In the group $\bigoplus_{j=-\infty}^{\infty} G$ does $a \cdot_k b = (a_{j+k} b_j)$ also form a group? The set $H := \bigoplus\limits_{j=-\infty}^{\infty} G$ forms a group under componentwise group law on $G$.  That is the direct product and that means essentially the same thing as $H' = \prod\limits_{j={-\infty}}^{\infty} G = \dots \times G \times \dots$ except there is only a finite number of non-identity entries in each element.  In either case, component-wise appylying $G$'s group law would form a group in these bi-infinite products.  
Clearly we can write componentwise $\cdot$ as $x \cdot y = (x_i \cdot y_i)$ where $x, y \in H$.  Then correspondingly we can write component-wise "shift second argument by $k$, and then perform $\cdot$" as $x \cdot_k y = (x_i\cdot y_{i-k})$  so was wondering if a group was also formed or if not, then is it at least associative and thus a monoid with identity $1_H = (\dots, 1, \dots)$ for any $k \in \Bbb{Z}$?  

What's the simplest proof you can give of associativity of $\cdot_k$, knowing that $\cdot \equiv \cdot_0$ is associative by definition of group $G$ and component-wise constructions?


If you need to, choose a basepoint, such as $0$, which would mean that $(a,b) := (\dots, 1, a, b, 1, \dots)$ has an $a$ at index $0$ and a $b$ at index $1$, and so on.
 A: No, the operation $\cdot_k$ for $k \gt 0$ is not generally associative.  Proof:
$a \cdot_k (b_i \cdot c_{i - k}) = (a_i \cdot b_{i-k} \cdot c_{i-2k}) \neq (a_i \cdot b_{i-k})\cdot_k c = (a_i \cdot b_{i-k} \cdot c_{i-k})$.
Notice the discrepancy in $c$'s indices.
A: The new operation is not associative. But in a way it fits into another associative operation, if we allow $k$ to vary and keep track of its value.

The mapping 
$$\phi:(y_i)\mapsto(y_{i-k})$$
is an automorphism of $H$. Your product is defined as
$$
x\cdot_k y=x\cdot \phi(y).
$$
As you observed, this is not associative, because
$$
(x\cdot_k y)\cdot_kz=(x\cdot\phi(y))\cdot\phi(z),
$$
but
$$
x\cdot_k(y\cdot_kz)=x\cdot(\phi(y)\cdot\phi(\phi(z))),
$$
and in the latter product $\phi$ is applied to $z$ twice.

But this is related to the concept of a semidirect product of groups. Whenever we have a homomorphism $\phi$ from a group $G$ to the group of automorphisms of another group $H$, we can form the semidirect product $H\rtimes_\phi G$. The underlying set is the usual Cartesian product $H\times G$, but the operation defined by the rule
$$
(h,g)*(h',g')=(h\phi(g)(h'),gg').
$$
In other words, we twist the $H$-component of the latter factor by applying the automorphism $\phi(g)$ to it before we multiply.

The rule $\phi(k):(x_i)\mapsto (x_{i-k})$ defines a homomorphism from the additive group $\Bbb{Z}$ to the group of automorphisms of $H$. In other words, for all $k\in\Bbb{Z}$, the shift mapping $\phi(k)$ is an automorphism of $G$, and furthermore
$$\phi(k)\circ\phi(\ell)=\phi(k+\ell)$$
for all integers $k,\ell$. This is related to your question as follows. In the semi-direct product $H\rtimes_\phi\Bbb{Z}$ we have
$$
(x,k)*(y,\ell)=(x\cdot_ky,k+\ell).
$$
You see that the group operation of the semidirect product keeps track of the "combined" shift (here $k+\ell$), and that makes the operation $*$ associative.
