Let $G \in Ab$ denote an abelian group and $X: \Delta^{op} \to Ab$ the functor such that $X([n]):=G^{n+1}$, the face maps $\delta_{n+1,i}:[n] \to [n+1]$ correspond to the projections $\pi_{n+1,i}:G^{n+1} \to G^{n}$ where the $i$-th term is omitted, and dually the degeneracy maps $\sigma_{n+1,i}:[n+1] \to [n]$ to inclusions $m_{n+1,i}:G^{n} \to G^{n+1}$ with trivial $i$-th term in the image.
My question: is $X$ a simplicial group? I know the definition and I have tried of prove the identities for hours but could't get my head around. I managed to prove from here https://en.wikipedia.org/wiki/Simplicial_set#Face_and_degeneracy_maps (3) and for relatively small $n$ some of the others, but nothing explicit.
No, this is not a simplicial group (unless $G$ is trivial). A quick way to see that it can't work is to remember that each degeneracy map must be split by two of the face maps (explicitly, $d_is_i$ and $d_{i+1}s_i$ are both the identity). But if $s_i:G^{n+1}\to G^{n+2}$ is $0$ on one coordinate, the only projection $G^{n+2}\to G^{n+1}$ which will be a left inverse for $s_i$ is the one that drops the $0$ coordinate. Very explicitly, in the case $n=0$, we have $s_0(g)=(0,g)$ so $d_1s_0(g)=0$ instead of $g$.
The natural construction that gives a simplicial group with projections as the face maps would be to make the degeneracy maps be the maps that repeat a coordinate, so for instance $s_0:G^2\to G^3$ would be $s_0(g,h)=(g,g,h)$. To verify that this works, you don't need to get your hands dirty with the simplicial identities: viewing $\Delta$ as the category of nonempty finite totally ordered sets, it's just the functor that takes a totally ordered set $S$ to $G^S$ and takes an order-preserving map $f:S\to T$ to the map $G^T\to G^S$ given by composition with $f$.