into Maschke's theorem.
Now all you have to do is calculate $\ker(A)$ in case $k=\mathbb{Q}$ which is $\mathbb{Q}\Delta G$. I leave that as an exercise.
Side note: I wonder if $k$ can be replaced with any unital ring having a solution to $|G|x=1$. It looks so. I don't see where the assumption about $k$ being a field (or even commutative) is relevant. Although this might not work for Maschke's theorem.
A: Hint: $\sum_g z_g g = (\sum_g z_g) 1 + \{\sum_{g} z_g g - [(\sum_g z_g) 1]\}$.
