What's the dimension of this vectorspace? Consider the concatenation operator, e.g. $(1,2)+(3,4)=(1,2,3,4)$. With the conceit that adjacent numbers can "cancel out", e.g. $(1,2)+(-2,3)=(1,3)$, this is a group. We can make it a vectorspace by the standard componentwise multiplication: $cx=(cx_1,cx_2,\dots)$. Note that any element in this vector space can be written as $\sum_i c_i(1)$ for some $c_i$. I'm wondering if this means that this vector space has dimension one, since a basis seems to be $\\{(1)\\}$? I have read that a vector space with a field $F$ and dimension $n$ is isomorphic to $F^n$ which would mean that my vectorspace is isomorphic to $\mathbb R$, which I don't think is true.

This is not a vector space. Vadim123 and Tobias Kildetoft already pointed out that the underlying additive group is not abelian. Also in a vector space the axiom $$ (a+b)X=aX+bX $$ is required to hold for all scalars $a,b$ and vectors $X$. But here if $X=(1,2)$, $a=3$, $b=4$, then $$ (a+b)X=7X=(7,14)\qquad\text{but}\qquad aX+bX=(3,6)+(4,8)=(3,6,4,8). $$ Allowing permutations won't help.